id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-019901 | Algebraic Foundations: Congruence Modulo m | 10 | Indicate where a theorem is used: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $12\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-17]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019902 | Foundations: Relations from Fibers of Maps | 10 | Compute the requested quantity: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $29\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-25]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 29$ be the canonical projection.",
"Step 2... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019903 | Algebraic Foundations: Congruence Modulo m | 10 | Provide a rigorous solution: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $164\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[56]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In pa... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019904 | Set Theory: Equivalence Relations — R/S/T | 10 | Answer with a short justification: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $179\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[17]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019905 | Foundations: Relations from Fibers of Maps | 10 | Challenge: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $168\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-40]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not sk... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019906 | Set Theory: Partitions and Classes | 10 | Indicate where a theorem is used: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $123\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[48]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 123$ be the canonical projection.",
"Step ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019907 | Set Theory: Equivalence Relations — R/S/T | 10 | Exercise: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $44\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-64]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019908 | Set Theory: Partitions and Classes | 10 | Write the solution set clearly: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $75\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[15]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In ... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 75$ be the canonical projection.",
"Step 2... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019909 | Algebraic Foundations: Congruence Modulo m | 10 | Explain each transformation: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $73\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-63]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In pa... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 73$ be the canonical projection.",
"Step 2... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019910 | Set Theory: Equivalence Relations — R/S/T | 10 | Warm-up: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $123\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[60]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip ... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019911 | Foundations: Relations from Fibers of Maps | 10 | Explain why your operations are valid: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $67\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-33]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 67$ be the canonical projection.",
"Step 2... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019912 | Algebraic Foundations: Congruence Modulo m | 10 | Complete the analysis: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $8\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-85]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a),... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019913 | Foundations: Relations from Fibers of Maps | 10 | Solve and include a self-check: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $55\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[51]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In ... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 55$ be the canonical projection.",
"Step 2... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019914 | Algebraic Foundations: Congruence Modulo m | 10 | Exercise: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $78\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-85]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019915 | Foundations: Relations from Fibers of Maps | 10 | Do not skip justification steps: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $142\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[30]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
I... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 142$ be the canonical projection.",
"Step ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019916 | Set Theory: Partitions and Classes | 10 | Solve and include a self-check: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $67\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-99]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 67$ be the canonical projection.",
"Step 2... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019917 | Foundations: Relations from Fibers of Maps | 10 | Keep the final answer in boxed form: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $83\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-73]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019918 | Foundations: Relations from Fibers of Maps | 10 | Answer with a short justification: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $131\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[23]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019919 | Set Theory: Equivalence Relations — R/S/T | 10 | Checkpoint: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $108\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[82]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not sk... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 108$ be the canonical projection.",
"Step ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019920 | Set Theory: Equivalence Relations — R/S/T | 10 | Answer with a short justification: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $181\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[84]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 181$ be the canonical projection.",
"Step ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019921 | Set Theory: Equivalence Relations — R/S/T | 10 | State any required conditions first: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $131\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-87]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 131$ be the canonical projection.",
"Step ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019922 | Foundations: Relations from Fibers of Maps | 10 | Solve and sanity-check: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $131\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[12]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 131$ be the canonical projection.",
"Step ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019923 | Foundations: Relations from Fibers of Maps | 10 | Give a theorem-based solution: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $46\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-47]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In ... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 46$ be the canonical projection.",
"Step 2... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019924 | Set Theory: Equivalence Relations — R/S/T | 10 | Give a fully justified solution: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $180\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-4]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
I... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019925 | Foundations: Relations from Fibers of Maps | 10 | Indicate where a theorem is used: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $26\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-88]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019926 | Algebraic Foundations: Congruence Modulo m | 10 | Explain what is being counted/optimized: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $193\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-7]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 193$ be the canonical projection.",
"Step ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019927 | Set Theory: Equivalence Relations — R/S/T | 10 | Problem: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $102\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-34]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 102$ be the canonical projection.",
"Step ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019928 | Foundations: Relations from Fibers of Maps | 10 | Derive the result step-by-step: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $123\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-27]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
I... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019929 | Foundations: Relations from Fibers of Maps | 10 | Problem: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $59\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[4]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip th... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019930 | Set Theory: Partitions and Classes | 10 | Provide both a computational and a conceptual explanation: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $5\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-90]$ explicitly as a set.
(c) Explain briefly how this relates to ... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 5$ be the canonical projection.",
"Step 2:... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019931 | Algebraic Foundations: Congruence Modulo m | 10 | Provide a rigorous solution: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $176\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-4]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In pa... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 176$ be the canonical projection.",
"Step ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019932 | Set Theory: Equivalence Relations — R/S/T | 10 | Try to avoid pattern-matching; explain why: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $185\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-14]$ explicitly as a set.
(c) Explain briefly how this relates to congruence mo... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019933 | Foundations: Relations from Fibers of Maps | 10 | Indicate where a theorem is used: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $140\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-96]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 140$ be the canonical projection.",
"Step ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019934 | Set Theory: Partitions and Classes | 10 | Use two approaches if possible: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $115\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[8]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In ... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019935 | Set Theory: Equivalence Relations — R/S/T | 10 | Find the exact value: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $17\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-62]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a),... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019936 | Set Theory: Equivalence Relations — R/S/T | 10 | Proceed methodically: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $21\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[46]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), ... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019937 | Algebraic Foundations: Congruence Modulo m | 10 | Work this out carefully: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $30\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[82]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 30$ be the canonical projection.",
"Step 2... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019938 | Algebraic Foundations: Congruence Modulo m | 10 | Answer with a short justification: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $126\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[92]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019939 | Set Theory: Equivalence Relations — R/S/T | 10 | Exercise: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $185\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-39]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not ski... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019940 | Set Theory: Equivalence Relations — R/S/T | 10 | Start by stating any domain restrictions: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $31\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[68]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 31$ be the canonical projection.",
"Step 2... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019941 | Set Theory: Partitions and Classes | 10 | Warm-up: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $29\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-88]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip ... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 29$ be the canonical projection.",
"Step 2... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019942 | Set Theory: Partitions and Classes | 10 | Solve and then verify: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $25\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[82]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a),... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019943 | Algebraic Foundations: Congruence Modulo m | 10 | Start by stating any domain restrictions: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $3\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-17]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019944 | Foundations: Relations from Fibers of Maps | 10 | Solve with verification: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $36\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[51]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 36$ be the canonical projection.",
"Step 2... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019945 | Set Theory: Equivalence Relations — R/S/T | 10 | State any required conditions first: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $97\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-77]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 97$ be the canonical projection.",
"Step 2... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019946 | Set Theory: Partitions and Classes | 10 | Where appropriate, name the theorem you use: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $60\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-88]$ explicitly as a set.
(c) Explain briefly how this relates to congruence mo... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019947 | Set Theory: Partitions and Classes | 10 | Answer with a short justification: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $93\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[54]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 93$ be the canonical projection.",
"Step 2... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019948 | Set Theory: Partitions and Classes | 10 | Make each step logically reversible (or explain if not): Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $196\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[35]$ explicitly as a set.
(c) Explain briefly how this relates to c... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019949 | Foundations: Relations from Fibers of Maps | 10 | Start by stating any domain restrictions: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $7\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[41]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo ... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019950 | Set Theory: Equivalence Relations — R/S/T | 10 | Keep the final answer in boxed form: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $192\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-10]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019951 | Set Theory: Partitions and Classes | 10 | Answer with a short justification: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $27\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-5]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019952 | Algebraic Foundations: Congruence Modulo m | 10 | Solve and then verify: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $3\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[72]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), ... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019953 | Algebraic Foundations: Congruence Modulo m | 10 | Challenge: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $6\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[63]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip ... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 6$ be the canonical projection.",
"Step 2:... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019954 | Set Theory: Equivalence Relations — R/S/T | 10 | Make each step logically reversible (or explain if not): Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $35\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[33]$ explicitly as a set.
(c) Explain briefly how this relates to co... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 35$ be the canonical projection.",
"Step 2... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019955 | Set Theory: Equivalence Relations — R/S/T | 10 | Problem: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $191\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-91]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 191$ be the canonical projection.",
"Step ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019956 | Set Theory: Equivalence Relations — R/S/T | 10 | State any required conditions first: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $6\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-6]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019957 | Foundations: Relations from Fibers of Maps | 10 | Keep the final answer in boxed form: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $34\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[26]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019958 | Algebraic Foundations: Congruence Modulo m | 10 | Task: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $77\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[88]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip the ... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 77$ be the canonical projection.",
"Step 2... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019959 | Algebraic Foundations: Congruence Modulo m | 10 | Where appropriate, name the theorem you use: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $45\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[93]$ explicitly as a set.
(c) Explain briefly how this relates to congruence mod... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019960 | Set Theory: Partitions and Classes | 10 | Compute the requested quantity: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $111\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-68]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
I... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019961 | Foundations: Relations from Fibers of Maps | 10 | Problem: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $21\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-83]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip ... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019962 | Set Theory: Equivalence Relations — R/S/T | 10 | Keep the final answer in boxed form: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $164\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-40]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019963 | Algebraic Foundations: Congruence Modulo m | 10 | Problem: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $164\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-17]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 164$ be the canonical projection.",
"Step ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019964 | Algebraic Foundations: Congruence Modulo m | 10 | Question: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $87\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[9]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip t... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019965 | Foundations: Relations from Fibers of Maps | 10 | Exercise: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $162\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[9]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip ... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019966 | Foundations: Relations from Fibers of Maps | 10 | Keep the final answer in boxed form: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $24\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[2]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019967 | Set Theory: Partitions and Classes | 10 | Where appropriate, name the theorem you use: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $61\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[86]$ explicitly as a set.
(c) Explain briefly how this relates to congruence mod... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019968 | Set Theory: Equivalence Relations — R/S/T | 10 | Where appropriate, name the theorem you use: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $135\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[84]$ explicitly as a set.
(c) Explain briefly how this relates to congruence mo... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019969 | Set Theory: Partitions and Classes | 10 | Determine the requested value: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $66\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-12]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In ... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019970 | Set Theory: Partitions and Classes | 10 | Carefully track domains: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $176\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[89]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019971 | Algebraic Foundations: Congruence Modulo m | 10 | State any required conditions first: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $190\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-51]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019972 | Foundations: Relations from Fibers of Maps | 10 | Indicate where a theorem is used: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $47\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[23]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
I... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 47$ be the canonical projection.",
"Step 2... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019973 | Foundations: Relations from Fibers of Maps | 10 | Give a fully justified solution: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $192\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-76]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019974 | Algebraic Foundations: Congruence Modulo m | 10 | Complete the analysis: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $36\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[26]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a),... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019975 | Foundations: Relations from Fibers of Maps | 10 | Use two approaches if possible: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $108\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-30]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
I... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019976 | Set Theory: Equivalence Relations — R/S/T | 10 | Solve and then verify: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $130\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[36]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a)... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 130$ be the canonical projection.",
"Step ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019977 | Set Theory: Partitions and Classes | 10 | Warm-up: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $55\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[57]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip t... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 55$ be the canonical projection.",
"Step 2... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019978 | Foundations: Relations from Fibers of Maps | 10 | Explain why your operations are valid: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $186\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-79]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo ... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 186$ be the canonical projection.",
"Step ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019979 | Set Theory: Partitions and Classes | 10 | Where appropriate, name the theorem you use: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $199\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[65]$ explicitly as a set.
(c) Explain briefly how this relates to congruence mo... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 199$ be the canonical projection.",
"Step ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019980 | Foundations: Relations from Fibers of Maps | 10 | Explain what is being counted/optimized: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $166\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[41]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 166$ be the canonical projection.",
"Step ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019981 | Set Theory: Partitions and Classes | 10 | Complete the analysis: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $183\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-95]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 183$ be the canonical projection.",
"Step ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019982 | Algebraic Foundations: Congruence Modulo m | 10 | Derive the result step-by-step: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $125\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[5]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In ... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019983 | Algebraic Foundations: Congruence Modulo m | 10 | Determine the requested value: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $156\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[15]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In ... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019984 | Foundations: Relations from Fibers of Maps | 10 | Track units/moduli carefully: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $174\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[63]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In p... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019985 | Set Theory: Equivalence Relations — R/S/T | 10 | Explain each transformation: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $25\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-2]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In par... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019986 | Foundations: Relations from Fibers of Maps | 10 | Exercise: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $190\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[30]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 190$ be the canonical projection.",
"Step ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019987 | Algebraic Foundations: Congruence Modulo m | 10 | Answer using clear logical steps: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $194\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-82]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019988 | Foundations: Relations from Fibers of Maps | 10 | Write the solution set clearly: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $174\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[20]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019989 | Set Theory: Equivalence Relations — R/S/T | 10 | Explain what is being counted/optimized: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $198\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-18]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modul... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 198$ be the canonical projection.",
"Step ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019990 | Algebraic Foundations: Congruence Modulo m | 10 | Solve and include a self-check: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $162\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-40]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
I... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019991 | Foundations: Relations from Fibers of Maps | 10 | Work carefully and justify each inference: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $83\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-64]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modu... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019992 | Set Theory: Partitions and Classes | 10 | Be explicit about assumptions: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $24\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-6]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In p... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019993 | Set Theory: Partitions and Classes | 10 | Explain why your operations are valid: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $200\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[32]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 200$ be the canonical projection.",
"Step ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-019994 | Set Theory: Equivalence Relations — R/S/T | 10 | Exercise: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $40\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-4]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip ... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019995 | Set Theory: Equivalence Relations — R/S/T | 10 | Be explicit about assumptions: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $110\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-23]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019996 | Algebraic Foundations: Congruence Modulo m | 10 | Give a theorem-based solution: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $119\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-81]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019997 | Set Theory: Equivalence Relations — R/S/T | 10 | Solve and sanity-check: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $79\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[7]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a),... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 79$ be the canonical projection.",
"Step 2... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019998 | Set Theory: Partitions and Classes | 10 | Carefully track domains: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $76\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[71]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-019999 | Algebraic Foundations: Congruence Modulo m | 10 | Give reasoning, not just computation: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $6\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-60]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-020000 | Set Theory: Equivalence Relations — R/S/T | 10 | Explain what is being counted/optimized: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $54\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-96]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
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