id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-013901 | Foundations: Two-Proof Reconciliation | 7 | Solve and justify each step: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-793)) = \frac{n(n+1)}{2}+(-793)n.$$
(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 $=1+(-793)$, RHS $=\\frac{1\\cdot 2}{2}+(-793)\\cdot 1=1+(-793)$.",
"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... | 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-013902 | Foundations: Relations from Fibers of Maps | 7 | Checkpoint: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $110\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-60]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not s... | [
{
"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-013903 | Proof Techniques: Induction — Base + Step | 7 | Answer with a short justification: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-655)) = \frac{n(n+1)}{2}+(-655)n.$$
(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-013904 | Foundations: Two-Proof Reconciliation | 7 | Task: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-106)) = \frac{n(n+1)}{2}+(-106)n.$$
(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 show... | [
{
"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-013905 | Discrete Math: Summation Linearity Cross-Check | 7 | Provide a rigorous solution: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(367)) = \frac{n(n+1)}{2}+(367)n.$$
(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. |
math-013906 | Algebraic Foundations: Congruence Modulo m | 7 | State any required conditions first: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $191\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[79]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-013907 | Discrete Math: Summation Linearity Cross-Check | 7 | Question: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(49)) = \frac{n(n+1)}{2}+(49)n.$$
(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 show... | [
{
"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-013908 | Set Theory: Partitions and Classes | 7 | Provide a rigorous solution: 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 $[36]$ 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 $\\pi(a)=a\\bmod 149$ be the canonical projection.",
"Step ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-013909 | Set Theory: Equivalence Relations — R/S/T | 7 | Question: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $166\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[99]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 166$ 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-013910 | Proof Techniques: Induction — Base + Step | 7 | Solve with verification: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(366)) = \frac{n(n+1)}{2}+(366)n.$$
(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... | [
{
"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... | 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-013911 | Set Theory: Partitions and Classes | 7 | Find the exact value: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $79\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[69]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), ... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 79$ be the canonical projection.",
"Step 2... | {
"consistency_check": "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-013912 | Proof Techniques: Induction — Base + Step | 7 | Warm-up: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-604)) = \frac{n(n+1)}{2}+(-604)n.$$
(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 $=1+(-604)$, RHS $=\\frac{1\\cdot 2}{2}+(-604)\\cdot 1=1+(-604)$.",
"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-013913 | Foundations: Relations from Fibers of Maps | 7 | Derive the result step-by-step: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $96\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-13]$ 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 $\\pi(a)=a\\bmod 96$ be the canonical projection.",
"Step 2... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-013914 | Set Theory: Partitions and Classes | 7 | Track units/moduli carefully: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $111\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[48]$ 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 $\\pi(a)=a\\bmod 111$ be the canonical projection.",
"Step ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-013915 | Proof Techniques: Induction — Avoiding Circularity | 7 | Give a fully justified solution: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(719)) = \frac{n(n+1)}{2}+(719)n.$$
(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 $=1+(719)$, RHS $=\\frac{1\\cdot 2}{2}+(719)\\cdot 1=1+(719)$.",
"Step 2: Inductive hypothesis: assume $\\sum_{k=1}^n (k+m)=\\frac... | {
"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-013916 | Algebraic Foundations: Congruence Modulo m | 7 | Carefully track domains: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $10\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[13]$ 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... | 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-013917 | Proof Techniques: Induction — Avoiding Circularity | 7 | Solve (and briefly cross-validate): Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(189)) = \frac{n(n+1)}{2}+(189)n.$$
(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 ind... | [
{
"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 $=1+(189)$, RHS $=\\frac{1\\cdot 2}{2}+(189)\\cdot 1=1+(189)$.",
"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... | 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-013918 | Set Theory: Partitions and Classes | 7 | Question: 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 $[4]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip ... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "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-013919 | Algebraic Foundations: Congruence Modulo m | 7 | Derive the result step-by-step: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $80\mid(a-b)$.
(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 ... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-013920 | Proof Techniques: Induction — Avoiding Circularity | 7 | Try to avoid pattern-matching; explain why: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-379)) = \frac{n(n+1)}{2}+(-379)n.$$
(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.
... | [
{
"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. |
math-013921 | Foundations: Relations from Fibers of Maps | 7 | Do not skip justification steps: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $112\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[44]$ 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 $\\pi(a)=a\\bmod 112$ 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-013922 | Foundations: Relations from Fibers of Maps | 7 | Problem: 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 $[-49]$ 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 $r+m\mathbb{Z}$. |
math-013923 | Algebraic Foundations: Congruence Modulo m | 7 | Problem: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $121\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[7]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip t... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 121$ be the canonical projection.",
"Step ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-013924 | Foundations: Relations from Fibers of Maps | 7 | Try to avoid pattern-matching; explain why: 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 $[-33]$ explicitly as a set.
(c) Explain briefly how this relates to congruence mo... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 123$ be the canonical projection.",
"Step ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-013925 | Proof Techniques: Induction — Avoiding Circularity | 7 | Prompt: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-1)) = \frac{n(n+1)}{2}+(-1)n.$$
(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 show h... | [
{
"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... | 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-013926 | Set Theory: Equivalence Relations — R/S/T | 7 | Question: 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 $[71]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 171$ be the canonical projection.",
"Step ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-013927 | Foundations: Relations from Fibers of Maps | 7 | Task: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $14\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[2]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip the t... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "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-013928 | Proof Techniques: Induction — Base + Step | 7 | Answer using clear logical steps: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-715)) = \frac{n(n+1)}{2}+(-715)n.$$
(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 ind... | [
{
"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 $=1+(-715)$, RHS $=\\frac{1\\cdot 2}{2}+(-715)\\cdot 1=1+(-715)$.",
"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... | 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-013929 | Algebraic Foundations: Congruence Modulo m | 7 | Explain why your operations are valid: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $14\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[44]$ 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 $\\pi(a)=a\\bmod 14$ be the canonical projection.",
"Step 2... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-013930 | Algebraic Foundations: Congruence Modulo m | 7 | Write the solution set clearly: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $141\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[32]$ 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 $\\pi(a)=a\\bmod 141$ be the canonical projection.",
"Step ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-013931 | Foundations: Relations from Fibers of Maps | 7 | Write the solution set clearly: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $177\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[87]$ 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 $\\pi(a)=a\\bmod 177$ be the canonical projection.",
"Step ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-013932 | Proof Techniques: Induction — Avoiding Circularity | 7 | Question: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(240)) = \frac{n(n+1)}{2}+(240)n.$$
(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": "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... | 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-013933 | Foundations: Two-Proof Reconciliation | 7 | Give an answer and a quick verification: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(275)) = \frac{n(n+1)}{2}+(275)n.$$
(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": "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. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-013934 | Foundations: Two-Proof Reconciliation | 7 | Solve and then verify: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(68)) = \frac{n(n+1)}{2}+(68)n.$$
(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, ex... | [
{
"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... | 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-013935 | Algebraic Foundations: Congruence Modulo m | 7 | State any required conditions first: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $43\mid(a-b)$.
(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 $m$... | [
{
"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-013936 | Foundations: Two-Proof Reconciliation | 7 | Give an answer and a quick verification: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(511)) = \frac{n(n+1)}{2}+(511)n.$$
(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": "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 $=1+(511)$, RHS $=\\frac{1\\cdot 2}{2}+(511)\\cdot 1=1+(511)$.",
"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. |
math-013937 | Set Theory: Equivalence Relations — R/S/T | 7 | Do not skip justification steps: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $93\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[91]$ 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 $\\pi(a)=a\\bmod 93$ 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-013938 | Set Theory: Partitions and Classes | 7 | Explain why your operations are valid: 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 $[85]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 75$ be the canonical projection.",
"Step 2... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-013939 | Set Theory: Partitions and Classes | 7 | Complete the analysis: 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 $[97]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a)... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-013940 | Algebraic Foundations: Congruence Modulo m | 7 | Write the solution set clearly: 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 $[23]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In ... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "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-013941 | Proof Techniques: Induction — Base + Step | 7 | Checkpoint: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(389)) = \frac{n(n+1)}{2}+(389)n.$$
(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": "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 $=1+(389)$, RHS $=\\frac{1\\cdot 2}{2}+(389)\\cdot 1=1+(389)$.",
"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. |
math-013942 | Set Theory: Equivalence Relations — R/S/T | 7 | Give reasoning, not just computation: 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 $[82]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-013943 | Algebraic Foundations: Congruence Modulo m | 7 | Solve (and briefly cross-validate): Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $47\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-71]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-013944 | Algebraic Foundations: Congruence Modulo m | 7 | Make each step logically reversible (or explain if not): Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $107\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-84]$ explicitly as a set.
(c) Explain briefly how this relates to ... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | 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-013945 | Discrete Math: Summation Linearity Cross-Check | 7 | Provide both a computational and a conceptual explanation: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-391)) = \frac{n(n+1)}{2}+(-391)n.$$
(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 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": "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-013946 | Foundations: Two-Proof Reconciliation | 7 | Exercise: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-189)) = \frac{n(n+1)}{2}+(-189)n.$$
(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": "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-013947 | Discrete Math: Summation Linearity Cross-Check | 7 | Keep the final answer in boxed form: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(473)) = \frac{n(n+1)}{2}+(473)n.$$
(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": "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-013948 | Proof Techniques: Induction — Avoiding Circularity | 7 | Proceed methodically: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-551)) = \frac{n(n+1)}{2}+(-551)n.$$
(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,... | [
{
"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-013949 | Proof Techniques: Induction — Avoiding Circularity | 7 | Explain why your operations are valid: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-549)) = \frac{n(n+1)}{2}+(-549)n.$$
(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": "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-013950 | Algebraic Foundations: Congruence Modulo m | 7 | Solve and justify each step: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $189\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[12]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In 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 $\\pi(a)=a\\bmod 189$ be the canonical projection.",
"Step ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Key idea: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-013951 | Proof Techniques: Induction — Avoiding Circularity | 7 | Solve with verification: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(654)) = \frac{n(n+1)}{2}+(654)n.$$
(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... | [
{
"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. |
math-013952 | Foundations: Relations from Fibers of Maps | 7 | Prompt: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $100\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-33]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip ... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "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-013953 | Set Theory: Equivalence Relations — R/S/T | 7 | Find the exact value: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $36\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-62]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a),... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 36$ be the canonical projection.",
"Step 2... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-013954 | Algebraic Foundations: Congruence Modulo m | 7 | Explain why your operations are valid: 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 $[-73]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 64$ be the canonical projection.",
"Step 2... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-013955 | Algebraic Foundations: Congruence Modulo m | 7 | State any required conditions first: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $108\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[22]$ 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 $\\pi(a)=a\\bmod 108$ be the canonical projection.",
"Step ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-013956 | Algebraic Foundations: Congruence Modulo m | 7 | Explain each transformation: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $71\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-21]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In pa... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-013957 | Foundations: Two-Proof Reconciliation | 7 | Proceed methodically: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(592)) = \frac{n(n+1)}{2}+(592)n.$$
(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 $=1+(592)$, RHS $=\\frac{1\\cdot 2}{2}+(592)\\cdot 1=1+(592)$.",
"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. |
math-013958 | Proof Techniques: Induction — Base + Step | 7 | Indicate where a theorem is used: 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 ind... | [
{
"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-013959 | Set Theory: Partitions and Classes | 7 | Prompt: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $93\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[89]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip th... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 93$ be the canonical projection.",
"Step 2... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-013960 | Foundations: Two-Proof Reconciliation | 7 | Be explicit about assumptions: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-726)) = \frac{n(n+1)}{2}+(-726)n.$$
(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": "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-013961 | Algebraic Foundations: Congruence Modulo m | 7 | Problem: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $5\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-74]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), do not skip t... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 5$ be the canonical projection.",
"Step 2:... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-013962 | Discrete Math: Summation Linearity Cross-Check | 7 | Give a fully justified solution: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-548)) = \frac{n(n+1)}{2}+(-548)n.$$
(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": "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. |
math-013963 | Foundations: Relations from Fibers of Maps | 7 | Answer using clear logical steps: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $176\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-27]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-013964 | Proof Techniques: Induction — Avoiding Circularity | 7 | Warm-up: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-200)) = \frac{n(n+1)}{2}+(-200)n.$$
(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 $=1+(-200)$, RHS $=\\frac{1\\cdot 2}{2}+(-200)\\cdot 1=1+(-200)$.",
"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-013965 | Discrete Math: Summation Linearity Cross-Check | 7 | Challenge: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(448)) = \frac{n(n+1)}{2}+(448)n.$$
(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... | 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-013966 | Set Theory: Equivalence Relations — R/S/T | 7 | Explain what is being counted/optimized: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $56\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-41]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 56$ be the canonical projection.",
"Step 2... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | 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-013967 | Foundations: Two-Proof Reconciliation | 7 | Answer with a short justification: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-192)) = \frac{n(n+1)}{2}+(-192)n.$$
(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": "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-013968 | Discrete Math: Summation Linearity Cross-Check | 7 | Do not skip justification steps: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-664)) = \frac{n(n+1)}{2}+(-664)n.$$
(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 $=1+(-664)$, RHS $=\\frac{1\\cdot 2}{2}+(-664)\\cdot 1=1+(-664)$.",
"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-013969 | Algebraic Foundations: Congruence Modulo m | 7 | Explain why your operations are valid: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $121\mid(a-b)$.
(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 ... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-013970 | Algebraic Foundations: Congruence Modulo m | 7 | Keep the final answer in boxed form: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $52\mid(a-b)$.
(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": "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-013971 | Set Theory: Equivalence Relations — R/S/T | 7 | Track units/moduli carefully: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $63\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-92]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In p... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-013972 | Foundations: Two-Proof Reconciliation | 7 | Problem: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-402)) = \frac{n(n+1)}{2}+(-402)n.$$
(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... | 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-013973 | Foundations: Relations from Fibers of Maps | 7 | Give a theorem-based solution: 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 $[87]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In p... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | 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-013974 | Algebraic Foundations: Congruence Modulo m | 7 | Answer using clear logical steps: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $189\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-50]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
... | [
{
"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 $\\pi(a)=a\\bmod 189$ 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-013975 | Proof Techniques: Induction — Avoiding Circularity | 7 | Keep the final answer in boxed form: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-310)) = \frac{n(n+1)}{2}+(-310)n.$$
(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 $=1+(-310)$, RHS $=\\frac{1\\cdot 2}{2}+(-310)\\cdot 1=1+(-310)$.",
"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-013976 | Proof Techniques: Induction — Avoiding Circularity | 7 | Solve with verification: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(706)) = \frac{n(n+1)}{2}+(706)n.$$
(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... | [
{
"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 $=1+(706)$, RHS $=\\frac{1\\cdot 2}{2}+(706)\\cdot 1=1+(706)$.",
"Step 2: Inductive hypothesis: assume $\\sum_{k=1}^n (k+m)=\\frac... | {
"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-013977 | Set Theory: Partitions and Classes | 7 | Indicate where a theorem is used: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $5\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[15]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "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-013978 | Foundations: Two-Proof Reconciliation | 7 | Explain each transformation: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(375)) = \frac{n(n+1)}{2}+(375)n.$$
(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 $=1+(375)$, RHS $=\\frac{1\\cdot 2}{2}+(375)\\cdot 1=1+(375)$.",
"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. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-013979 | Algebraic Foundations: Congruence Modulo m | 7 | Question: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $156\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-86]$ 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 $\\pi(a)=a\\bmod 156$ be the canonical projection.",
"Step ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-013980 | Set Theory: Equivalence Relations — R/S/T | 7 | Solve with verification: 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 $[5]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a)... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-013981 | Set Theory: Partitions and Classes | 7 | Derive the result step-by-step: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $135\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[62]$ 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 $\\pi(a)=a\\bmod 135$ be the canonical projection.",
"Step ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Remember: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-013982 | Discrete Math: Summation Linearity Cross-Check | 7 | Complete the analysis: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-536)) = \frac{n(n+1)}{2}+(-536)n.$$
(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... | [
{
"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-013983 | Discrete Math: Summation Linearity Cross-Check | 7 | Prompt: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(-179)) = \frac{n(n+1)}{2}+(-179)n.$$
(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 $=1+(-179)$, RHS $=\\frac{1\\cdot 2}{2}+(-179)\\cdot 1=1+(-179)$.",
"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... | 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-013984 | Foundations: Relations from Fibers of Maps | 7 | Work carefully and justify each inference: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $58\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[48]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modul... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 58$ 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-013985 | Foundations: Two-Proof Reconciliation | 7 | Make each step logically reversible (or explain if not): Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(125)) = \frac{n(n+1)}{2}+(125)n.$$
(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... | [
{
"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 $=1+(125)$, RHS $=\\frac{1\\cdot 2}{2}+(125)\\cdot 1=1+(125)$.",
"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-013986 | Proof Techniques: Induction — Avoiding Circularity | 7 | Track units/moduli carefully: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(449)) = \frac{n(n+1)}{2}+(449)n.$$
(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... | 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-013987 | Proof Techniques: Induction — Base + Step | 7 | Prompt: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(246)) = \frac{n(n+1)}{2}+(246)n.$$
(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 show... | [
{
"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. |
math-013988 | Foundations: Two-Proof Reconciliation | 7 | Solve and sanity-check: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(415)) = \frac{n(n+1)}{2}+(415)n.$$
(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,... | [
{
"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 $=1+(415)$, RHS $=\\frac{1\\cdot 2}{2}+(415)\\cdot 1=1+(415)$.",
"Step 2: Inductive hypothesis: assume $\\sum_{k=1}^n (k+m)=\\frac... | {
"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. (Here the result is $\boxed{\frac{n(n+1)}$.) |
math-013989 | Foundations: Two-Proof Reconciliation | 7 | Solve and sanity-check: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(653)) = \frac{n(n+1)}{2}+(653)n.$$
(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,... | [
{
"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-013990 | Proof Techniques: Induction — Avoiding Circularity | 7 | Problem: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(450)) = \frac{n(n+1)}{2}+(450)n.$$
(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 $=1+(450)$, RHS $=\\frac{1\\cdot 2}{2}+(450)\\cdot 1=1+(450)$.",
"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... | 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-013991 | Algebraic Foundations: Congruence Modulo m | 7 | Give reasoning, not just computation: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $186\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-40]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 186$ be the canonical projection.",
"Step ... | {
"consistency_check": "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-013992 | Set Theory: Partitions and Classes | 7 | Solve (and briefly cross-validate): Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $13\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-31]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an equiv... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Takeaway: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-013993 | Proof Techniques: Induction — Base + Step | 7 | Give reasoning, not just computation: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(599)) = \frac{n(n+1)}{2}+(599)n.$$
(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": "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-013994 | Foundations: Relations from Fibers of Maps | 7 | Answer with a short justification: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $71\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-76]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
... | [
{
"method_name": "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 $\\pi(a)=a\\bmod 71$ 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-013995 | Algebraic Foundations: Congruence Modulo m | 7 | Work carefully and justify each inference: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $116\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[90]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modu... | [
{
"method_name": "Projection Map / Fibers",
"approach": "Use the map $\\pi:\\mathbb{Z}\\to\\mathbb{Z}/m\\mathbb{Z}$; $a\\sim b$ iff $\\pi(a)=\\pi(b)$, and equality of images defines an equivalence relation.",
"steps": [
"Step 1: Let $\\pi(a)=a\\bmod 116$ 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-013996 | Set Theory: Partitions and Classes | 7 | Make each step logically reversible (or explain if not): Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $128\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[89]$ explicitly as a set.
(c) Explain briefly how this relates to c... | [
{
"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 $\\pi(a)=a\\bmod 128$ be the canonical projection.",
"Step ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues, which is automatically an... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. |
math-013997 | Proof Techniques: Induction — Base + Step | 7 | Challenge: Prove by induction that for all integers $n\ge 1$,
$$\sum_{k=1}^n (k+(515)) = \frac{n(n+1)}{2}+(515)n.$$
(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... | 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-013998 | Set Theory: Partitions and Classes | 7 | Solve and include a self-check: 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 $[-90]$ 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 $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-013999 | Algebraic Foundations: Congruence Modulo m | 7 | Show all reasoning: Define a relation $\sim$ on $\mathbb{Z}$ by $a\sim b$ iff $49\mid(a-b)$.
(a) Prove that $\sim$ is an equivalence relation (reflexive, symmetric, transitive).
(b) Describe the equivalence class $[-82]$ explicitly as a set.
(c) Explain briefly how this relates to congruence modulo $m$.
In part (a), d... | [
{
"method_name": "Direct R/S/T Verification",
"approach": "Check reflexivity, symmetry, transitivity directly from the divisibility definition.",
"steps": [
"Step 1: Reflexive: $a-a=0$ and every integer divides 0, so $a\\sim a$.",
"Step 2: Symmetric: if $m\\mid(a-b)$ then $m\\mid-(a-b)=(b-a)... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Equivalence; }$.\nThe direct R/S/T proof shows the relation satisfies the axioms. The projection-map proof identifies the same relation as congruence mod $m$ and uses equality of residues,... | [
{
"error_description": "Treated '$m\\mid(a-b)$' like an equation and tried to 'divide by $m$' directly.",
"why_plausible": "Divisibility notation resembles equality and invites cancellation.",
"why_wrong": "Divisibility means existence of an integer $k$ with $a-b=mk$; you cannot cancel without introduci... | Core principle: Equivalence relations partition a set into classes. Congruence mod $m$ partitions $\mathbb{Z}$ into infinite arithmetic progressions $r+m\mathbb{Z}$. (Here the result is $\boxed{\text{Equivalence; }$.) |
math-014000 | Set Theory: Equivalence Relations — R/S/T | 7 | Give a theorem-based solution: 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 $[1]$ 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": "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 $r+m\mathbb{Z}$. |
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