id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-015901 | Foundations: Relations from Fibers of Maps | 8 | Question: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $102\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[81]$ 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... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-015902 | Set Theory: Equivalence Relations — R/S/T | 8 | Problem: 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 $[-77]$ 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 $\\pi(a)=a\\bmod 179$ 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 $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-015903 | Set Theory: Partitions and Classes | 8 | Explain what is being counted/optimized: 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 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 $\\pi(a)=a\\bmod 131$ be the canonical projection.",
"Step ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-015904 | Set Theory: Partitions and Classes | 8 | Make each step logically reversible (or explain if not): Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $5\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-14]$ explicitly as a set.
(c) Explain briefly how this relates to co... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 5$ be the canonical projection.",
"Step 2:... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-015905 | Set Theory: Partitions and Classes | 8 | Keep the final answer in boxed form: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $62\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-88]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "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-015906 | Algebraic Foundations: Congruence Modulo m | 8 | Work this out carefully: 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 $[40]$ 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 $r+m\mathbb{Z}$. |
math-015907 | Algebraic Foundations: Congruence Modulo m | 8 | Explain each transformation: 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 $[57]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In pa... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 200$ 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-015908 | Set Theory: Equivalence Relations — R/S/T | 8 | Solve with 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 $[-34]$ 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 $\\pi(a)=a\\bmod 89$ 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-015909 | Algebraic Foundations: Congruence Modulo m | 8 | Prompt: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $120\mid(a-b)$.
(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 (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... | 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-015910 | Set Theory: Equivalence Relations — R/S/T | 8 | Problem: 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 $[33]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip ... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-015911 | Foundations: Relations from Fibers of Maps | 8 | Explain why your operations are valid: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $104\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[42]$ 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 $\\pi(a)=a\\bmod 104$ be the canonical projection.",
"Step ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-015912 | Set Theory: Equivalence Relations — R/S/T | 8 | Try to avoid pattern-matching; explain why: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $114\mid(a-b)$.
(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": "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-015913 | Algebraic Foundations: Congruence Modulo m | 8 | State any required conditions first: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $5\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[17]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "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-015914 | Algebraic Foundations: Congruence Modulo m | 8 | Compute the requested quantity: 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 $[-22]$ 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 $\\pi(a)=a\\bmod 131$ 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... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-015915 | Set Theory: Partitions and Classes | 8 | Track quantifiers carefully: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $177\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-44]$ 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... | 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-015916 | Algebraic Foundations: Congruence Modulo m | 8 | Provide a rigorous solution: 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 $[-99]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In pa... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 27$ 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-015917 | Set Theory: Equivalence Relations — R/S/T | 8 | Provide both a computational and a conceptual explanation: 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 $[-62]$ explicitly as a set.
(c) Explain briefly how this relates t... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "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-015918 | Algebraic Foundations: Congruence Modulo m | 8 | Give a fully justified solution: 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 $[57]$ 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 $r+m\mathbb{Z}$. |
math-015919 | Set Theory: Partitions and Classes | 8 | Challenge: 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 $[85]$ 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": "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-015920 | Foundations: Relations from Fibers of Maps | 8 | State any required conditions first: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $68\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-60]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$... | [
{
"method_name": "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 $\\pi(a)=a\\bmod 68$ 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... | 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-015921 | Set Theory: Equivalence Relations — R/S/T | 8 | Try to avoid pattern-matching; explain why: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $65\mid(a-b)$.
(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 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 $\\pi(a)=a\\bmod 65$ be the canonical projection.",
"Step 2... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-015922 | Set Theory: Equivalence Relations — R/S/T | 8 | Track quantifiers carefully: 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 $[-98]$ 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": "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-015923 | Algebraic Foundations: Congruence Modulo m | 8 | Question: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $92\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[56]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In 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... | 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-015924 | Set Theory: Partitions and Classes | 8 | Explain why your operations are valid: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $114\mid(a-b)$.
(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 ... | [
{
"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 $\\pi(a)=a\\bmod 114$ 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-015925 | Set Theory: Partitions and Classes | 8 | Answer with a short justification: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $155\mid(a-b)$.
(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 $m$.... | [
{
"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 $\\pi(a)=a\\bmod 155$ be the canonical projection.",
"Step ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-015926 | Foundations: Relations from Fibers of Maps | 8 | Solve and justify each step: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $3\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-18]$ 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... | 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-015927 | Set Theory: Partitions and Classes | 8 | Try to avoid pattern-matching; explain why: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $132\mid(a-b)$.
(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 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 $\\pi(a)=a\\bmod 132$ 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-015928 | Algebraic Foundations: Congruence Modulo m | 8 | Complete the analysis: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $5\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-41]$ 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... | 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-015929 | Set Theory: Equivalence Relations — R/S/T | 8 | Solve with verification: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $130\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-57]$ 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": "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-015930 | Algebraic Foundations: Congruence Modulo m | 8 | Challenge: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $123\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-57]$ 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 $\\pi(a)=a\\bmod 123$ be the canonical projection.",
"Step ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-015931 | Algebraic Foundations: Congruence Modulo m | 8 | Solve and justify each step: 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 $[-2]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In pa... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | 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-015932 | Set Theory: Equivalence Relations — R/S/T | 8 | Compute the requested quantity: 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 $[56]$ 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-015933 | Set Theory: Equivalence Relations — R/S/T | 8 | Give reasoning, not just computation: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $61\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[5]$ 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 $\\pi(a)=a\\bmod 61$ 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... | 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-015934 | Set Theory: Equivalence Relations — R/S/T | 8 | Determine the requested value: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $16\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[44]$ 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 $\\pi(a)=a\\bmod 16$ 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-015935 | Foundations: Relations from Fibers of Maps | 8 | Solve and sanity-check: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $63\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[73]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
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 $\\pi(a)=a\\bmod 63$ 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-015936 | Set Theory: Equivalence Relations — R/S/T | 8 | Answer with a short justification: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $16\mid(a-b)$.
(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 $m$.
... | [
{
"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 $\\pi(a)=a\\bmod 16$ be the canonical projection.",
"Step 2... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-015937 | Foundations: Relations from Fibers of Maps | 8 | Give an answer and a quick verification: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $143\mid(a-b)$.
(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 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": "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-015938 | Foundations: Relations from Fibers of Maps | 8 | Solve with verification: 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 $[-92]$ 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": "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-015939 | Foundations: Relations from Fibers of Maps | 8 | State any required conditions first: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $163\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-10]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m... | [
{
"method_name": "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 $\\pi(a)=a\\bmod 163$ 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... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-015940 | Set Theory: Partitions and Classes | 8 | Provide a rigorous solution: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $14\mid(a-b)$.
(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 pa... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 14$ 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-015941 | Foundations: Relations from Fibers of Maps | 8 | Carefully track domains: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $199\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-76]$ 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 $\\pi(a)=a\\bmod 199$ 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-015942 | Set Theory: Equivalence Relations — R/S/T | 8 | Complete the analysis: 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 $[-29]$ 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 $\\pi(a)=a\\bmod 190$ 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... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-015943 | Foundations: Relations from Fibers of Maps | 8 | Indicate where a theorem is used: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $148\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[64]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-015944 | Algebraic Foundations: Congruence Modulo m | 8 | Write the solution set clearly: 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 $[63]$ 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-015945 | Set Theory: Equivalence Relations — R/S/T | 8 | Problem: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $3\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[49]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip th... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | 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-015946 | Algebraic Foundations: Congruence Modulo m | 8 | Work carefully and justify each inference: 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 $[-78]$ 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 $\\pi(a)=a\\bmod 8$ 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-015947 | Algebraic Foundations: Congruence Modulo m | 8 | Write the solution set clearly: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $150\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[52]$ 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 $\\pi(a)=a\\bmod 150$ 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-015948 | Foundations: Relations from Fibers of Maps | 8 | Solve (and briefly cross-validate): 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 $[-15]$ 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 $\\pi(a)=a\\bmod 113$ 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 $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-015949 | Algebraic Foundations: Congruence Modulo m | 8 | Be explicit about assumptions: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $199\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[47]$ 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-015950 | Set Theory: Equivalence Relations — R/S/T | 8 | Track quantifiers carefully: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $199\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-26]$ 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-015951 | Algebraic Foundations: Congruence Modulo m | 8 | Give an answer and a quick verification: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $145\mid(a-b)$.
(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... | [
{
"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 $\\pi(a)=a\\bmod 145$ 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-015952 | Algebraic Foundations: Congruence Modulo m | 8 | Provide both a computational and a conceptual explanation: 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 $[-72]$ explicitly as a set.
(c) Explain briefly how this relates 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 $\\pi(a)=a\\bmod 193$ be the canonical projection.",
"Step ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-015953 | Foundations: Relations from Fibers of Maps | 8 | Solve with verification: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $175\mid(a-b)$.
(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": "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 $r+m\mathbb{Z}$. |
math-015954 | Foundations: Relations from Fibers of Maps | 8 | Checkpoint: 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 $[-86]$ 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 $\\pi(a)=a\\bmod 44$ 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-015955 | Set Theory: Partitions and Classes | 8 | Where appropriate, name the theorem you use: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $4\mid(a-b)$.
(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 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 $\\pi(a)=a\\bmod 4$ be the canonical projection.",
"Step 2:... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-015956 | Foundations: Relations from Fibers of Maps | 8 | Track quantifiers carefully: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $115\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-92]$ 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 $\\pi(a)=a\\bmod 115$ be the canonical projection.",
"Step ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-015957 | Set Theory: Partitions and Classes | 8 | Proceed methodically: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $63\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-2]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In 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 $\\pi(a)=a\\bmod 63$ 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-015958 | Set Theory: Partitions and Classes | 8 | Prompt: 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 $[-92]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip ... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-015959 | Foundations: Relations from Fibers of Maps | 8 | Warm-up: 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 $[-8]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip t... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "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-015960 | Set Theory: Equivalence Relations — R/S/T | 8 | Give a fully justified solution: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $38\mid(a-b)$.
(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... | [
{
"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-015961 | Algebraic Foundations: Congruence Modulo m | 8 | Solve and include a self-check: 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 $[-27]$ 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 $\\pi(a)=a\\bmod 27$ 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... | 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-015962 | Set Theory: Equivalence Relations — R/S/T | 8 | Write the solution set clearly: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $164\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-24]$ 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... | 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-015963 | Set Theory: Partitions and Classes | 8 | Use two approaches if possible: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $25\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-61]$ 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-015964 | Set Theory: Partitions and Classes | 8 | Complete the analysis: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $51\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-72]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a)... | [
{
"method_name": "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 $\\pi(a)=a\\bmod 51$ be the canonical projection.",
"Step 2... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-015965 | Algebraic Foundations: Congruence Modulo m | 8 | Prompt: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $175\mid(a-b)$.
(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 (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 $\boxed{\text{Equivalence; }$.) |
math-015966 | Set Theory: Equivalence Relations — R/S/T | 8 | Solve and then verify: 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 $[-88]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a),... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-015967 | Foundations: Relations from Fibers of Maps | 8 | Work carefully and justify each inference: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $60\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[0]$ 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 $\\pi(a)=a\\bmod 60$ 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... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-015968 | Set Theory: Equivalence Relations — R/S/T | 8 | Try to avoid pattern-matching; explain why: 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 $[3]$ 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 $\\pi(a)=a\\bmod 146$ 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-015969 | Set Theory: Partitions and Classes | 8 | Give a theorem-based solution: 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 $[3]$ 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": "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-015970 | Algebraic Foundations: Congruence Modulo m | 8 | Show all reasoning: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $61\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-45]$ 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 $\\pi(a)=a\\bmod 61$ 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 $r+m\mathbb{Z}$. |
math-015971 | Set Theory: Equivalence Relations — R/S/T | 8 | Answer with a short justification: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $148\mid(a-b)$.
(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 $m$.... | [
{
"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 $\\pi(a)=a\\bmod 148$ 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-015972 | Algebraic Foundations: Congruence Modulo m | 8 | Give a fully justified solution: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $105\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-17]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-015973 | Foundations: Relations from Fibers of Maps | 8 | Derive the result step-by-step: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $114\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[51]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 114$ 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-015974 | Set Theory: Partitions and Classes | 8 | Find the exact value: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $155\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-5]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In 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 $r+m\mathbb{Z}$. |
math-015975 | Set Theory: Equivalence Relations — R/S/T | 8 | 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 $[-81]$ 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... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-015976 | Algebraic Foundations: Congruence Modulo m | 8 | Provide a rigorous solution: 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 $[-36]$ 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... | 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-015977 | Algebraic Foundations: Congruence Modulo m | 8 | Track quantifiers carefully: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $194\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[56]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In pa... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-015978 | Foundations: Relations from Fibers of Maps | 8 | Carefully track domains: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $137\mid(a-b)$.
(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 $m$.
In part (... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | 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-015979 | Set Theory: Partitions and Classes | 8 | Track quantifiers carefully: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $46\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[15]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In par... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-015980 | Algebraic Foundations: Congruence Modulo m | 8 | Keep the final answer in boxed form: 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 $[-94]$ 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 $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-015981 | Algebraic Foundations: Congruence Modulo m | 8 | Explain why your operations are valid: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $46\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-88]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | 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-015982 | Foundations: Relations from Fibers of Maps | 8 | Make each step logically reversible (or explain if not): 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 $[-7]$ explicitly as a set.
(c) Explain briefly how this relates to co... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 81$ 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-015983 | Set Theory: Partitions and Classes | 8 | Work carefully and justify each inference: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $93\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-55]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modu... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-015984 | Set Theory: Equivalence Relations — R/S/T | 8 | Indicate where a theorem is used: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $139\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-30]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
... | [
{
"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-015985 | Foundations: Relations from Fibers of Maps | 8 | Answer using clear logical steps: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $186\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[21]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-015986 | Algebraic Foundations: Congruence Modulo m | 8 | Explain each transformation: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $155\mid(a-b)$.
(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 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... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-015987 | Set Theory: Equivalence Relations — R/S/T | 8 | Provide both a computational and a conceptual explanation: 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 $[43]$ 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 $\\pi(a)=a\\bmod 56$ 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-015988 | Set Theory: Equivalence Relations — R/S/T | 8 | Warm-up: 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 $[63]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip t... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "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-015989 | Algebraic Foundations: Congruence Modulo m | 8 | Exercise: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $119\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[3]$ 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 $\\pi(a)=a\\bmod 119$ 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-015990 | Algebraic Foundations: Congruence Modulo m | 8 | Challenge: 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 $[-12]$ 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 $\\pi(a)=a\\bmod 167$ 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-015991 | Foundations: Relations from Fibers of Maps | 8 | Solve and sanity-check: 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 $[-87]$ 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-015992 | Algebraic Foundations: Congruence Modulo m | 8 | Solve and then verify: 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 $[18]$ 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... | 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-015993 | Set Theory: Equivalence Relations — R/S/T | 8 | Make each step logically reversible (or explain if not): 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 $[-58]$ 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": "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-015994 | Set Theory: Partitions and Classes | 8 | Keep the final answer in boxed form: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $166\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-72]$ 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 $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-015995 | Foundations: Relations from Fibers of Maps | 8 | Compute the requested quantity: 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 $[-2]$ 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 $\\pi(a)=a\\bmod 50$ be the canonical projection.",
"Step 2... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-015996 | Set Theory: Partitions and Classes | 8 | Explain what is being counted/optimized: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $13\mid(a-b)$.
(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 ... | [
{
"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-015997 | Set Theory: Partitions and Classes | 8 | Explain each transformation: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $199\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[77]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In pa... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 199$ 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-015998 | Set Theory: Equivalence Relations — R/S/T | 8 | Give a theorem-based solution: 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 $[-24]$ 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 $\\pi(a)=a\\bmod 50$ 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-015999 | Set Theory: Equivalence Relations — R/S/T | 8 | Explain why your operations are valid: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $157\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-42]$ 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... | 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-016000 | Foundations: Relations from Fibers of Maps | 8 | Solve with verification: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $78\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[76]$ 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... | 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.