id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-017901 | Algebraic Foundations: Congruence Modulo m | 9 | Question: 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 $[-10]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not ski... | [
{
"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": "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-017902 | Algebraic Foundations: Congruence Modulo m | 9 | Carefully track domains: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $195\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[98]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (... | [
{
"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 $\\pi(a)=a\\bmod 195$ 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... | 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-017903 | Set Theory: Partitions and Classes | 9 | Solve with verification: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $84\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[45]$ 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 $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-017904 | Algebraic Foundations: Congruence Modulo m | 9 | 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 $[-83]$ 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 $\\pi(a)=a\\bmod 126$ 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... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-017905 | Foundations: Relations from Fibers of Maps | 9 | Start by stating any domain restrictions: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $113\mid(a-b)$.
(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 modul... | [
{
"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-017906 | Set Theory: Equivalence Relations — R/S/T | 9 | Solve (and briefly cross-validate): Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $2\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-69]$ 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... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-017907 | Set Theory: Equivalence Relations — R/S/T | 9 | Solve and then verify: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $169\mid(a-b)$.
(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 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 $\\pi(a)=a\\bmod 169$ 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-017908 | Algebraic Foundations: Congruence Modulo m | 9 | Make each step logically reversible (or explain if not): 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 $[-62]$ 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 $\\pi(a)=a\\bmod 181$ 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-017909 | Foundations: Relations from Fibers of Maps | 9 | Proceed methodically: 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 $[-84]$ 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-017910 | Algebraic Foundations: Congruence Modulo m | 9 | Give a theorem-based solution: 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 $[65]$ 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... | 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-017911 | Set Theory: Partitions and Classes | 9 | Exercise: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $90\mid(a-b)$.
(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 $m$.
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 $\\pi(a)=a\\bmod 90$ 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-017912 | Set Theory: Equivalence Relations — R/S/T | 9 | Explain why your operations are valid: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $10\mid(a-b)$.
(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... | [
{
"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-017913 | Foundations: Relations from Fibers of Maps | 9 | Indicate where a theorem is used: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $11\mid(a-b)$.
(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 $m$.
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 $\\pi(a)=a\\bmod 11$ 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-017914 | Algebraic Foundations: Congruence Modulo m | 9 | Write the solution set clearly: 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 $[24]$ 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 $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-017915 | Set Theory: Equivalence Relations — R/S/T | 9 | Problem: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $10\mid(a-b)$.
(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 $\\pi(a)=a\\bmod 10$ 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-017916 | Algebraic Foundations: Congruence Modulo m | 9 | Explain each transformation: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $64\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[11]$ 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": "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 $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-017917 | Algebraic Foundations: Congruence Modulo m | 9 | Explain what is being counted/optimized: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $56\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-1]$ 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 $\\pi(a)=a\\bmod 56$ 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... | 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-017918 | Foundations: Relations from Fibers of Maps | 9 | Warm-up: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $28\mid(a-b)$.
(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 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 $\\pi(a)=a\\bmod 28$ 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... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-017919 | Algebraic Foundations: Congruence Modulo m | 9 | Checkpoint: 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 $[50]$ 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 $\\pi(a)=a\\bmod 196$ 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-017920 | Foundations: Relations from Fibers of Maps | 9 | Solve (and briefly cross-validate): Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $134\mid(a-b)$.
(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": "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-017921 | Set Theory: Partitions and Classes | 9 | Question: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $37\mid(a-b)$.
(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 $m$.
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 $\\pi(a)=a\\bmod 37$ 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 $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-017922 | Algebraic Foundations: Congruence Modulo m | 9 | Provide both a computational and a conceptual explanation: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $158\mid(a-b)$.
(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... | [
{
"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-017923 | Algebraic Foundations: Congruence Modulo m | 9 | Complete the analysis: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $81\mid(a-b)$.
(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 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... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-017924 | Set Theory: Partitions and Classes | 9 | Solve (and briefly cross-validate): Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $28\mid(a-b)$.
(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... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-017925 | Foundations: Relations from Fibers of Maps | 9 | Determine the requested value: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $106\mid(a-b)$.
(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 $m$.
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-017926 | Set Theory: Equivalence Relations — R/S/T | 9 | Give a theorem-based solution: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $58\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[94]$ 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": "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-017927 | Set Theory: Equivalence Relations — R/S/T | 9 | Solve with verification: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $144\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[45]$ 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": "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-017928 | Foundations: Relations from Fibers of Maps | 9 | Explain what is being counted/optimized: 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 $[47]$ 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": "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-017929 | Set Theory: Partitions and Classes | 9 | Question: 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 $[-14]$ 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-017930 | Algebraic Foundations: Congruence Modulo m | 9 | Start by stating any domain restrictions: 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 $[37]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modul... | [
{
"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-017931 | Set Theory: Partitions and Classes | 9 | Solve and justify each step: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $64\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[53]$ 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": "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-017932 | Algebraic Foundations: Congruence Modulo m | 9 | Give reasoning, not just computation: 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 $[-5]$ 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": "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-017933 | Set Theory: Equivalence Relations — R/S/T | 9 | Solve and include a self-check: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $19\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-16]$ 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 $\\pi(a)=a\\bmod 19$ 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... | 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-017934 | Set Theory: Equivalence Relations — R/S/T | 9 | Give a fully justified solution: 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 $[14]$ 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 $\\pi(a)=a\\bmod 140$ 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-017935 | Set Theory: Partitions and Classes | 9 | Use two approaches if possible: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $167\mid(a-b)$.
(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 $m$.
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... | 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-017936 | Foundations: Relations from Fibers of Maps | 9 | Solve (and briefly cross-validate): Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $85\mid(a-b)$.
(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 $m$.
... | [
{
"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 $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-017937 | Foundations: Relations from Fibers of Maps | 9 | Give an answer and a quick verification: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $89\mid(a-b)$.
(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 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": "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 $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-017938 | Set Theory: Partitions and Classes | 9 | Answer using clear logical 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 $[-62]$ 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 $\\pi(a)=a\\bmod 142$ 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... | 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-017939 | Set Theory: Partitions and Classes | 9 | Track units/moduli carefully: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $74\mid(a-b)$.
(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": "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-017940 | Set Theory: Equivalence Relations — R/S/T | 9 | Find the exact value: 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 $[-81]$ 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... | 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-017941 | Algebraic Foundations: Congruence Modulo m | 9 | Work carefully and justify each inference: 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 $[-16]$ 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": "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-017942 | Foundations: Relations from Fibers of Maps | 9 | Checkpoint: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $116\mid(a-b)$.
(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 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": "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-017943 | Algebraic Foundations: Congruence Modulo m | 9 | Give a fully justified solution: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $112\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[50]$ 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-017944 | Set Theory: Partitions and Classes | 9 | Be explicit about assumptions: 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 $[99]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In p... | [
{
"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": "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-017945 | Foundations: Relations from Fibers of Maps | 9 | Keep the final answer in boxed form: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $64\mid(a-b)$.
(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 $m$... | [
{
"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 $\\pi(a)=a\\bmod 64$ 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-017946 | Set Theory: Equivalence Relations — R/S/T | 9 | Solve with verification: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $100\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[38]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (... | [
{
"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 $\\pi(a)=a\\bmod 100$ 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-017947 | Set Theory: Partitions and Classes | 9 | Show all reasoning: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $151\mid(a-b)$.
(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 $m$.
In part (a), d... | [
{
"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 $\\pi(a)=a\\bmod 151$ 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-017948 | Foundations: Relations from Fibers of Maps | 9 | Solve and sanity-check: 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 $[-29]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (... | [
{
"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": "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-017949 | Foundations: Relations from Fibers of Maps | 9 | Give an answer and a quick verification: 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 $[-1]$ 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 $\\pi(a)=a\\bmod 44$ 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... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-017950 | Set Theory: Equivalence Relations — R/S/T | 9 | Give an answer and a quick verification: 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 $[-79]$ 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 $\\pi(a)=a\\bmod 87$ 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-017951 | Set Theory: Partitions and Classes | 9 | Solve with verification: 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 $[95]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (... | [
{
"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": "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 $r+m\mathbb{Z}$. |
math-017952 | Algebraic Foundations: Congruence Modulo m | 9 | Keep the final answer in boxed form: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $106\mid(a-b)$.
(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 $m$... | [
{
"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-017953 | Set Theory: Equivalence Relations — R/S/T | 9 | Proceed methodically: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $50\mid(a-b)$.
(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), ... | [
{
"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... | 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-017954 | Foundations: Relations from Fibers of Maps | 9 | Use two approaches if possible: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $167\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-80]$ 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 $\\pi(a)=a\\bmod 167$ 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 $r+m\mathbb{Z}$. |
math-017955 | Set Theory: Equivalence Relations — R/S/T | 9 | Solve and include a self-check: 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 $[89]$ 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... | 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-017956 | Set Theory: Equivalence Relations — R/S/T | 9 | Solve with verification: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $2\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-75]$ 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 $\\pi(a)=a\\bmod 2$ 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... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-017957 | Set Theory: Equivalence Relations — R/S/T | 9 | Warm-up: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $149\mid(a-b)$.
(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 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 $\\pi(a)=a\\bmod 149$ 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-017958 | Set Theory: Equivalence Relations — R/S/T | 9 | Track units/moduli carefully: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $91\mid(a-b)$.
(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 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... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-017959 | Foundations: Relations from Fibers of Maps | 9 | Determine the requested value: 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 $[-85]$ 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 $\\pi(a)=a\\bmod 111$ 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-017960 | Set Theory: Equivalence Relations — R/S/T | 9 | Warm-up: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $161\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-100]$ 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": "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-017961 | Set Theory: Equivalence Relations — R/S/T | 9 | Explain each transformation: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $138\mid(a-b)$.
(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 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": "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-017962 | Set Theory: Equivalence Relations — R/S/T | 9 | Solve and then verify: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $170\mid(a-b)$.
(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 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 $\\pi(a)=a\\bmod 170$ 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-017963 | Algebraic Foundations: Congruence Modulo m | 9 | Try to avoid pattern-matching; explain why: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $33\mid(a-b)$.
(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 congruence modu... | [
{
"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 $\\pi(a)=a\\bmod 33$ 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... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-017964 | Foundations: Relations from Fibers of Maps | 9 | Try to avoid pattern-matching; explain why: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $158\mid(a-b)$.
(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 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... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-017965 | Foundations: Relations from Fibers of Maps | 9 | Solve with verification: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $2\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-70]$ 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 $\\pi(a)=a\\bmod 2$ 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 $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-017966 | Set Theory: Equivalence Relations — R/S/T | 9 | Write the solution set clearly: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $116\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[66]$ 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 $\\pi(a)=a\\bmod 116$ 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-017967 | Foundations: Relations from Fibers of Maps | 9 | Answer using clear logical steps: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $81\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[97]$ 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": "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-017968 | Set Theory: Equivalence Relations — R/S/T | 9 | Challenge: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $116\mid(a-b)$.
(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 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": "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-017969 | Set Theory: Equivalence Relations — R/S/T | 9 | Explain what is being counted/optimized: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $37\mid(a-b)$.
(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 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": "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-017970 | Algebraic Foundations: Congruence Modulo m | 9 | Track units/moduli carefully: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $187\mid(a-b)$.
(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 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... | 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-017971 | Foundations: Relations from Fibers of Maps | 9 | Start by stating any domain restrictions: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $96\mid(a-b)$.
(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... | [
{
"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 $\\pi(a)=a\\bmod 96$ 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-017972 | Foundations: Relations from Fibers of Maps | 9 | Solve and include a self-check: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $188\mid(a-b)$.
(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 $m$.
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 $\\pi(a)=a\\bmod 188$ 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-017973 | Foundations: Relations from Fibers of Maps | 9 | Solve with verification: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $127\mid(a-b)$.
(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 $m$.
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": "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-017974 | Algebraic Foundations: Congruence Modulo m | 9 | Do not skip justification steps: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $154\mid(a-b)$.
(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 $m$.
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": "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-017975 | Foundations: Relations from Fibers of Maps | 9 | Where appropriate, name the theorem you use: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $101\mid(a-b)$.
(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 mod... | [
{
"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 $\\pi(a)=a\\bmod 101$ 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 $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-017976 | Set Theory: Partitions and Classes | 9 | Give reasoning, not just computation: 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 $[-33]$ 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": "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-017977 | Foundations: Relations from Fibers of Maps | 9 | Make each step logically reversible (or explain if not): Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $188\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[97]$ 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-017978 | Set Theory: Partitions and Classes | 9 | Task: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $122\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[94]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip the... | [
{
"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-017979 | Set Theory: Partitions and Classes | 9 | Explain what is being counted/optimized: 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 $[-66]$ 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": "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-017980 | Algebraic Foundations: Congruence Modulo m | 9 | Determine the requested value: 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 $[78]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In p... | [
{
"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 $\\pi(a)=a\\bmod 26$ 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-017981 | Foundations: Relations from Fibers of Maps | 9 | Determine the requested value: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $172\mid(a-b)$.
(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 ... | [
{
"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-017982 | Set Theory: Equivalence Relations — R/S/T | 9 | Start by stating any domain restrictions: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $39\mid(a-b)$.
(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 $\\pi(a)=a\\bmod 39$ 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-017983 | Set Theory: Partitions and Classes | 9 | Complete the analysis: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $158\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[50]$ 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 $\\pi(a)=a\\bmod 158$ 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-017984 | Set Theory: Partitions and Classes | 9 | Start by stating any domain restrictions: 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 $[-96]$ 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 $\\pi(a)=a\\bmod 47$ 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 $r+m\mathbb{Z}$. |
math-017985 | Algebraic Foundations: Congruence Modulo m | 9 | Solve with verification: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $58\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-53]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (... | [
{
"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 $\\pi(a)=a\\bmod 58$ 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-017986 | Set Theory: Equivalence Relations — R/S/T | 9 | Solve (and briefly cross-validate): Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $141\mid(a-b)$.
(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 $m$... | [
{
"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 $\\pi(a)=a\\bmod 141$ 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... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-017987 | Set Theory: Equivalence Relations — R/S/T | 9 | Start by stating any domain restrictions: 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 $[0]$ 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": "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-017988 | Algebraic Foundations: Congruence Modulo m | 9 | Solve and sanity-check: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $134\mid(a-b)$.
(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 $m$.
In part (... | [
{
"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 $\\pi(a)=a\\bmod 134$ 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 $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-017989 | Algebraic Foundations: Congruence Modulo m | 9 | Challenge: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $117\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[24]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not ski... | [
{
"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 $\\pi(a)=a\\bmod 117$ 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... | 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-017990 | Algebraic Foundations: Congruence Modulo m | 9 | Solve and justify each step: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $195\mid(a-b)$.
(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 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... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-017991 | Foundations: Relations from Fibers of Maps | 9 | Work carefully and justify each inference: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $138\mid(a-b)$.
(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 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": "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-017992 | Set Theory: Equivalence Relations — R/S/T | 9 | Provide a rigorous solution: 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 $[-28]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In p... | [
{
"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": "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-017993 | Set Theory: Partitions and Classes | 9 | Where appropriate, name the theorem you use: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $182\mid(a-b)$.
(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 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... | 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-017994 | Set Theory: Equivalence Relations — R/S/T | 9 | Do not skip justification steps: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $107\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-37]$ 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... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-017995 | Set Theory: Partitions and Classes | 9 | Problem: 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 $[-35]$ 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 $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-017996 | Set Theory: Equivalence Relations — R/S/T | 9 | Find the exact value: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $146\mid(a-b)$.
(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 $m$.
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 $\\pi(a)=a\\bmod 146$ 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-017997 | Set Theory: Equivalence Relations — R/S/T | 9 | Explain what is being counted/optimized: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $170\mid(a-b)$.
(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 modul... | [
{
"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 $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-017998 | Set Theory: Partitions and Classes | 9 | Make each step logically reversible (or explain if not): 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 $[-96]$ 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 $\\pi(a)=a\\bmod 181$ 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... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-017999 | Set Theory: Equivalence Relations — R/S/T | 9 | Challenge: 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 $[48]$ 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 $\\pi(a)=a\\bmod 27$ 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-018000 | Foundations: Relations from Fibers of Maps | 9 | Prompt: 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 $[-22]$ 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": "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 $\boxed{\text{Equivalence; }$.) |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.