id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-018601 | Real Analysis: Uniform Continuity (Variant C) | 10 | Make each step logically reversible (or explain if not): Let $f:(0,1414)\to\mathbb{R}$ be $f(x)=\frac{1}{(-24)x}$. Is $f$ uniformly continuous on $(0,1414)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018602 | Real Analysis: Uniform Continuity (Variant B) | 10 | Prompt: Let $f:(0,1847)\to\mathbb{R}$ be $f(x)=\frac{1}{(-27)x}$. Is $f$ uniformly continuous on $(0,1847)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}}.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018603 | Real Analysis: Uniform Continuity (Variant A) | 10 | Derive the result step-by-step: Let $f:[-1851,1851]\to\mathbb{R}$ be $f(x)=(-1)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1851,1851]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Heine–Cantor",
"approach": "Continuous functions on compact metric spaces are uniformly continuous.",
"steps": [
"Step 1: The interval $[-1851,1851]$ is compact in $\\mathbb{R}$.",
"Step 2: The function $x\\mapsto x^2$ is continuous on $\\mathbb{R}$, hence on the compact se... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same conclusion.",
"rob... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018604 | Real Analysis: Uniform Continuity (Variant B) | 10 | Answer using clear logical steps: Let $f:(0,364)\to\mathbb{R}$ be $f(x)=\frac{1}{(10)x}$. Is $f$ uniformly continuous on $(0,364)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{364}{n}$ and $y_n=\\frac{364}{2n}$ in $(0,364)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018605 | Real Analysis: Uniform Continuity (Variant B) | 10 | Solve (and briefly cross-validate): Let $f:(0,1316)\to\mathbb{R}$ be $f(x)=\ln(45x)$. Is $f$ uniformly continuous on $(0,1316)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{1316}{n}$ and $y_n=\\frac{1316}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\l... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018606 | Real Analysis: Uniform Continuity (Variant A) | 10 | Indicate where a theorem is used: Let $f:(0,1052)\to\mathbb{R}$ be $f(x)=\frac{1}{(3)x}$. Is $f$ uniformly continuous on $(0,1052)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{1052}{n}$ and $y_n=\\frac{1052}{2n}$ in $(0,1052)$.",
"Step 2: Then $|x_n-y_n|=\\frac{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018607 | Real Analysis: Uniform Continuity (Variant B) | 10 | Give a fully justified solution: Let $f:(0,348)\to\mathbb{R}$ be $f(x)=\ln(33x)$. Is $f$ uniformly continuous on $(0,348)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018608 | Real Analysis: Uniform Continuity (Core) | 10 | Work carefully and justify each inference: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(-17x)$. Prove that $f$ is uniformly continuous on $\mathbb{R}$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz via Mean Value Theorem",
"approach": "Use MVT to get a global Lipschitz bound $|\\sin x-\\sin y|\\le |x-y|$.",
"steps": [
"Step 1: Apply MVT to $g(t)=\\sin(ct)$: there exists $\\xi$ between $x$ and $y$ with $g(x)-g(y)=g'(\\xi)(x-y)$.",
"Step 2: Since $g'(t)=(-17)\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Yes}$.\nThe MVT proof gives an explicit global Lipschitz modulus, which immediately implies the Cauchy-sequence property. Both are equivalent characterizations of uniform continuity.",
"... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018609 | Real Analysis: Uniform Continuity (Core) | 10 | Derive the result step-by-step: Let $f:[-1767,1767]\to\mathbb{R}$ be $f(x)=(2)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1767,1767]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Heine–Cantor",
"approach": "Continuous functions on compact metric spaces are uniformly continuous.",
"steps": [
"Step 1: The interval $[-1767,1767]$ is compact in $\\mathbb{R}$.",
"Step 2: The function $x\\mapsto x^2$ is continuous on $\\mathbb{R}$, hence on the compact se... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same conclusion.",
"rob... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018610 | Real Analysis: Uniform Continuity (Core) | 10 | Indicate where a theorem is used: Let $f:(0,1811)\to\mathbb{R}$ be $f(x)=\ln(25x)$. Is $f$ uniformly continuous on $(0,1811)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": "Robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018611 | Real Analysis: Uniform Continuity (Variant A) | 10 | Give a theorem-based solution: Let $f:(0,1838)\to\mathbb{R}$ be $f(x)=\frac{1}{(15)x}$. Is $f$ uniformly continuous on $(0,1838)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{1838}{n}$ and $y_n=\\frac{1838}{2n}$ in $(0,1838)$.",
"Step 2: Then $|x_n-y_n|=\\frac{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018612 | Real Analysis: Uniform Continuity (Variant C) | 10 | Do not skip justification steps: Let $f:(0,1550)\to\mathbb{R}$ be $f(x)=\sqrt{17x}$. Prove that $f$ is uniformly continuous on $(0,1550)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Direct Inequality",
"approach": "Bound $|\\sqrt{x}-\\sqrt{y}|$ in terms of $|x-y|$ uniformly using rationalization.",
"steps": [
"Step 1: For $x,y>0$, $|\\sqrt{x}-\\sqrt{y}|=\\frac{|x-y|}{\\sqrt{x}+\\sqrt{y}}$.",
"Step 2: Since $\\sqrt{x}+\\sqrt{y}\\ge \\sqrt{|x-y|}$, we ge... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conclusion.",
"robustness_an... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018613 | Real Analysis: Uniform Continuity (Variant A) | 10 | Derive the result step-by-step: Let $f:(0,132)\to\mathbb{R}$ be $f(x)=\ln(38x)$. Is $f$ uniformly continuous on $(0,132)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{132}{n}$ and $y_n=\\frac{132}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\ln(... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018614 | Real Analysis: Uniform Continuity (Variant B) | 10 | Indicate where a theorem is used: Let $f:(0,901)\to\mathbb{R}$ be $f(x)=\sqrt{14x}$. Prove that $f$ is uniformly continuous on $(0,901)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Compact Extension + Heine–Cantor",
"approach": "Extend continuously to a compact interval and invoke Heine–Cantor, then restrict back.",
"steps": [
"Step 1: Define $g:[0,901]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,901]$.",
"Step 2: The interval $[0,9... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conclusion.",... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018615 | Real Analysis: Uniform Continuity (Variant C) | 10 | Show all reasoning: Let $f:(0,229)\to\mathbb{R}$ be $f(x)=\sqrt{52x}$. Prove that $f$ is uniformly continuous on $(0,229)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Compact Extension + Heine–Cantor",
"approach": "Extend continuously to a compact interval and invoke Heine–Cantor, then restrict back.",
"steps": [
"Step 1: Define $g:[0,229]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,229]$.",
"Step 2: The interval $[0,2... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conclusion.",... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. |
math-018616 | Real Analysis: Uniform Continuity (Variant C) | 10 | Use two approaches if possible: Let $f:(0,250)\to\mathbb{R}$ be $f(x)=\ln(29x)$. Is $f$ uniformly continuous on $(0,250)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{250}{n}$ and $y_n=\\frac{250}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\ln(... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018617 | Real Analysis: Uniform Continuity (Variant B) | 10 | Track quantifiers carefully: Let $f:(0,1089)\to\mathbb{R}$ be $f(x)=\frac{1}{(-25)x}$. Is $f$ uniformly continuous on $(0,1089)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018618 | Real Analysis: Uniform Continuity (Variant A) | 10 | Answer with a short justification: Let $f:(0,1341)\to\mathbb{R}$ be $f(x)=\ln(52x)$. Is $f$ uniformly continuous on $(0,1341)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{1341}{n}$ and $y_n=\\frac{1341}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\l... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018619 | Real Analysis: Uniform Continuity (Variant A) | 10 | Give a fully justified solution: Let $f:(0,46)\to\mathbb{R}$ be $f(x)=\ln(31x)$. Is $f$ uniformly continuous on $(0,46)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{46}{n}$ and $y_n=\\frac{46}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\ln(c ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018620 | Real Analysis: Uniform Continuity (Core) | 10 | Solve (and briefly cross-validate): Let $f:[-1621,1621]\to\mathbb{R}$ be $f(x)=(1)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1621,1621]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz on Bounded Domain",
"approach": "On a bounded interval, $|x^2-y^2|=|x-y||x+y|$ and $|x+y|$ is uniformly bounded, giving a global Lipschitz constant.",
"steps": [
"Step 1: For $x,y\\in[-1621,1621]$, $|f(x)-f(y)|=|1|\\,|x^2-y^2|=|1|\\,|x-y||x+y|$.",
"Step 2: Bound $... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018621 | Real Analysis: Uniform Continuity (Variant B) | 10 | Solve (and briefly cross-validate): Let $f:(0,728)\to\mathbb{R}$ be $f(x)=\frac{1}{(-16)x}$. Is $f$ uniformly continuous on $(0,728)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{728}{n}$ and $y_n=\\frac{728}{2n}$ in $(0,728)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018622 | Real Analysis: Uniform Continuity (Variant C) | 10 | Provide a rigorous solution: Let $f:[-111,111]\to\mathbb{R}$ be $f(x)=(20)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-111,111]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Heine–Cantor",
"approach": "Continuous functions on compact metric spaces are uniformly continuous.",
"steps": [
"Step 1: The interval $[-111,111]$ is compact in $\\mathbb{R}$.",
"Step 2: The function $x\\mapsto x^2$ is continuous on $\\mathbb{R}$, hence on the compact set.... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same con... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. |
math-018623 | Real Analysis: Uniform Continuity (Variant B) | 10 | Try to avoid pattern-matching; explain why: Let $f:[-809,809]\to\mathbb{R}$ be $f(x)=(-25)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-809,809]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz on Bounded Domain",
"approach": "On a bounded interval, $|x^2-y^2|=|x-y||x+y|$ and $|x+y|$ is uniformly bounded, giving a global Lipschitz constant.",
"steps": [
"Step 1: For $x,y\\in[-809,809]$, $|f(x)-f(y)|=|-25|\\,|x^2-y^2|=|-25|\\,|x-y||x+y|$.",
"Step 2: Bound... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same con... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018624 | Real Analysis: Uniform Continuity (Variant B) | 10 | Warm-up: Let $f:(0,1183)\to\mathbb{R}$ be $f(x)=\ln(3x)$. Is $f$ uniformly continuous on $(0,1183)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018625 | Real Analysis: Uniform Continuity (Core) | 10 | Explain why your operations are valid: Let $f:(0,1493)\to\mathbb{R}$ be $f(x)=\frac{1}{(-9)x}$. Is $f$ uniformly continuous on $(0,1493)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{1493}{n}$ and $y_n=\\frac{1493}{2n}$ in $(0,1493)$.",
"Step 2: Then $|x_n-y_n|=\\frac{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both c... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018626 | Real Analysis: Uniform Continuity (Variant B) | 10 | Solve with verification: Let $f:(0,592)\to\mathbb{R}$ be $f(x)=\frac{1}{(-15)x}$. Is $f$ uniformly continuous on $(0,592)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{592}{n}$ and $y_n=\\frac{592}{2n}$ in $(0,592)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both c... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018627 | Real Analysis: Uniform Continuity (Variant C) | 10 | Start by stating any domain restrictions: Let $f:(0,1564)\to\mathbb{R}$ be $f(x)=\ln(43x)$. Is $f$ uniformly continuous on $(0,1564)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{1564}{n}$ and $y_n=\\frac{1564}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\l... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018628 | Real Analysis: Uniform Continuity (Core) | 10 | Solve and sanity-check: Let $f:(0,1747)\to\mathbb{R}$ be $f(x)=\frac{1}{(9)x}$. Is $f$ uniformly continuous on $(0,1747)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018629 | Real Analysis: Uniform Continuity (Variant B) | 10 | Answer using clear logical steps: Let $f:(0,1360)\to\mathbb{R}$ be $f(x)=\ln(50x)$. Is $f$ uniformly continuous on $(0,1360)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{1360}{n}$ and $y_n=\\frac{1360}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\l... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018630 | Real Analysis: Uniform Continuity (Core) | 10 | Solve and sanity-check: Let $f:(0,649)\to\mathbb{R}$ be $f(x)=\frac{1}{(-7)x}$. Is $f$ uniformly continuous on $(0,649)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{649}{n}$ and $y_n=\\frac{649}{2n}$ in $(0,649)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018631 | Real Analysis: Uniform Continuity (Variant B) | 10 | Provide a rigorous solution: Let $f:(0,978)\to\mathbb{R}$ be $f(x)=\ln(48x)$. Is $f$ uniformly continuous on $(0,978)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018632 | Real Analysis: Uniform Continuity (Variant C) | 10 | Complete the analysis: Let $f:(0,212)\to\mathbb{R}$ be $f(x)=\frac{1}{(2)x}$. Is $f$ uniformly continuous on $(0,212)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018633 | Real Analysis: Uniform Continuity (Variant A) | 10 | Find the exact value: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(12)x^2$ with $c\ne 0$. Is $f$ uniformly continuous on $\mathbb{R}$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Mean Value Theorem Blow-Up",
"approach": "Use MVT to show that on unbounded domains, the derivative growth makes a uniform modulus impossible.",
"steps": [
"Step 1: Suppose uniform continuity holds. Take $\\varepsilon=1$ and let $\\delta>0$ correspond.",
"Step 2: Choose $x$... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nThe sequence method explicitly exhibits $|x_n-y_n|\\to 0$ while $|f(x_n)-f(y_n)|\\not\\to 0$. The MVT method formalizes the same idea via unbounded derivative growth. Both conclude non-uniform continuity.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018634 | Real Analysis: Uniform Continuity (Variant C) | 10 | Give a fully justified solution: Let $f:(0,1474)\to\mathbb{R}$ be $f(x)=\sqrt{53x}$. Prove that $f$ is uniformly continuous on $(0,1474)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Compact Extension + Heine–Cantor",
"approach": "Extend continuously to a compact interval and invoke Heine–Cantor, then restrict back.",
"steps": [
"Step 1: Define $g:[0,1474]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,1474]$.",
"Step 2: The interval $[0... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conclusion.",
"robustness_an... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018635 | Real Analysis: Uniform Continuity (Variant A) | 10 | Solve with verification: Let $f:(0,1616)\to\mathbb{R}$ be $f(x)=\ln(41x)$. Is $f$ uniformly continuous on $(0,1616)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{1616}{n}$ and $y_n=\\frac{1616}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\l... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018636 | Real Analysis: Uniform Continuity (Core) | 10 | Complete the analysis: Let $f:(0,775)\to\mathbb{R}$ be $f(x)=\ln(21x)$. Is $f$ uniformly continuous on $(0,775)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018637 | Real Analysis: Uniform Continuity (Variant B) | 10 | Indicate where a theorem is used: Let $f:(0,1794)\to\mathbb{R}$ be $f(x)=\ln(44x)$. Is $f$ uniformly continuous on $(0,1794)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{1794}{n}$ and $y_n=\\frac{1794}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\l... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018638 | Real Analysis: Uniform Continuity (Core) | 10 | Warm-up: Let $f:(0,870)\to\mathbb{R}$ be $f(x)=\frac{1}{(19)x}$. Is $f$ uniformly continuous on $(0,870)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018639 | Real Analysis: Uniform Continuity (Variant B) | 10 | Make each step logically reversible (or explain if not): Let $f:(0,825)\to\mathbb{R}$ be $f(x)=\frac{1}{(-8)x}$. Is $f$ uniformly continuous on $(0,825)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{825}{n}$ and $y_n=\\frac{825}{2n}$ in $(0,825)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018640 | Real Analysis: Uniform Continuity (Variant B) | 10 | Do not skip justification steps: Let $f:(0,460)\to\mathbb{R}$ be $f(x)=\sqrt{37x}$. Prove that $f$ is uniformly continuous on $(0,460)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Compact Extension + Heine–Cantor",
"approach": "Extend continuously to a compact interval and invoke Heine–Cantor, then restrict back.",
"steps": [
"Step 1: Define $g:[0,460]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,460]$.",
"Step 2: The interval $[0,4... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conclusion.",
"robustness_analysis... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018641 | Real Analysis: Uniform Continuity (Variant B) | 10 | Solve (and briefly cross-validate): Let $f:(0,271)\to\mathbb{R}$ be $f(x)=\frac{1}{(-4)x}$. Is $f$ uniformly continuous on $(0,271)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018642 | Real Analysis: Uniform Continuity (Variant C) | 10 | Exercise: Let $f:(0,916)\to\mathbb{R}$ be $f(x)=\frac{1}{(-26)x}$. Is $f$ uniformly continuous on $(0,916)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018643 | Real Analysis: Uniform Continuity (Variant A) | 10 | Proceed methodically: Let $f:(0,1945)\to\mathbb{R}$ be $f(x)=\sqrt{39x}$. Prove that $f$ is uniformly continuous on $(0,1945)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Compact Extension + Heine–Cantor",
"approach": "Extend continuously to a compact interval and invoke Heine–Cantor, then restrict back.",
"steps": [
"Step 1: Define $g:[0,1945]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,1945]$.",
"Step 2: The interval $[0... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conclusion.",
"robustness_an... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018644 | Real Analysis: Uniform Continuity (Variant C) | 10 | Give reasoning, not just computation: Let $f:[-1893,1893]\to\mathbb{R}$ be $f(x)=(22)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1893,1893]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz on Bounded Domain",
"approach": "On a bounded interval, $|x^2-y^2|=|x-y||x+y|$ and $|x+y|$ is uniformly bounded, giving a global Lipschitz constant.",
"steps": [
"Step 1: For $x,y\\in[-1893,1893]$, $|f(x)-f(y)|=|22|\\,|x^2-y^2|=|22|\\,|x-y||x+y|$.",
"Step 2: Bound... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same con... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. |
math-018645 | Real Analysis: Uniform Continuity (Variant B) | 10 | Solve and then verify: Let $f:(0,347)\to\mathbb{R}$ be $f(x)=\frac{1}{(-5)x}$. Is $f$ uniformly continuous on $(0,347)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018646 | Real Analysis: Uniform Continuity (Core) | 10 | Do not skip justification steps: Let $f:(0,1954)\to\mathbb{R}$ be $f(x)=\frac{1}{(-19)x}$. Is $f$ uniformly continuous on $(0,1954)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018647 | Real Analysis: Uniform Continuity (Variant B) | 10 | Exercise: Let $f:(0,618)\to\mathbb{R}$ be $f(x)=\sqrt{14x}$. Prove that $f$ is uniformly continuous on $(0,618)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Compact Extension + Heine–Cantor",
"approach": "Extend continuously to a compact interval and invoke Heine–Cantor, then restrict back.",
"steps": [
"Step 1: Define $g:[0,618]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,618]$.",
"Step 2: The interval $[0,6... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conclusion.",
"robustness_analysis... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018648 | Real Analysis: Uniform Continuity (Variant C) | 10 | Solve and then verify: Let $f:(0,1136)\to\mathbb{R}$ be $f(x)=\ln(41x)$. Is $f$ uniformly continuous on $(0,1136)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018649 | Real Analysis: Uniform Continuity (Variant C) | 10 | Solve (and briefly cross-validate): Let $f:(0,1854)\to\mathbb{R}$ be $f(x)=\frac{1}{(24)x}$. Is $f$ uniformly continuous on $(0,1854)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018650 | Real Analysis: Uniform Continuity (Variant A) | 10 | Track quantifiers carefully: Let $f:(0,1567)\to\mathbb{R}$ be $f(x)=\frac{1}{(-13)x}$. Is $f$ uniformly continuous on $(0,1567)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{1567}{n}$ and $y_n=\\frac{1567}{2n}$ in $(0,1567)$.",
"Step 2: Then $|x_n-y_n|=\\frac{... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}}.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018651 | Real Analysis: Uniform Continuity (Core) | 10 | Find the exact value: Let $f:(0,1553)\to\mathbb{R}$ be $f(x)=\ln(18x)$. Is $f$ uniformly continuous on $(0,1553)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{1553}{n}$ and $y_n=\\frac{1553}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\l... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": "Gener... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018652 | Real Analysis: Uniform Continuity (Variant B) | 10 | Problem: Let $f:[-1666,1666]\to\mathbb{R}$ be $f(x)=(1)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1666,1666]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Heine–Cantor",
"approach": "Continuous functions on compact metric spaces are uniformly continuous.",
"steps": [
"Step 1: The interval $[-1666,1666]$ is compact in $\\mathbb{R}$.",
"Step 2: The function $x\\mapsto x^2$ is continuous on $\\mathbb{R}$, hence on the compact se... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same conclusion.",
"robustnes... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018653 | Real Analysis: Uniform Continuity (Core) | 10 | Exercise: Let $f:(0,688)\to\mathbb{R}$ be $f(x)=\frac{1}{(-20)x}$. Is $f$ uniformly continuous on $(0,688)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018654 | Real Analysis: Uniform Continuity (Variant B) | 10 | Where appropriate, name the theorem you use: Let $f:(0,815)\to\mathbb{R}$ be $f(x)=\frac{1}{(-14)x}$. Is $f$ uniformly continuous on $(0,815)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{815}{n}$ and $y_n=\\frac{815}{2n}$ in $(0,815)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both c... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018655 | Real Analysis: Uniform Continuity (Variant B) | 10 | Answer with a short justification: Let $f:[-862,862]\to\mathbb{R}$ be $f(x)=(-13)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-862,862]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Heine–Cantor",
"approach": "Continuous functions on compact metric spaces are uniformly continuous.",
"steps": [
"Step 1: The interval $[-862,862]$ is compact in $\\mathbb{R}$.",
"Step 2: The function $x\\mapsto x^2$ is continuous on $\\mathbb{R}$, hence on the compact set.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same conclusion.",
"robustnes... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. |
math-018656 | Real Analysis: Uniform Continuity (Core) | 10 | Keep the final answer in boxed form: Let $f:(0,1522)\to\mathbb{R}$ be $f(x)=\ln(37x)$. Is $f$ uniformly continuous on $(0,1522)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018657 | Real Analysis: Uniform Continuity (Core) | 10 | Be explicit about assumptions: Let $f:[-1934,1934]\to\mathbb{R}$ be $f(x)=(25)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1934,1934]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Heine–Cantor",
"approach": "Continuous functions on compact metric spaces are uniformly continuous.",
"steps": [
"Step 1: The interval $[-1934,1934]$ is compact in $\\mathbb{R}$.",
"Step 2: The function $x\\mapsto x^2$ is continuous on $\\mathbb{R}$, hence on the compact se... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same con... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018658 | Real Analysis: Uniform Continuity (Variant B) | 10 | Use two approaches if possible: Let $f:[-1259,1259]\to\mathbb{R}$ be $f(x)=(-20)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1259,1259]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz on Bounded Domain",
"approach": "On a bounded interval, $|x^2-y^2|=|x-y||x+y|$ and $|x+y|$ is uniformly bounded, giving a global Lipschitz constant.",
"steps": [
"Step 1: For $x,y\\in[-1259,1259]$, $|f(x)-f(y)|=|-20|\\,|x^2-y^2|=|-20|\\,|x-y||x+y|$.",
"Step 2: Bou... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same conclusion.",
"robustnes... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. |
math-018659 | Real Analysis: Uniform Continuity (Variant B) | 10 | Question: Let $f:[-1207,1207]\to\mathbb{R}$ be $f(x)=(9)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1207,1207]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz on Bounded Domain",
"approach": "On a bounded interval, $|x^2-y^2|=|x-y||x+y|$ and $|x+y|$ is uniformly bounded, giving a global Lipschitz constant.",
"steps": [
"Step 1: For $x,y\\in[-1207,1207]$, $|f(x)-f(y)|=|9|\\,|x^2-y^2|=|9|\\,|x-y||x+y|$.",
"Step 2: Bound $... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same conclusion.",
"rob... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018660 | Real Analysis: Uniform Continuity (Core) | 10 | Explain why your operations are valid: Let $f:[-971,971]\to\mathbb{R}$ be $f(x)=(-9)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-971,971]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz on Bounded Domain",
"approach": "On a bounded interval, $|x^2-y^2|=|x-y||x+y|$ and $|x+y|$ is uniformly bounded, giving a global Lipschitz constant.",
"steps": [
"Step 1: For $x,y\\in[-971,971]$, $|f(x)-f(y)|=|-9|\\,|x^2-y^2|=|-9|\\,|x-y||x+y|$.",
"Step 2: Bound $... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same conclusion.",
"robustnes... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018661 | Real Analysis: Uniform Continuity (Variant B) | 10 | Provide both a computational and a conceptual explanation: Let $f:(0,1477)\to\mathbb{R}$ be $f(x)=\sqrt{22x}$. Prove that $f$ is uniformly continuous on $(0,1477)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Compact Extension + Heine–Cantor",
"approach": "Extend continuously to a compact interval and invoke Heine–Cantor, then restrict back.",
"steps": [
"Step 1: Define $g:[0,1477]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,1477]$.",
"Step 2: The interval $[0... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conc... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. |
math-018662 | Real Analysis: Uniform Continuity (Variant A) | 10 | Where appropriate, name the theorem you use: Let $f:(0,742)\to\mathbb{R}$ be $f(x)=\sqrt{24x}$. Prove that $f$ is uniformly continuous on $(0,742)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Compact Extension + Heine–Cantor",
"approach": "Extend continuously to a compact interval and invoke Heine–Cantor, then restrict back.",
"steps": [
"Step 1: Define $g:[0,742]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,742]$.",
"Step 2: The interval $[0,7... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conclusion.",
"robustness_an... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. |
math-018663 | Real Analysis: Uniform Continuity (Variant B) | 10 | Give a fully justified solution: Let $f:[-1454,1454]\to\mathbb{R}$ be $f(x)=(23)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1454,1454]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz on Bounded Domain",
"approach": "On a bounded interval, $|x^2-y^2|=|x-y||x+y|$ and $|x+y|$ is uniformly bounded, giving a global Lipschitz constant.",
"steps": [
"Step 1: For $x,y\\in[-1454,1454]$, $|f(x)-f(y)|=|23|\\,|x^2-y^2|=|23|\\,|x-y||x+y|$.",
"Step 2: Bound... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018664 | Real Analysis: Uniform Continuity (Core) | 10 | Exercise: Let $f:(0,1718)\to\mathbb{R}$ be $f(x)=\frac{1}{(10)x}$. Is $f$ uniformly continuous on $(0,1718)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018665 | Real Analysis: Uniform Continuity (Core) | 10 | Task: Let $f:(0,123)\to\mathbb{R}$ be $f(x)=\frac{1}{(4)x}$. Is $f$ uniformly continuous on $(0,123)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{123}{n}$ and $y_n=\\frac{123}{2n}$ in $(0,123)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018666 | Real Analysis: Uniform Continuity (Variant C) | 10 | Provide a rigorous solution: Let $f:(0,1556)\to\mathbb{R}$ be $f(x)=\frac{1}{(-20)x}$. Is $f$ uniformly continuous on $(0,1556)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both c... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018667 | Real Analysis: Uniform Continuity (Variant C) | 10 | Determine the requested value: Let $f:(0,1565)\to\mathbb{R}$ be $f(x)=\ln(41x)$. Is $f$ uniformly continuous on $(0,1565)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{1565}{n}$ and $y_n=\\frac{1565}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\l... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018668 | Real Analysis: Uniform Continuity (Variant B) | 10 | Where appropriate, name the theorem you use: Let $f:(0,1325)\to\mathbb{R}$ be $f(x)=\sqrt{36x}$. Prove that $f$ is uniformly continuous on $(0,1325)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Compact Extension + Heine–Cantor",
"approach": "Extend continuously to a compact interval and invoke Heine–Cantor, then restrict back.",
"steps": [
"Step 1: Define $g:[0,1325]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,1325]$.",
"Step 2: The interval $[0... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conc... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. |
math-018669 | Real Analysis: Uniform Continuity (Variant C) | 10 | State any required conditions first: Let $f:(0,1982)\to\mathbb{R}$ be $f(x)=\frac{1}{(1)x}$. Is $f$ uniformly continuous on $(0,1982)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{1982}{n}$ and $y_n=\\frac{1982}{2n}$ in $(0,1982)$.",
"Step 2: Then $|x_n-y_n|=\\frac{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018670 | Real Analysis: Uniform Continuity (Variant A) | 10 | Work carefully and justify each inference: Let $f:[-353,353]\to\mathbb{R}$ be $f(x)=(5)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-353,353]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz on Bounded Domain",
"approach": "On a bounded interval, $|x^2-y^2|=|x-y||x+y|$ and $|x+y|$ is uniformly bounded, giving a global Lipschitz constant.",
"steps": [
"Step 1: For $x,y\\in[-353,353]$, $|f(x)-f(y)|=|5|\\,|x^2-y^2|=|5|\\,|x-y||x+y|$.",
"Step 2: Bound $|x... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same con... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. |
math-018671 | Real Analysis: Uniform Continuity (Core) | 10 | Answer with a short justification: Let $f:(0,339)\to\mathbb{R}$ be $f(x)=\frac{1}{(6)x}$. Is $f$ uniformly continuous on $(0,339)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018672 | Real Analysis: Uniform Continuity (Core) | 10 | Complete the analysis: Let $f:[-1805,1805]\to\mathbb{R}$ be $f(x)=(-5)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1805,1805]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz on Bounded Domain",
"approach": "On a bounded interval, $|x^2-y^2|=|x-y||x+y|$ and $|x+y|$ is uniformly bounded, giving a global Lipschitz constant.",
"steps": [
"Step 1: For $x,y\\in[-1805,1805]$, $|f(x)-f(y)|=|-5|\\,|x^2-y^2|=|-5|\\,|x-y||x+y|$.",
"Step 2: Bound... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same con... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. |
math-018673 | Real Analysis: Uniform Continuity (Core) | 10 | Track quantifiers carefully: Let $f:[-890,890]\to\mathbb{R}$ be $f(x)=(10)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-890,890]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Heine–Cantor",
"approach": "Continuous functions on compact metric spaces are uniformly continuous.",
"steps": [
"Step 1: The interval $[-890,890]$ is compact in $\\mathbb{R}$.",
"Step 2: The function $x\\mapsto x^2$ is continuous on $\\mathbb{R}$, hence on the compact set.... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same conclusion.",
"rob... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018674 | Real Analysis: Uniform Continuity (Variant B) | 10 | Problem: Let $f:(0,774)\to\mathbb{R}$ be $f(x)=\ln(35x)$. Is $f$ uniformly continuous on $(0,774)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018675 | Real Analysis: Uniform Continuity (Variant B) | 10 | Solve and include a self-check: Let $f:(0,1467)\to\mathbb{R}$ be $f(x)=\sqrt{55x}$. Prove that $f$ is uniformly continuous on $(0,1467)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Direct Inequality",
"approach": "Bound $|\\sqrt{x}-\\sqrt{y}|$ in terms of $|x-y|$ uniformly using rationalization.",
"steps": [
"Step 1: For $x,y>0$, $|\\sqrt{x}-\\sqrt{y}|=\\frac{|x-y|}{\\sqrt{x}+\\sqrt{y}}$.",
"Step 2: Since $\\sqrt{x}+\\sqrt{y}\\ge \\sqrt{|x-y|}$, we ge... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conclusion.",
"robustness_an... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018676 | Real Analysis: Uniform Continuity (Core) | 10 | Complete the analysis: Let $f:(0,905)\to\mathbb{R}$ be $f(x)=\frac{1}{(20)x}$. Is $f$ uniformly continuous on $(0,905)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{905}{n}$ and $y_n=\\frac{905}{2n}$ in $(0,905)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both c... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018677 | Real Analysis: Uniform Continuity (Variant B) | 10 | Explain why your operations are valid: Let $f:(0,338)\to\mathbb{R}$ be $f(x)=\ln(10x)$. Is $f$ uniformly continuous on $(0,338)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{338}{n}$ and $y_n=\\frac{338}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\ln(... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018678 | Real Analysis: Uniform Continuity (Variant A) | 10 | Explain what is being counted/optimized: Let $f:(0,109)\to\mathbb{R}$ be $f(x)=\ln(40x)$. Is $f$ uniformly continuous on $(0,109)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": "Robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018679 | Real Analysis: Uniform Continuity (Variant A) | 10 | Checkpoint: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(22x)$. Prove that $f$ is uniformly continuous on $\mathbb{R}$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Characterization",
"approach": "A function is uniformly continuous iff it sends Cauchy sequences to Cauchy sequences; use the global inequality $|\\sin x-\\sin y|\\le |x-y|$.",
"steps": [
"Step 1: For all real $x,y$, $|\\sin x-\\sin y|\\le |x-y|$ (e.g., by MVT).",... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Yes}$.\nThe MVT proof gives an explicit global Lipschitz modulus, which immediately implies the Cauchy-sequence property. Both are equivalent characterizations of uniform continuity.",
"robustness_analysis": "Sen... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018680 | Real Analysis: Uniform Continuity (Variant C) | 10 | Show all reasoning: Let $f:(0,70)\to\mathbb{R}$ be $f(x)=\frac{1}{(-7)x}$. Is $f$ uniformly continuous on $(0,70)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{70}{n}$ and $y_n=\\frac{70}{2n}$ in $(0,70)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}}.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018681 | Real Analysis: Uniform Continuity (Variant A) | 10 | Task: Let $f:[-1924,1924]\to\mathbb{R}$ be $f(x)=(-6)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1924,1924]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Heine–Cantor",
"approach": "Continuous functions on compact metric spaces are uniformly continuous.",
"steps": [
"Step 1: The interval $[-1924,1924]$ is compact in $\\mathbb{R}$.",
"Step 2: The function $x\\mapsto x^2$ is continuous on $\\mathbb{R}$, hence on the compact se... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same con... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. |
math-018682 | Real Analysis: Uniform Continuity (Variant B) | 10 | Carefully track domains: Let $f:(0,1906)\to\mathbb{R}$ be $f(x)=\frac{1}{(26)x}$. Is $f$ uniformly continuous on $(0,1906)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}}.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018683 | Real Analysis: Uniform Continuity (Core) | 10 | Track units/moduli carefully: Let $f:(0,333)\to\mathbb{R}$ be $f(x)=\frac{1}{(1)x}$. Is $f$ uniformly continuous on $(0,333)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018684 | Real Analysis: Uniform Continuity (Variant B) | 10 | Challenge: Let $f:(0,1846)\to\mathbb{R}$ be $f(x)=\ln(22x)$. Is $f$ uniformly continuous on $(0,1846)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{1846}{n}$ and $y_n=\\frac{1846}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\l... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018685 | Real Analysis: Uniform Continuity (Variant A) | 10 | Determine the requested value: Let $f:(0,881)\to\mathbb{R}$ be $f(x)=\ln(59x)$. Is $f$ uniformly continuous on $(0,881)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": "Gener... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018686 | Real Analysis: Uniform Continuity (Variant A) | 10 | Challenge: Let $f:[-289,289]\to\mathbb{R}$ be $f(x)=(-12)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-289,289]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Heine–Cantor",
"approach": "Continuous functions on compact metric spaces are uniformly continuous.",
"steps": [
"Step 1: The interval $[-289,289]$ is compact in $\\mathbb{R}$.",
"Step 2: The function $x\\mapsto x^2$ is continuous on $\\mathbb{R}$, hence on the compact set.... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same con... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018687 | Real Analysis: Uniform Continuity (Variant A) | 10 | Checkpoint: Let $f:[-1199,1199]\to\mathbb{R}$ be $f(x)=(-30)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1199,1199]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Heine–Cantor",
"approach": "Continuous functions on compact metric spaces are uniformly continuous.",
"steps": [
"Step 1: The interval $[-1199,1199]$ is compact in $\\mathbb{R}$.",
"Step 2: The function $x\\mapsto x^2$ is continuous on $\\mathbb{R}$, hence on the compact se... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same con... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018688 | Real Analysis: Uniform Continuity (Variant C) | 10 | Find the exact value: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(8)x^2$ with $c\ne 0$. Is $f$ uniformly continuous on $\mathbb{R}$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Explicit Sequence Counterexample",
"approach": "Pick sequences approaching each other at infinity but whose squares stay separated.",
"steps": [
"Step 1: Let $x_n=n$ and $y_n=n+\\frac{1}{n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{1}{n}\\to 0$.",
"Final step: But $|f(x_n)-... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nThe sequence method explicitly exhibits $|x_n-y_n|\\to 0$ while $|f(x_n)-f(y_n)|\\not\\to 0$. The MVT method formalizes the same idea via unbounded derivative growth. Both conclude non-uniform continuity.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018689 | Real Analysis: Uniform Continuity (Core) | 10 | Work this out carefully: Let $f:(0,1074)\to\mathbb{R}$ be $f(x)=\sqrt{6x}$. Prove that $f$ is uniformly continuous on $(0,1074)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Compact Extension + Heine–Cantor",
"approach": "Extend continuously to a compact interval and invoke Heine–Cantor, then restrict back.",
"steps": [
"Step 1: Define $g:[0,1074]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,1074]$.",
"Step 2: The interval $[0... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conc... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018690 | Real Analysis: Uniform Continuity (Variant C) | 10 | Indicate where a theorem is used: Let $f:(0,552)\to\mathbb{R}$ be $f(x)=\ln(50x)$. Is $f$ uniformly continuous on $(0,552)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{552}{n}$ and $y_n=\\frac{552}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\ln(... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": "Gener... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018691 | Real Analysis: Uniform Continuity (Variant C) | 10 | Solve (and briefly cross-validate): Let $f:(0,708)\to\mathbb{R}$ be $f(x)=\ln(49x)$. Is $f$ uniformly continuous on $(0,708)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018692 | Real Analysis: Uniform Continuity (Variant B) | 10 | Work this out carefully: Let $f:(0,435)\to\mathbb{R}$ be $f(x)=\frac{1}{(-29)x}$. Is $f$ uniformly continuous on $(0,435)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018693 | Real Analysis: Uniform Continuity (Variant A) | 10 | Track units/moduli carefully: Let $f:(0,1983)\to\mathbb{R}$ be $f(x)=\ln(27x)$. Is $f$ uniformly continuous on $(0,1983)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018694 | Real Analysis: Uniform Continuity (Core) | 10 | Start by stating any domain restrictions: Let $f:(0,1938)\to\mathbb{R}$ be $f(x)=\ln(57x)$. Is $f$ uniformly continuous on $(0,1938)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{1938}{n}$ and $y_n=\\frac{1938}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\l... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": "Robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018695 | Real Analysis: Uniform Continuity (Variant A) | 10 | Track quantifiers carefully: Let $f:[-1313,1313]\to\mathbb{R}$ be $f(x)=(19)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1313,1313]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Heine–Cantor",
"approach": "Continuous functions on compact metric spaces are uniformly continuous.",
"steps": [
"Step 1: The interval $[-1313,1313]$ is compact in $\\mathbb{R}$.",
"Step 2: The function $x\\mapsto x^2$ is continuous on $\\mathbb{R}$, hence on the compact se... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same con... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018696 | Real Analysis: Uniform Continuity (Variant C) | 10 | Question: Let $f:(0,1281)\to\mathbb{R}$ be $f(x)=\ln(24x)$. Is $f$ uniformly continuous on $(0,1281)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{1281}{n}$ and $y_n=\\frac{1281}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\l... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": "If th... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018697 | Real Analysis: Uniform Continuity (Variant B) | 10 | Compute the requested quantity: Let $f:(0,1510)\to\mathbb{R}$ be $f(x)=\frac{1}{(-15)x}$. Is $f$ uniformly continuous on $(0,1510)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{1510}{n}$ and $y_n=\\frac{1510}{2n}$ in $(0,1510)$.",
"Step 2: Then $|x_n-y_n|=\\frac{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both c... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018698 | Real Analysis: Uniform Continuity (Core) | 10 | Work carefully and justify each inference: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(-4)x^2$ with $c\ne 0$. Is $f$ uniformly continuous on $\mathbb{R}$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Mean Value Theorem Blow-Up",
"approach": "Use MVT to show that on unbounded domains, the derivative growth makes a uniform modulus impossible.",
"steps": [
"Step 1: Suppose uniform continuity holds. Take $\\varepsilon=1$ and let $\\delta>0$ correspond.",
"Step 2: Choose $x$... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nThe sequence method explicitly exhibits $|x_n-y_n|\\to 0$ while $|f(x_n)-f(y_n)|\\not\\to 0$. The MVT method formalizes the same idea via unbounded derivative growth. Both conclude ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018699 | Real Analysis: Uniform Continuity (Variant C) | 10 | Complete the analysis: Let $f:[-1005,1005]\to\mathbb{R}$ be $f(x)=(20)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1005,1005]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz on Bounded Domain",
"approach": "On a bounded interval, $|x^2-y^2|=|x-y||x+y|$ and $|x+y|$ is uniformly bounded, giving a global Lipschitz constant.",
"steps": [
"Step 1: For $x,y\\in[-1005,1005]$, $|f(x)-f(y)|=|20|\\,|x^2-y^2|=|20|\\,|x-y||x+y|$.",
"Step 2: Bound... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same con... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018700 | Real Analysis: Uniform Continuity (Variant A) | 10 | Warm-up: Let $f:[-565,565]\to\mathbb{R}$ be $f(x)=(-24)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-565,565]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz on Bounded Domain",
"approach": "On a bounded interval, $|x^2-y^2|=|x-y||x+y|$ and $|x+y|$ is uniformly bounded, giving a global Lipschitz constant.",
"steps": [
"Step 1: For $x,y\\in[-565,565]$, $|f(x)-f(y)|=|-24|\\,|x^2-y^2|=|-24|\\,|x-y||x+y|$.",
"Step 2: Bound... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
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