id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-011901 | Proof Techniques: Induction — Avoiding Circularity | 6 | Solve and include a self-check: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(787)) = \frac{n(n+1)}{2}+(787)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the inducti... | [
{
"method_name": "Linearity of Summation",
"approach": "Split $\\sum(k+m)=\\sum k + \\sum m$ and use known formulas.",
"steps": [
"Step 1: Use linearity:\n$$\\sum_{k=1}^n (k+m)=\\sum_{k=1}^n k + \\sum_{k=1}^n m.$$",
"Step 2: Since $m$ is constant, $\\sum_{k=1}^n m = mn$.",
"Step 3: Use... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, they must agree.",
... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Key idea: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. |
math-011902 | Proof Techniques: Induction — Avoiding Circularity | 6 | Give an answer and a quick verification: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-798)) = \frac{n(n+1)}{2}+(-798)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In ... | [
{
"method_name": "Induction",
"approach": "Prove the base case, assume the statement for $n$, and deduce it for $n+1$.",
"steps": [
"Step 1: Base case $n=1$: LHS $=1+(-798)$, RHS $=\\frac{1\\cdot 2}{2}+(-798)\\cdot 1=1+(-798)$.",
"Step 2: Inductive hypothesis: assume $\\sum_{k=1}^n (k+m)=\\f... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, the... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Core principle: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. |
math-011903 | Foundations: Two-Proof Reconciliation | 6 | Warm-up: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-254)) = \frac{n(n+1)}{2}+(-254)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the induction step, explicitly s... | [
{
"method_name": "Linearity of Summation",
"approach": "Split $\\sum(k+m)=\\sum k + \\sum m$ and use known formulas.",
"steps": [
"Step 1: Use linearity:\n$$\\sum_{k=1}^n (k+m)=\\sum_{k=1}^n k + \\sum_{k=1}^n m.$$",
"Step 2: Since $m$ is constant, $\\sum_{k=1}^n m = mn$.",
"Step 3: Use... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, the... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Takeaway: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-011904 | Discrete Math: Summation Linearity Cross-Check | 6 | Do not skip justification steps: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-238)) = \frac{n(n+1)}{2}+(-238)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the indu... | [
{
"method_name": "Linearity of Summation",
"approach": "Split $\\sum(k+m)=\\sum k + \\sum m$ and use known formulas.",
"steps": [
"Step 1: Use linearity:\n$$\\sum_{k=1}^n (k+m)=\\sum_{k=1}^n k + \\sum_{k=1}^n m.$$",
"Step 2: Since $m$ is constant, $\\sum_{k=1}^n m = mn$.",
"Step 3: Use... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, they must agree.",
"robu... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Takeaway: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. |
math-011905 | Discrete Math: Summation Linearity Cross-Check | 6 | Find the exact value: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(478)) = \frac{n(n+1)}{2}+(478)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the induction step, e... | [
{
"method_name": "Linearity of Summation",
"approach": "Split $\\sum(k+m)=\\sum k + \\sum m$ and use known formulas.",
"steps": [
"Step 1: Use linearity:\n$$\\sum_{k=1}^n (k+m)=\\sum_{k=1}^n k + \\sum_{k=1}^n m.$$",
"Step 2: Since $m$ is constant, $\\sum_{k=1}^n m = mn$.",
"Step 3: Use... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, the... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Key idea: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-011906 | Discrete Math: Summation Linearity Cross-Check | 6 | Provide a rigorous solution: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(510)) = \frac{n(n+1)}{2}+(510)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the induction ... | [
{
"method_name": "Induction",
"approach": "Prove the base case, assume the statement for $n$, and deduce it for $n+1$.",
"steps": [
"Step 1: Base case $n=1$: LHS $=1+(510)$, RHS $=\\frac{1\\cdot 2}{2}+(510)\\cdot 1=1+(510)$.",
"Step 2: Inductive hypothesis: assume $\\sum_{k=1}^n (k+m)=\\frac... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quan... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Takeaway: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-011907 | Algebraic Foundations: Congruence Modulo m | 6 | Give a theorem-based solution: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $159\mid(a-b)$.
(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": "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-011908 | Discrete Math: Summation Linearity Cross-Check | 6 | Explain why your operations are valid: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(19)) = \frac{n(n+1)}{2}+(19)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the in... | [
{
"method_name": "Linearity of Summation",
"approach": "Split $\\sum(k+m)=\\sum k + \\sum m$ and use known formulas.",
"steps": [
"Step 1: Use linearity:\n$$\\sum_{k=1}^n (k+m)=\\sum_{k=1}^n k + \\sum_{k=1}^n m.$$",
"Step 2: Since $m$ is constant, $\\sum_{k=1}^n m = mn$.",
"Step 3: Use... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quan... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Takeaway: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-011909 | Proof Techniques: Induction — Avoiding Circularity | 6 | Provide a rigorous solution: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(742)) = \frac{n(n+1)}{2}+(742)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the induction ... | [
{
"method_name": "Induction",
"approach": "Prove the base case, assume the statement for $n$, and deduce it for $n+1$.",
"steps": [
"Step 1: Base case $n=1$: LHS $=1+(742)$, RHS $=\\frac{1\\cdot 2}{2}+(742)\\cdot 1=1+(742)$.",
"Step 2: Inductive hypothesis: assume $\\sum_{k=1}^n (k+m)=\\frac... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, they must agree.",
"robu... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Remember: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. |
math-011910 | Discrete Math: Summation Linearity Cross-Check | 6 | Give a fully justified solution: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-431)) = \frac{n(n+1)}{2}+(-431)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the indu... | [
{
"method_name": "Induction",
"approach": "Prove the base case, assume the statement for $n$, and deduce it for $n+1$.",
"steps": [
"Step 1: Base case $n=1$: LHS $=1+(-431)$, RHS $=\\frac{1\\cdot 2}{2}+(-431)\\cdot 1=1+(-431)$.",
"Step 2: Inductive hypothesis: assume $\\sum_{k=1}^n (k+m)=\\f... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, they must agree.",
"robu... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Remember: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. |
math-011911 | Algebraic Foundations: Congruence Modulo m | 6 | Carefully track domains: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $75\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-55]$ 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": "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-011912 | Set Theory: Partitions and Classes | 6 | Solve and include a self-check: 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 $[-51]$ 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 $\\pi(a)=a\\bmod 146$ be the canonical projection.",
"Step ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-011913 | Set Theory: Partitions and Classes | 6 | Solve and sanity-check: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $77\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[70]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a)... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-011914 | Proof Techniques: Induction — Avoiding Circularity | 6 | Find the exact value: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(582)) = \frac{n(n+1)}{2}+(582)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the induction step, e... | [
{
"method_name": "Induction",
"approach": "Prove the base case, assume the statement for $n$, and deduce it for $n+1$.",
"steps": [
"Step 1: Base case $n=1$: LHS $=1+(582)$, RHS $=\\frac{1\\cdot 2}{2}+(582)\\cdot 1=1+(582)$.",
"Step 2: Inductive hypothesis: assume $\\sum_{k=1}^n (k+m)=\\frac... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, they must agree.",
"robu... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Key idea: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. |
math-011915 | Proof Techniques: Induction — Avoiding Circularity | 6 | Give a theorem-based solution: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(46)) = \frac{n(n+1)}{2}+(46)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the induction ... | [
{
"method_name": "Linearity of Summation",
"approach": "Split $\\sum(k+m)=\\sum k + \\sum m$ and use known formulas.",
"steps": [
"Step 1: Use linearity:\n$$\\sum_{k=1}^n (k+m)=\\sum_{k=1}^n k + \\sum_{k=1}^n m.$$",
"Step 2: Since $m$ is constant, $\\sum_{k=1}^n m = mn$.",
"Step 3: Use... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, they must agree.",
... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Core principle: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-011916 | Foundations: Relations from Fibers of Maps | 6 | Prompt: 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 $[-80]$ 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-011917 | Set Theory: Equivalence Relations — R/S/T | 6 | Find the exact value: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $165\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[10]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a),... | [
{
"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-011918 | Foundations: Two-Proof Reconciliation | 6 | Find the exact value: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(460)) = \frac{n(n+1)}{2}+(460)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the induction step, e... | [
{
"method_name": "Linearity of Summation",
"approach": "Split $\\sum(k+m)=\\sum k + \\sum m$ and use known formulas.",
"steps": [
"Step 1: Use linearity:\n$$\\sum_{k=1}^n (k+m)=\\sum_{k=1}^n k + \\sum_{k=1}^n m.$$",
"Step 2: Since $m$ is constant, $\\sum_{k=1}^n m = mn$.",
"Step 3: Use... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quan... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Key idea: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. |
math-011919 | Proof Techniques: Induction — Base + Step | 6 | Explain why your operations are valid: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-74)) = \frac{n(n+1)}{2}+(-74)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the ... | [
{
"method_name": "Induction",
"approach": "Prove the base case, assume the statement for $n$, and deduce it for $n+1$.",
"steps": [
"Step 1: Base case $n=1$: LHS $=1+(-74)$, RHS $=\\frac{1\\cdot 2}{2}+(-74)\\cdot 1=1+(-74)$.",
"Step 2: Inductive hypothesis: assume $\\sum_{k=1}^n (k+m)=\\frac... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quan... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Key idea: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. |
math-011920 | Set Theory: Equivalence Relations — R/S/T | 6 | Warm-up: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $32\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[1]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
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": "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-011921 | Set Theory: Equivalence Relations — R/S/T | 6 | Solve and justify each step: 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 $[-69]$ 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 $\\pi(a)=a\\bmod 38$ 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-011922 | Set Theory: Equivalence Relations — R/S/T | 6 | Solve and then verify: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $171\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-39]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a... | [
{
"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-011923 | Proof Techniques: Induction — Avoiding Circularity | 6 | Warm-up: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-667)) = \frac{n(n+1)}{2}+(-667)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the induction step, explicitly s... | [
{
"method_name": "Induction",
"approach": "Prove the base case, assume the statement for $n$, and deduce it for $n+1$.",
"steps": [
"Step 1: Base case $n=1$: LHS $=1+(-667)$, RHS $=\\frac{1\\cdot 2}{2}+(-667)\\cdot 1=1+(-667)$.",
"Step 2: Inductive hypothesis: assume $\\sum_{k=1}^n (k+m)=\\f... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, the... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Remember: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-011924 | Set Theory: Equivalence Relations — R/S/T | 6 | Complete the analysis: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $53\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-87]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a)... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 53$ 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-011925 | Proof Techniques: Induction — Base + Step | 6 | Explain each transformation: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(268)) = \frac{n(n+1)}{2}+(268)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the induction ... | [
{
"method_name": "Linearity of Summation",
"approach": "Split $\\sum(k+m)=\\sum k + \\sum m$ and use known formulas.",
"steps": [
"Step 1: Use linearity:\n$$\\sum_{k=1}^n (k+m)=\\sum_{k=1}^n k + \\sum_{k=1}^n m.$$",
"Step 2: Since $m$ is constant, $\\sum_{k=1}^n m = mn$.",
"Step 3: Use... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, they must agree.",
"robu... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Remember: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. |
math-011926 | Foundations: Two-Proof Reconciliation | 6 | Track quantifiers carefully: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(277)) = \frac{n(n+1)}{2}+(277)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the induction ... | [
{
"method_name": "Induction",
"approach": "Prove the base case, assume the statement for $n$, and deduce it for $n+1$.",
"steps": [
"Step 1: Base case $n=1$: LHS $=1+(277)$, RHS $=\\frac{1\\cdot 2}{2}+(277)\\cdot 1=1+(277)$.",
"Step 2: Inductive hypothesis: assume $\\sum_{k=1}^n (k+m)=\\frac... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, they must agree.",
"robu... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Core principle: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-011927 | Algebraic Foundations: Congruence Modulo m | 6 | Compute the requested quantity: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $149\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[18]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "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-011928 | Set Theory: Partitions and Classes | 6 | Challenge: 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 $[-3]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not ski... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 113$ 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-011929 | Foundations: Relations from Fibers of Maps | 6 | Solve with verification: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $83\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[71]$ 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 $\\pi(a)=a\\bmod 83$ be the canonical projection.",
"Step 2... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-011930 | Foundations: Two-Proof Reconciliation | 6 | Solve (and briefly cross-validate): Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-351)) = \frac{n(n+1)}{2}+(-351)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the i... | [
{
"method_name": "Linearity of Summation",
"approach": "Split $\\sum(k+m)=\\sum k + \\sum m$ and use known formulas.",
"steps": [
"Step 1: Use linearity:\n$$\\sum_{k=1}^n (k+m)=\\sum_{k=1}^n k + \\sum_{k=1}^n m.$$",
"Step 2: Since $m$ is constant, $\\sum_{k=1}^n m = mn$.",
"Step 3: Use... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, they must agree.",
... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Key idea: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. |
math-011931 | Set Theory: Partitions and Classes | 6 | Solve and justify each step: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $18\mid(a-b)$.
(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 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": "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-011932 | Set Theory: Partitions and Classes | 6 | Provide a rigorous solution: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $182\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[90]$ 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 $\\pi(a)=a\\bmod 182$ 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-011933 | Proof Techniques: Induction — Base + Step | 6 | Explain why your operations are valid: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-759)) = \frac{n(n+1)}{2}+(-759)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In th... | [
{
"method_name": "Linearity of Summation",
"approach": "Split $\\sum(k+m)=\\sum k + \\sum m$ and use known formulas.",
"steps": [
"Step 1: Use linearity:\n$$\\sum_{k=1}^n (k+m)=\\sum_{k=1}^n k + \\sum_{k=1}^n m.$$",
"Step 2: Since $m$ is constant, $\\sum_{k=1}^n m = mn$.",
"Step 3: Use... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, they must agree.",
"robu... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Remember: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-011934 | Discrete Math: Summation Linearity Cross-Check | 6 | Give a theorem-based solution: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-191)) = \frac{n(n+1)}{2}+(-191)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the induct... | [
{
"method_name": "Induction",
"approach": "Prove the base case, assume the statement for $n$, and deduce it for $n+1$.",
"steps": [
"Step 1: Base case $n=1$: LHS $=1+(-191)$, RHS $=\\frac{1\\cdot 2}{2}+(-191)\\cdot 1=1+(-191)$.",
"Step 2: Inductive hypothesis: assume $\\sum_{k=1}^n (k+m)=\\f... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, they must agree.",
"robu... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Core principle: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. |
math-011935 | Set Theory: Partitions and Classes | 6 | Explain each transformation: 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 $[-75]$ 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 $\\pi(a)=a\\bmod 104$ 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 $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-011936 | Discrete Math: Summation Linearity Cross-Check | 6 | Keep the final answer in boxed form: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(150)) = \frac{n(n+1)}{2}+(150)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the in... | [
{
"method_name": "Linearity of Summation",
"approach": "Split $\\sum(k+m)=\\sum k + \\sum m$ and use known formulas.",
"steps": [
"Step 1: Use linearity:\n$$\\sum_{k=1}^n (k+m)=\\sum_{k=1}^n k + \\sum_{k=1}^n m.$$",
"Step 2: Since $m$ is constant, $\\sum_{k=1}^n m = mn$.",
"Step 3: Use... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, they must agree.",
... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Remember: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-011937 | Proof Techniques: Induction — Base + Step | 6 | Keep the final answer in boxed form: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(786)) = \frac{n(n+1)}{2}+(786)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the in... | [
{
"method_name": "Linearity of Summation",
"approach": "Split $\\sum(k+m)=\\sum k + \\sum m$ and use known formulas.",
"steps": [
"Step 1: Use linearity:\n$$\\sum_{k=1}^n (k+m)=\\sum_{k=1}^n k + \\sum_{k=1}^n m.$$",
"Step 2: Since $m$ is constant, $\\sum_{k=1}^n m = mn$.",
"Step 3: Use... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quan... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Core principle: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-011938 | Algebraic Foundations: Congruence Modulo m | 6 | Answer with a short justification: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $94\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[68]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
... | [
{
"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-011939 | Discrete Math: Summation Linearity Cross-Check | 6 | Solve and justify each step: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(453)) = \frac{n(n+1)}{2}+(453)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the induction ... | [
{
"method_name": "Induction",
"approach": "Prove the base case, assume the statement for $n$, and deduce it for $n+1$.",
"steps": [
"Step 1: Base case $n=1$: LHS $=1+(453)$, RHS $=\\frac{1\\cdot 2}{2}+(453)\\cdot 1=1+(453)$.",
"Step 2: Inductive hypothesis: assume $\\sum_{k=1}^n (k+m)=\\frac... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, the... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Takeaway: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-011940 | Algebraic Foundations: Congruence Modulo m | 6 | Solve and sanity-check: 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 $[71]$ 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 $\\pi(a)=a\\bmod 78$ 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-011941 | Discrete Math: Summation Linearity Cross-Check | 6 | Carefully track domains: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-649)) = \frac{n(n+1)}{2}+(-649)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the induction st... | [
{
"method_name": "Induction",
"approach": "Prove the base case, assume the statement for $n$, and deduce it for $n+1$.",
"steps": [
"Step 1: Base case $n=1$: LHS $=1+(-649)$, RHS $=\\frac{1\\cdot 2}{2}+(-649)\\cdot 1=1+(-649)$.",
"Step 2: Inductive hypothesis: assume $\\sum_{k=1}^n (k+m)=\\f... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, they must agree.",
"robu... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Key idea: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. |
math-011942 | Algebraic Foundations: Congruence Modulo m | 6 | Find the exact value: 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 $[17]$ 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 $\\pi(a)=a\\bmod 137$ 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-011943 | Proof Techniques: Induction — Base + Step | 6 | Do not skip justification steps: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-754)) = \frac{n(n+1)}{2}+(-754)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the indu... | [
{
"method_name": "Linearity of Summation",
"approach": "Split $\\sum(k+m)=\\sum k + \\sum m$ and use known formulas.",
"steps": [
"Step 1: Use linearity:\n$$\\sum_{k=1}^n (k+m)=\\sum_{k=1}^n k + \\sum_{k=1}^n m.$$",
"Step 2: Since $m$ is constant, $\\sum_{k=1}^n m = mn$.",
"Step 3: Use... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, they must agree.",
"robu... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Remember: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-011944 | Discrete Math: Summation Linearity Cross-Check | 6 | Determine the requested value: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(670)) = \frac{n(n+1)}{2}+(670)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the inductio... | [
{
"method_name": "Linearity of Summation",
"approach": "Split $\\sum(k+m)=\\sum k + \\sum m$ and use known formulas.",
"steps": [
"Step 1: Use linearity:\n$$\\sum_{k=1}^n (k+m)=\\sum_{k=1}^n k + \\sum_{k=1}^n m.$$",
"Step 2: Since $m$ is constant, $\\sum_{k=1}^n m = mn$.",
"Step 3: Use... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, the... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Key idea: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-011945 | Foundations: Two-Proof Reconciliation | 6 | Give an answer and a quick verification: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-396)) = \frac{n(n+1)}{2}+(-396)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In ... | [
{
"method_name": "Linearity of Summation",
"approach": "Split $\\sum(k+m)=\\sum k + \\sum m$ and use known formulas.",
"steps": [
"Step 1: Use linearity:\n$$\\sum_{k=1}^n (k+m)=\\sum_{k=1}^n k + \\sum_{k=1}^n m.$$",
"Step 2: Since $m$ is constant, $\\sum_{k=1}^n m = mn$.",
"Step 3: Use... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, they must agree.",
... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Core principle: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. |
math-011946 | Foundations: Two-Proof Reconciliation | 6 | Use two approaches if possible: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-703)) = \frac{n(n+1)}{2}+(-703)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the induc... | [
{
"method_name": "Linearity of Summation",
"approach": "Split $\\sum(k+m)=\\sum k + \\sum m$ and use known formulas.",
"steps": [
"Step 1: Use linearity:\n$$\\sum_{k=1}^n (k+m)=\\sum_{k=1}^n k + \\sum_{k=1}^n m.$$",
"Step 2: Since $m$ is constant, $\\sum_{k=1}^n m = mn$.",
"Step 3: Use... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, they must agree.",
... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Takeaway: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-011947 | Set Theory: Partitions and Classes | 6 | Give a theorem-based solution: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $165\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[90]$ explicitly as a set.
(c) Explain briefly how this relates to congruence 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 $\\pi(a)=a\\bmod 165$ 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-011948 | Proof Techniques: Induction — Avoiding Circularity | 6 | Track quantifiers carefully: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-32)) = \frac{n(n+1)}{2}+(-32)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the induction ... | [
{
"method_name": "Induction",
"approach": "Prove the base case, assume the statement for $n$, and deduce it for $n+1$.",
"steps": [
"Step 1: Base case $n=1$: LHS $=1+(-32)$, RHS $=\\frac{1\\cdot 2}{2}+(-32)\\cdot 1=1+(-32)$.",
"Step 2: Inductive hypothesis: assume $\\sum_{k=1}^n (k+m)=\\frac... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quan... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Key idea: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-011949 | Set Theory: Equivalence Relations — R/S/T | 6 | Work this out carefully: 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 $[-15]$ 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 $\\pi(a)=a\\bmod 50$ be the canonical projection.",
"Step 2... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-011950 | Algebraic Foundations: Congruence Modulo m | 6 | Give reasoning, not just computation: 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 $[19]$ 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-011951 | Set Theory: Equivalence Relations — R/S/T | 6 | Find the exact value: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $64\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[39]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), ... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-011952 | Algebraic Foundations: Congruence Modulo m | 6 | Proceed methodically: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $54\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[62]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), ... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "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-011953 | Algebraic Foundations: Congruence Modulo m | 6 | Show all reasoning: 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 $[85]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-011954 | Set Theory: Equivalence Relations — R/S/T | 6 | Task: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $136\mid(a-b)$.
(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 $m$.
In part (a), do not skip the... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 136$ 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 $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-011955 | Discrete Math: Summation Linearity Cross-Check | 6 | Compute the requested quantity: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-602)) = \frac{n(n+1)}{2}+(-602)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the induc... | [
{
"method_name": "Induction",
"approach": "Prove the base case, assume the statement for $n$, and deduce it for $n+1$.",
"steps": [
"Step 1: Base case $n=1$: LHS $=1+(-602)$, RHS $=\\frac{1\\cdot 2}{2}+(-602)\\cdot 1=1+(-602)$.",
"Step 2: Inductive hypothesis: assume $\\sum_{k=1}^n (k+m)=\\f... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, they must agree.",
"robu... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Key idea: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-011956 | Set Theory: Partitions and Classes | 6 | Provide both a computational and a conceptual explanation: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $69\mid(a-b)$.
(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 ... | [
{
"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 $\\pi(a)=a\\bmod 69$ be the canonical projection.",
"Step 2... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-011957 | Algebraic Foundations: Congruence Modulo m | 6 | Solve and sanity-check: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $151\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-28]$ 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... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-011958 | Proof Techniques: Induction — Base + Step | 6 | Determine the requested value: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(522)) = \frac{n(n+1)}{2}+(522)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the inductio... | [
{
"method_name": "Induction",
"approach": "Prove the base case, assume the statement for $n$, and deduce it for $n+1$.",
"steps": [
"Step 1: Base case $n=1$: LHS $=1+(522)$, RHS $=\\frac{1\\cdot 2}{2}+(522)\\cdot 1=1+(522)$.",
"Step 2: Inductive hypothesis: assume $\\sum_{k=1}^n (k+m)=\\frac... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, they must agree.",
"robu... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Key idea: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. |
math-011959 | Discrete Math: Summation Linearity Cross-Check | 6 | Explain each transformation: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(675)) = \frac{n(n+1)}{2}+(675)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the induction ... | [
{
"method_name": "Linearity of Summation",
"approach": "Split $\\sum(k+m)=\\sum k + \\sum m$ and use known formulas.",
"steps": [
"Step 1: Use linearity:\n$$\\sum_{k=1}^n (k+m)=\\sum_{k=1}^n k + \\sum_{k=1}^n m.$$",
"Step 2: Since $m$ is constant, $\\sum_{k=1}^n m = mn$.",
"Step 3: Use... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, the... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Takeaway: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-011960 | Algebraic Foundations: Congruence Modulo m | 6 | Use two approaches if possible: 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 $[-20]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-011961 | Algebraic Foundations: Congruence Modulo m | 6 | Make each step logically reversible (or explain if not): Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $98\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-25]$ explicitly as a set.
(c) Explain briefly how this relates to 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-011962 | Proof Techniques: Induction — Base + Step | 6 | Where appropriate, name the theorem you use: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(290)) = \frac{n(n+1)}{2}+(290)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
I... | [
{
"method_name": "Linearity of Summation",
"approach": "Split $\\sum(k+m)=\\sum k + \\sum m$ and use known formulas.",
"steps": [
"Step 1: Use linearity:\n$$\\sum_{k=1}^n (k+m)=\\sum_{k=1}^n k + \\sum_{k=1}^n m.$$",
"Step 2: Since $m$ is constant, $\\sum_{k=1}^n m = mn$.",
"Step 3: Use... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quan... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Takeaway: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-011963 | Set Theory: Partitions and Classes | 6 | Answer using clear logical steps: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $41\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[23]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
I... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 41$ 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... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-011964 | Algebraic Foundations: Congruence Modulo m | 6 | Exercise: 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 $[46]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | 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-011965 | Set Theory: Partitions and Classes | 6 | Provide both a computational and a conceptual explanation: 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 $[-53]$ 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 $\\pi(a)=a\\bmod 131$ 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... | 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-011966 | Foundations: Relations from Fibers of Maps | 6 | Be explicit about assumptions: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $192\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-58]$ 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 $\\pi(a)=a\\bmod 192$ 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-011967 | Set Theory: Equivalence Relations — R/S/T | 6 | Find the exact value: 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 $[24]$ 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... | 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-011968 | Proof Techniques: Induction — Avoiding Circularity | 6 | Determine the requested value: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(748)) = \frac{n(n+1)}{2}+(748)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the inductio... | [
{
"method_name": "Linearity of Summation",
"approach": "Split $\\sum(k+m)=\\sum k + \\sum m$ and use known formulas.",
"steps": [
"Step 1: Use linearity:\n$$\\sum_{k=1}^n (k+m)=\\sum_{k=1}^n k + \\sum_{k=1}^n m.$$",
"Step 2: Since $m$ is constant, $\\sum_{k=1}^n m = mn$.",
"Step 3: Use... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, the... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Core principle: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-011969 | Foundations: Two-Proof Reconciliation | 6 | Challenge: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-218)) = \frac{n(n+1)}{2}+(-218)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the induction step, explicitly... | [
{
"method_name": "Linearity of Summation",
"approach": "Split $\\sum(k+m)=\\sum k + \\sum m$ and use known formulas.",
"steps": [
"Step 1: Use linearity:\n$$\\sum_{k=1}^n (k+m)=\\sum_{k=1}^n k + \\sum_{k=1}^n m.$$",
"Step 2: Since $m$ is constant, $\\sum_{k=1}^n m = mn$.",
"Step 3: Use... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, they must agree.",
... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Takeaway: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-011970 | Algebraic Foundations: Congruence Modulo m | 6 | Solve and justify each step: 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 $[-77]$ 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 $\\pi(a)=a\\bmod 194$ be the canonical projection.",
"Step ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-011971 | Foundations: Two-Proof Reconciliation | 6 | Answer with a short justification: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(178)) = \frac{n(n+1)}{2}+(178)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the indu... | [
{
"method_name": "Linearity of Summation",
"approach": "Split $\\sum(k+m)=\\sum k + \\sum m$ and use known formulas.",
"steps": [
"Step 1: Use linearity:\n$$\\sum_{k=1}^n (k+m)=\\sum_{k=1}^n k + \\sum_{k=1}^n m.$$",
"Step 2: Since $m$ is constant, $\\sum_{k=1}^n m = mn$.",
"Step 3: Use... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, they must agree.",
"robu... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Remember: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-011972 | Proof Techniques: Induction — Avoiding Circularity | 6 | Warm-up: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(117)) = \frac{n(n+1)}{2}+(117)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the induction step, explicitly sho... | [
{
"method_name": "Induction",
"approach": "Prove the base case, assume the statement for $n$, and deduce it for $n+1$.",
"steps": [
"Step 1: Base case $n=1$: LHS $=1+(117)$, RHS $=\\frac{1\\cdot 2}{2}+(117)\\cdot 1=1+(117)$.",
"Step 2: Inductive hypothesis: assume $\\sum_{k=1}^n (k+m)=\\frac... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, they must agree.",
"robu... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Core principle: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-011973 | Proof Techniques: Induction — Avoiding Circularity | 6 | Solve and sanity-check: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-692)) = \frac{n(n+1)}{2}+(-692)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the induction ste... | [
{
"method_name": "Induction",
"approach": "Prove the base case, assume the statement for $n$, and deduce it for $n+1$.",
"steps": [
"Step 1: Base case $n=1$: LHS $=1+(-692)$, RHS $=\\frac{1\\cdot 2}{2}+(-692)\\cdot 1=1+(-692)$.",
"Step 2: Inductive hypothesis: assume $\\sum_{k=1}^n (k+m)=\\f... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quan... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Takeaway: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-011974 | Foundations: Relations from Fibers of Maps | 6 | Be explicit about assumptions: 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 $[93]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In p... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-011975 | Algebraic Foundations: Congruence Modulo m | 6 | Problem: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $6\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[53]$ 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": "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-011976 | Foundations: Relations from Fibers of Maps | 6 | Solve (and briefly cross-validate): 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 $[-93]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-011977 | Foundations: Relations from Fibers of Maps | 6 | Answer using clear logical steps: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $99\mid(a-b)$.
(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": "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 $\\pi(a)=a\\bmod 99$ be the canonical projection.",
"Step 2... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-011978 | Proof Techniques: Induction — Avoiding Circularity | 6 | Write the solution set clearly: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-175)) = \frac{n(n+1)}{2}+(-175)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the induc... | [
{
"method_name": "Induction",
"approach": "Prove the base case, assume the statement for $n$, and deduce it for $n+1$.",
"steps": [
"Step 1: Base case $n=1$: LHS $=1+(-175)$, RHS $=\\frac{1\\cdot 2}{2}+(-175)\\cdot 1=1+(-175)$.",
"Step 2: Inductive hypothesis: assume $\\sum_{k=1}^n (k+m)=\\f... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, they must agree.",
... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Key idea: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. |
math-011979 | Foundations: Relations from Fibers of Maps | 6 | Solve and include a self-check: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $160\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[22]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In... | [
{
"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 $\\pi(a)=a\\bmod 160$ be the canonical projection.",
"Step ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-011980 | Foundations: Relations from Fibers of Maps | 6 | Answer using clear logical steps: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $98\mid(a-b)$.
(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 $m$.
I... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-011981 | Foundations: Relations from Fibers of Maps | 6 | Proceed methodically: 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 $[33]$ 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... | 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-011982 | Foundations: Two-Proof Reconciliation | 6 | Explain each transformation: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(308)) = \frac{n(n+1)}{2}+(308)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the induction ... | [
{
"method_name": "Induction",
"approach": "Prove the base case, assume the statement for $n$, and deduce it for $n+1$.",
"steps": [
"Step 1: Base case $n=1$: LHS $=1+(308)$, RHS $=\\frac{1\\cdot 2}{2}+(308)\\cdot 1=1+(308)$.",
"Step 2: Inductive hypothesis: assume $\\sum_{k=1}^n (k+m)=\\frac... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, they must agree.",
"robu... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Takeaway: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-011983 | Set Theory: Equivalence Relations — R/S/T | 6 | Give a theorem-based solution: 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 $[94]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In p... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-011984 | Algebraic Foundations: Congruence Modulo m | 6 | Give a fully justified solution: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $197\mid(a-b)$.
(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 modulo $m$.
I... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-011985 | Set Theory: Equivalence Relations — R/S/T | 6 | Write the solution set clearly: 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 $[44]$ 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 $\\pi(a)=a\\bmod 61$ 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... | 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-011986 | Set Theory: Partitions and Classes | 6 | Provide both a computational and a conceptual explanation: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $31\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[75]$ explicitly as a set.
(c) Explain briefly how this relates to ... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "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-011987 | Foundations: Relations from Fibers of Maps | 6 | Solve and include a self-check: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $17\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-61]$ 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": "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-011988 | Discrete Math: Summation Linearity Cross-Check | 6 | Where appropriate, name the theorem you use: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(613)) = \frac{n(n+1)}{2}+(613)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
I... | [
{
"method_name": "Induction",
"approach": "Prove the base case, assume the statement for $n$, and deduce it for $n+1$.",
"steps": [
"Step 1: Base case $n=1$: LHS $=1+(613)$, RHS $=\\frac{1\\cdot 2}{2}+(613)\\cdot 1=1+(613)$.",
"Step 2: Inductive hypothesis: assume $\\sum_{k=1}^n (k+m)=\\frac... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quan... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Remember: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-011989 | Set Theory: Partitions and Classes | 6 | Try to avoid pattern-matching; explain why: 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 $[24]$ 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": "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-011990 | Foundations: Relations from Fibers of Maps | 6 | Provide a rigorous solution: 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 $[60]$ 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 $\\pi(a)=a\\bmod 8$ 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-011991 | Foundations: Two-Proof Reconciliation | 6 | Exercise: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(319)) = \frac{n(n+1)}{2}+(319)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the induction step, explicitly sh... | [
{
"method_name": "Induction",
"approach": "Prove the base case, assume the statement for $n$, and deduce it for $n+1$.",
"steps": [
"Step 1: Base case $n=1$: LHS $=1+(319)$, RHS $=\\frac{1\\cdot 2}{2}+(319)\\cdot 1=1+(319)$.",
"Step 2: Inductive hypothesis: assume $\\sum_{k=1}^n (k+m)=\\frac... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quantity, the... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Key idea: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. |
math-011992 | Algebraic Foundations: Congruence Modulo m | 6 | Solve and justify each step: 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 $[4]$ 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 $\\pi(a)=a\\bmod 68$ 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... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-011993 | Algebraic Foundations: Congruence Modulo m | 6 | Be explicit about assumptions: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $64\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-81]$ 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 $\\pi(a)=a\\bmod 64$ be the canonical projection.",
"Step 2... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-011994 | Set Theory: Equivalence Relations — R/S/T | 6 | Determine the requested value: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $23\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-32]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In ... | [
{
"method_name": "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 $\\pi(a)=a\\bmod 23$ 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... | 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-011995 | Foundations: Relations from Fibers of Maps | 6 | Where appropriate, name the theorem you use: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $9\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[53]$ explicitly as a set.
(c) Explain briefly how this relates to congruence 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": "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-011996 | Set Theory: Equivalence Relations — R/S/T | 6 | Prompt: 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 $[80]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip the... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 3$ 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-011997 | Algebraic Foundations: Congruence Modulo m | 6 | Carefully track domains: 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 $[-79]$ 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": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-011998 | Foundations: Two-Proof Reconciliation | 6 | Work this out carefully: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-476)) = \frac{n(n+1)}{2}+(-476)n.$$
(a) Give a full induction proof.
(b) Give a second proof using linearity of summation.
(c) Explain why (b) can be viewed as a 'sanity check' on the induction algebra.
In the induction st... | [
{
"method_name": "Linearity of Summation",
"approach": "Split $\\sum(k+m)=\\sum k + \\sum m$ and use known formulas.",
"steps": [
"Step 1: Use linearity:\n$$\\sum_{k=1}^n (k+m)=\\sum_{k=1}^n k + \\sum_{k=1}^n m.$$",
"Step 2: Since $m$ is constant, $\\sum_{k=1}^n m = mn$.",
"Step 3: Use... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\frac{n(n+1)}$.\nInduction proves the identity for all $n$ by logic. The linearity method derives the same closed form directly from known sums. Since both are valid derivations of the same quan... | [
{
"error_description": "Used circular reasoning by assuming the $(n+1)$ case in the inductive step.",
"why_plausible": "It is tempting to rewrite expressions into the target form by 'recognizing' it prematurely.",
"why_wrong": "Induction requires deriving $P(n+1)$ from $P(n)$ only; assuming $P(n+1)$ inv... | Key idea: Induction is a rigorous engine for 'for all $n$' claims, while linearity/sum-splitting offers a quick independent check when the expression decomposes into known sums. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-011999 | Set Theory: Equivalence Relations — R/S/T | 6 | Compute the requested quantity: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $198\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-97]$ 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 $\\pi(a)=a\\bmod 198$ 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 $r+m\mathbb{Z}$. |
math-012000 | Set Theory: Equivalence Relations — R/S/T | 6 | Solve (and briefly cross-validate): 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 $[79]$ 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 $\\pi(a)=a\\bmod 4$ 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 $r+m\mathbb{Z}$. |
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