id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-018501 | Real Analysis: Uniform Continuity (Variant C) | 10 | Use two approaches if possible: Let $f:(0,1963)\to\mathbb{R}$ be $f(x)=\frac{1}{(-18)x}$. Is $f$ uniformly continuous on $(0,1963)$? 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-018502 | Real Analysis: Uniform Continuity (Core) | 10 | Give an answer and a quick verification: Let $f:(0,408)\to\mathbb{R}$ be $f(x)=\frac{1}{(27)x}$. Is $f$ uniformly continuous on $(0,408)$? 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 $x_n=\\frac{408}{n}$ and $y_n=\\frac{408}{2n}$ in $(0,408)$.",
"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-018503 | Real Analysis: Uniform Continuity (Core) | 10 | Solve with verification: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(-29)x^2$ 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": "Both approaches agree after simplification. 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.",
"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 No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018504 | Real Analysis: Uniform Continuity (Core) | 10 | Indicate where a theorem is used: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(3)x^2$ 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 ... | 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-018505 | Real Analysis: Uniform Continuity (Variant A) | 10 | Compute the requested quantity: Let $f:(0,193)\to\mathbb{R}$ be $f(x)=\frac{1}{(-10)x}$. Is $f$ uniformly continuous on $(0,193)$? 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 $x_n=\\frac{193}{n}$ and $y_n=\\frac{193}{2n}$ in $(0,193)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"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 ... | 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-018506 | Real Analysis: Uniform Continuity (Variant B) | 10 | Explain each transformation: Let $f:(0,794)\to\mathbb{R}$ be $f(x)=\frac{1}{(-8)x}$. Is $f$ uniformly continuous on $(0,794)$? 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. (Here the result is $\boxed{\text{No}$.) |
math-018507 | Real Analysis: Uniform Continuity (Variant A) | 10 | Work carefully and justify each inference: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(-11x)$. 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)=(-11)\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. 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.",
"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. |
math-018508 | Real Analysis: Uniform Continuity (Core) | 10 | Make each step logically reversible (or explain if not): Let $f:[-1185,1185]\to\mathbb{R}$ be $f(x)=(-6)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1185,1185]$.
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 $x,y\\in[-1185,1185]$, $|f(x)-f(y)|=|-6|\\,|x^2-y^2|=|-6|\\,|x-y||x+y|$.",
"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 ... | 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-018509 | Real Analysis: Uniform Continuity (Core) | 10 | Exercise: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(6x)$. 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": "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 ... | 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-018510 | Real Analysis: Uniform Continuity (Variant B) | 10 | Do not skip justification steps: Let $f:(0,232)\to\mathbb{R}$ be $f(x)=\ln(54x)$. Is $f$ uniformly continuous on $(0,232)$? 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": "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. (Here the result is $\boxed{\text{No}$.) |
math-018511 | Real Analysis: Uniform Continuity (Variant C) | 10 | Work carefully and justify each inference: Let $f:(0,733)\to\mathbb{R}$ be $f(x)=\sqrt{30x}$. Prove that $f$ is uniformly continuous on $(0,733)$.
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": "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 ... | 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-018512 | Real Analysis: Uniform Continuity (Variant A) | 10 | State any required conditions first: 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": "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. |
math-018513 | Real Analysis: Uniform Continuity (Variant B) | 10 | Solve and include a self-check: Let $f:(0,1972)\to\mathbb{R}$ be $f(x)=\sqrt{14x}$. Prove that $f$ is uniformly continuous on $(0,1972)$.
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 $g:[0,1972]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,1972]$.",
"Step 2: The interval $[0... | {
"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 ... | 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-018514 | Real Analysis: Uniform Continuity (Variant A) | 10 | Track quantifiers carefully: Let $f:(0,782)\to\mathbb{R}$ be $f(x)=\frac{1}{(-22)x}$. Is $f$ uniformly continuous on $(0,782)$? 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 ... | 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-018515 | Real Analysis: Uniform Continuity (Variant A) | 10 | Solve and justify each step: Let $f:(0,791)\to\mathbb{R}$ be $f(x)=\ln(52x)$. Is $f$ uniformly continuous on $(0,791)$? 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 $x_n=\\frac{791}{n}$ and $y_n=\\frac{791}{2n}$.",
"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": "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-018516 | Real Analysis: Uniform Continuity (Variant A) | 10 | Exercise: Let $f:(0,1564)\to\mathbb{R}$ be $f(x)=\frac{1}{(14)x}$. Is $f$ uniformly continuous on $(0,1564)$? 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 $x_n=\\frac{1564}{n}$ and $y_n=\\frac{1564}{2n}$ in $(0,1564)$.",
"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 ... | 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-018517 | Real Analysis: Uniform Continuity (Variant B) | 10 | Work carefully and justify each inference: Let $f:(0,928)\to\mathbb{R}$ be $f(x)=\frac{1}{(25)x}$. Is $f$ uniformly continuous on $(0,928)$? 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 ... | 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-018518 | Real Analysis: Uniform Continuity (Variant C) | 10 | Do not skip justification steps: Let $f:(0,387)\to\mathbb{R}$ be $f(x)=\ln(48x)$. Is $f$ uniformly continuous on $(0,387)$? 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": "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 ... | 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-018519 | Real Analysis: Uniform Continuity (Variant B) | 10 | Solve (and briefly cross-validate): Let $f:[-1628,1628]\to\mathbb{R}$ be $f(x)=(-7)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1628,1628]$.
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 $x,y\\in[-1628,1628]$, $|f(x)-f(y)|=|-7|\\,|x^2-y^2|=|-7|\\,|x-y||x+y|$.",
"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 ... | 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-018520 | Real Analysis: Uniform Continuity (Variant C) | 10 | Solve and include a self-check: Let $f:(0,1866)\to\mathbb{R}$ be $f(x)=\frac{1}{(10)x}$. Is $f$ uniformly continuous on $(0,1866)$? 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 $x_n=\\frac{1866}{n}$ and $y_n=\\frac{1866}{2n}$ in $(0,1866)$.",
"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 ... | 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-018521 | Real Analysis: Uniform Continuity (Variant B) | 10 | Prompt: Let $f:(0,625)\to\mathbb{R}$ be $f(x)=\sqrt{27x}$. Prove that $f$ is uniformly continuous on $(0,625)$.
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-018522 | Real Analysis: Uniform Continuity (Core) | 10 | Solve with verification: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(-4x)$. 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": "If ... | [
{
"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-018523 | Real Analysis: Uniform Continuity (Core) | 10 | Question: Let $f:(0,1103)\to\mathbb{R}$ be $f(x)=\ln(7x)$. Is $f$ uniformly continuous on $(0,1103)$? 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 $x_n=\\frac{1103}{n}$ and $y_n=\\frac{1103}{2n}$.",
"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 ... | 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-018524 | Real Analysis: Uniform Continuity (Variant C) | 10 | Compute the requested quantity: Let $f:(0,1041)\to\mathbb{R}$ be $f(x)=\ln(57x)$. Is $f$ uniformly continuous on $(0,1041)$? 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 ... | 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-018525 | Real Analysis: Uniform Continuity (Core) | 10 | Show all reasoning: Let $f:(0,1404)\to\mathbb{R}$ be $f(x)=\frac{1}{(-3)x}$. Is $f$ uniformly continuous on $(0,1404)$? 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-018526 | Real Analysis: Uniform Continuity (Variant B) | 10 | Complete the analysis: Let $f:(0,321)\to\mathbb{R}$ be $f(x)=\frac{1}{(10)x}$. Is $f$ uniformly continuous on $(0,321)$? 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-018527 | Real Analysis: Uniform Continuity (Core) | 10 | Task: Let $f:(0,283)\to\mathbb{R}$ be $f(x)=\ln(21x)$. Is $f$ uniformly continuous on $(0,283)$? 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 $x_n=\\frac{283}{n}$ and $y_n=\\frac{283}{2n}$.",
"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": "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. |
math-018528 | Real Analysis: Uniform Continuity (Variant A) | 10 | Complete the analysis: Let $f:(0,461)\to\mathbb{R}$ be $f(x)=\ln(10x)$. Is $f$ uniformly continuous on $(0,461)$? 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 $x_n=\\frac{461}{n}$ and $y_n=\\frac{461}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\ln(... | {
"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. (Here the result is $\boxed{\text{No}$.) |
math-018529 | Real Analysis: Uniform Continuity (Variant B) | 10 | Explain why your operations are valid: Let $f:[-172,172]\to\mathbb{R}$ be $f(x)=(8)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-172,172]$.
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 $x,y\\in[-172,172]$, $|f(x)-f(y)|=|8|\\,|x^2-y^2|=|8|\\,|x-y||x+y|$.",
"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 ... | 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-018530 | Real Analysis: Uniform Continuity (Core) | 10 | Solve and include a self-check: Let $f:(0,471)\to\mathbb{R}$ be $f(x)=\ln(56x)$. Is $f$ uniformly continuous on $(0,471)$? 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-018531 | Real Analysis: Uniform Continuity (Variant C) | 10 | Start by stating any domain restrictions: Let $f:(0,1925)\to\mathbb{R}$ be $f(x)=\frac{1}{(-19)x}$. Is $f$ uniformly continuous on $(0,1925)$? 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 $x_n=\\frac{1925}{n}$ and $y_n=\\frac{1925}{2n}$ in $(0,1925)$.",
"Step 2: Then $|x_n-y_n|=\\frac{... | {
"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-018532 | Real Analysis: Uniform Continuity (Variant A) | 10 | Checkpoint: Let $f:[-902,902]\to\mathbb{R}$ be $f(x)=(-19)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-902,902]$.
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 $[-902,902]$ 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": "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 ... | 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-018533 | Real Analysis: Uniform Continuity (Variant B) | 10 | Determine the requested value: Let $f:(0,1508)\to\mathbb{R}$ be $f(x)=\ln(6x)$. Is $f$ uniformly continuous on $(0,1508)$? 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": "Sensi... | [
{
"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-018534 | Real Analysis: Uniform Continuity (Core) | 10 | Prompt: Let $f:(0,187)\to\mathbb{R}$ be $f(x)=\ln(54x)$. Is $f$ uniformly continuous on $(0,187)$? 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 $x_n=\\frac{187}{n}$ and $y_n=\\frac{187}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\ln(... | {
"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 ... | 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-018535 | Real Analysis: Uniform Continuity (Variant C) | 10 | Answer with a short justification: Let $f:(0,114)\to\mathbb{R}$ be $f(x)=\ln(30x)$. Is $f$ uniformly continuous on $(0,114)$? 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 ... | 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-018536 | Real Analysis: Uniform Continuity (Core) | 10 | Show all reasoning: Let $f:(0,1936)\to\mathbb{R}$ be $f(x)=\frac{1}{(2)x}$. Is $f$ uniformly continuous on $(0,1936)$? 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 $x_n=\\frac{1936}{n}$ and $y_n=\\frac{1936}{2n}$ in $(0,1936)$.",
"Step 2: Then $|x_n-y_n|=\\frac{... | {
"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. (Here the result is $\boxed{\text{No}$.) |
math-018537 | Real Analysis: Uniform Continuity (Variant B) | 10 | Make each step logically reversible (or explain if not): Let $f:(0,1646)\to\mathbb{R}$ be $f(x)=\ln(23x)$. Is $f$ uniformly continuous on $(0,1646)$? 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 $x_n=\\frac{1646}{n}$ and $y_n=\\frac{1646}{2n}$.",
"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-018538 | Real Analysis: Uniform Continuity (Variant C) | 10 | Question: Let $f:(0,621)\to\mathbb{R}$ be $f(x)=\sqrt{17x}$. Prove that $f$ is uniformly continuous on $(0,621)$.
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 $g:[0,621]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,621]$.",
"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-018539 | Real Analysis: Uniform Continuity (Variant B) | 10 | Work carefully and justify each inference: Let $f:(0,1300)\to\mathbb{R}$ be $f(x)=\sqrt{25x}$. Prove that $f$ is uniformly continuous on $(0,1300)$.
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 $g:[0,1300]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,1300]$.",
"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 ... | 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-018540 | Real Analysis: Uniform Continuity (Core) | 10 | Question: Let $f:(0,324)\to\mathbb{R}$ be $f(x)=\frac{1}{(-11)x}$. Is $f$ uniformly continuous on $(0,324)$? 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. |
math-018541 | Real Analysis: Uniform Continuity (Variant A) | 10 | State any required conditions first: Let $f:(0,1543)\to\mathbb{R}$ be $f(x)=\ln(19x)$. Is $f$ uniformly continuous on $(0,1543)$? 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 ... | 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-018542 | Real Analysis: Uniform Continuity (Variant B) | 10 | Solve and sanity-check: Let $f:(0,1682)\to\mathbb{R}$ be $f(x)=\frac{1}{(8)x}$. Is $f$ uniformly continuous on $(0,1682)$? 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 $x_n=\\frac{1682}{n}$ and $y_n=\\frac{1682}{2n}$ in $(0,1682)$.",
"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-018543 | Real Analysis: Uniform Continuity (Core) | 10 | Work carefully and justify each inference: Let $f:(0,149)\to\mathbb{R}$ be $f(x)=\frac{1}{(28)x}$. Is $f$ uniformly continuous on $(0,149)$? 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 $x_n=\\frac{149}{n}$ and $y_n=\\frac{149}{2n}$ in $(0,149)$.",
"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. (Here the result is $\boxed{\text{No}$.) |
math-018544 | Real Analysis: Uniform Continuity (Core) | 10 | Solve with verification: Let $f:(0,1276)\to\mathbb{R}$ be $f(x)=\ln(8x)$. Is $f$ uniformly continuous on $(0,1276)$? 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. (Here the result is $\boxed{\text{No}$.) |
math-018545 | Real Analysis: Uniform Continuity (Variant B) | 10 | Provide a rigorous solution: Let $f:(0,1328)\to\mathbb{R}$ be $f(x)=\ln(37x)$. Is $f$ uniformly continuous on $(0,1328)$? 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. |
math-018546 | Real Analysis: Uniform Continuity (Variant C) | 10 | Track units/moduli carefully: Let $f:[-1286,1286]\to\mathbb{R}$ be $f(x)=(19)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1286,1286]$.
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 $x,y\\in[-1286,1286]$, $|f(x)-f(y)|=|19|\\,|x^2-y^2|=|19|\\,|x-y||x+y|$.",
"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 ... | 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-018547 | Real Analysis: Uniform Continuity (Variant A) | 10 | Work carefully and justify each inference: Let $f:(0,1831)\to\mathbb{R}$ be $f(x)=\ln(29x)$. Is $f$ uniformly continuous on $(0,1831)$? 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. |
math-018548 | Real Analysis: Uniform Continuity (Variant A) | 10 | Answer using clear logical steps: Let $f:(0,620)\to\mathbb{R}$ be $f(x)=\frac{1}{(3)x}$. Is $f$ uniformly continuous on $(0,620)$? 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-018549 | Real Analysis: Uniform Continuity (Variant C) | 10 | Use two approaches if possible: Let $f:(0,1230)\to\mathbb{R}$ be $f(x)=\sqrt{10x}$. Prove that $f$ is uniformly continuous on $(0,1230)$.
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 $g:[0,1230]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,1230]$.",
"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 ... | 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-018550 | Real Analysis: Uniform Continuity (Variant A) | 10 | Prompt: Let $f:[-129,129]\to\mathbb{R}$ be $f(x)=(26)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-129,129]$.
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 $x,y\\in[-129,129]$, $|f(x)-f(y)|=|26|\\,|x^2-y^2|=|26|\\,|x-y||x+y|$.",
"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-018551 | Real Analysis: Uniform Continuity (Variant B) | 10 | Explain each transformation: Let $f:[-1392,1392]\to\mathbb{R}$ be $f(x)=(3)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1392,1392]$.
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 $[-1392,1392]$ 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 ... | 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-018552 | Real Analysis: Uniform Continuity (Variant C) | 10 | State any required conditions first: Let $f:(0,358)\to\mathbb{R}$ be $f(x)=\frac{1}{(-2)x}$. Is $f$ uniformly continuous on $(0,358)$? 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 $x_n=\\frac{358}{n}$ and $y_n=\\frac{358}{2n}$ in $(0,358)$.",
"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-018553 | Real Analysis: Uniform Continuity (Variant B) | 10 | Solve with verification: Let $f:(0,963)\to\mathbb{R}$ be $f(x)=\sqrt{1x}$. Prove that $f$ is uniformly continuous on $(0,963)$.
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": "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-018554 | Real Analysis: Uniform Continuity (Variant B) | 10 | Solve and sanity-check: Let $f:(0,1454)\to\mathbb{R}$ be $f(x)=\frac{1}{(-9)x}$. Is $f$ uniformly continuous on $(0,1454)$? 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-018555 | Real Analysis: Uniform Continuity (Core) | 10 | Complete the analysis: Let $f:(0,1462)\to\mathbb{R}$ be $f(x)=\ln(6x)$. Is $f$ uniformly continuous on $(0,1462)$? 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 $x_n=\\frac{1462}{n}$ and $y_n=\\frac{1462}{2n}$.",
"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 ... | 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-018556 | Real Analysis: Uniform Continuity (Variant B) | 10 | Answer using clear logical steps: Let $f:(0,1474)\to\mathbb{R}$ be $f(x)=\ln(46x)$. Is $f$ uniformly continuous on $(0,1474)$? 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": "Sensi... | [
{
"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-018557 | Real Analysis: Uniform Continuity (Variant B) | 10 | Make each step logically reversible (or explain if not): Let $f:(0,1603)\to\mathbb{R}$ be $f(x)=\frac{1}{(14)x}$. Is $f$ uniformly continuous on $(0,1603)$? 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 $x_n=\\frac{1603}{n}$ and $y_n=\\frac{1603}{2n}$ in $(0,1603)$.",
"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-018558 | Real Analysis: Uniform Continuity (Core) | 10 | Solve and justify each step: Let $f:[-397,397]\to\mathbb{R}$ be $f(x)=(-10)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-397,397]$.
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 $x,y\\in[-397,397]$, $|f(x)-f(y)|=|-10|\\,|x^2-y^2|=|-10|\\,|x-y||x+y|$.",
"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. |
math-018559 | Real Analysis: Uniform Continuity (Core) | 10 | Complete the analysis: Let $f:(0,1581)\to\mathbb{R}$ be $f(x)=\ln(46x)$. Is $f$ uniformly continuous on $(0,1581)$? 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 $x_n=\\frac{1581}{n}$ and $y_n=\\frac{1581}{2n}$.",
"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 ... | 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-018560 | Real Analysis: Uniform Continuity (Core) | 10 | Solve (and briefly cross-validate): Let $f:(0,1109)\to\mathbb{R}$ be $f(x)=\ln(10x)$. Is $f$ uniformly continuous on $(0,1109)$? 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-018561 | Real Analysis: Uniform Continuity (Variant B) | 10 | Solve and justify each step: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(-6)x^2$ 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 ... | 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-018562 | Real Analysis: Uniform Continuity (Variant A) | 10 | Solve and sanity-check: Let $f:(0,1437)\to\mathbb{R}$ be $f(x)=\ln(18x)$. Is $f$ uniformly continuous on $(0,1437)$? 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-018563 | Real Analysis: Uniform Continuity (Variant C) | 10 | Complete the analysis: Let $f:(0,414)\to\mathbb{R}$ be $f(x)=\frac{1}{(-29)x}$. Is $f$ uniformly continuous on $(0,414)$? 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 $x_n=\\frac{414}{n}$ and $y_n=\\frac{414}{2n}$ in $(0,414)$.",
"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-018564 | Real Analysis: Uniform Continuity (Variant C) | 10 | Explain each transformation: Let $f:(0,1029)\to\mathbb{R}$ be $f(x)=\sqrt{55x}$. Prove that $f$ is uniformly continuous on $(0,1029)$.
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": "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 ... | 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-018565 | Real Analysis: Uniform Continuity (Variant C) | 10 | Provide both a computational and a conceptual explanation: Let $f:(0,1369)\to\mathbb{R}$ be $f(x)=\sqrt{40x}$. Prove that $f$ is uniformly continuous on $(0,1369)$.
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": "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 ... | 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-018566 | Real Analysis: Uniform Continuity (Variant A) | 10 | Carefully track domains: Let $f:(0,1624)\to\mathbb{R}$ be $f(x)=\ln(47x)$. Is $f$ uniformly continuous on $(0,1624)$? 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 ... | 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-018567 | Real Analysis: Uniform Continuity (Variant B) | 10 | Question: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(19x)$. 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": "If ... | [
{
"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-018568 | Real Analysis: Uniform Continuity (Variant C) | 10 | Solve and then verify: Let $f:(0,1831)\to\mathbb{R}$ be $f(x)=\sqrt{8x}$. Prove that $f$ is uniformly continuous on $(0,1831)$.
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": "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. (Here the result is $\boxed{\text{Yes}$.) |
math-018569 | Real Analysis: Uniform Continuity (Variant A) | 10 | Solve and sanity-check: Let $f:(0,423)\to\mathbb{R}$ be $f(x)=\ln(50x)$. Is $f$ uniformly continuous on $(0,423)$? 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 $x_n=\\frac{423}{n}$ and $y_n=\\frac{423}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\ln(... | {
"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 ... | 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-018570 | Real Analysis: Uniform Continuity (Variant A) | 10 | Start by stating any domain restrictions: Let $f:[-1644,1644]\to\mathbb{R}$ be $f(x)=(-15)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1644,1644]$.
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 $[-1644,1644]$ 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 ... | 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-018571 | Real Analysis: Uniform Continuity (Variant B) | 10 | Answer with a short justification: Let $f:(0,1426)\to\mathbb{R}$ be $f(x)=\ln(54x)$. Is $f$ uniformly continuous on $(0,1426)$? 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 ... | 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-018572 | Real Analysis: Uniform Continuity (Variant B) | 10 | Exercise: Let $f:(0,592)\to\mathbb{R}$ be $f(x)=\frac{1}{(22)x}$. 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": "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 ... | 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-018573 | Real Analysis: Uniform Continuity (Variant C) | 10 | Explain why your operations are valid: Let $f:(0,205)\to\mathbb{R}$ be $f(x)=\ln(27x)$. Is $f$ uniformly continuous on $(0,205)$? 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-018574 | Real Analysis: Uniform Continuity (Variant A) | 10 | Work carefully and justify each inference: Let $f:(0,1003)\to\mathbb{R}$ be $f(x)=\ln(18x)$. Is $f$ uniformly continuous on $(0,1003)$? 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 ... | 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-018575 | Real Analysis: Uniform Continuity (Core) | 10 | Carefully track domains: Let $f:(0,1881)\to\mathbb{R}$ be $f(x)=\sqrt{39x}$. Prove that $f$ is uniformly continuous on $(0,1881)$.
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": "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 ... | 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-018576 | Real Analysis: Uniform Continuity (Variant A) | 10 | Explain each transformation: Let $f:(0,1185)\to\mathbb{R}$ be $f(x)=\ln(3x)$. Is $f$ uniformly continuous on $(0,1185)$? 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 $x_n=\\frac{1185}{n}$ and $y_n=\\frac{1185}{2n}$.",
"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 ... | 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-018577 | Real Analysis: Uniform Continuity (Variant A) | 10 | Provide both a computational and a conceptual explanation: Let $f:[-173,173]\to\mathbb{R}$ be $f(x)=(-14)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-173,173]$.
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 $[-173,173]$ 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": "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 ... | 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-018578 | Real Analysis: Uniform Continuity (Core) | 10 | Track quantifiers carefully: Let $f:(0,16)\to\mathbb{R}$ be $f(x)=\sqrt{44x}$. Prove that $f$ is uniformly continuous on $(0,16)$.
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": "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-018579 | Real Analysis: Uniform Continuity (Variant C) | 10 | State any required conditions first: Let $f:(0,481)\to\mathbb{R}$ be $f(x)=\ln(38x)$. Is $f$ uniformly continuous on $(0,481)$? 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. |
math-018580 | Real Analysis: Uniform Continuity (Variant A) | 10 | Work this out carefully: Let $f:(0,138)\to\mathbb{R}$ be $f(x)=\ln(19x)$. Is $f$ uniformly continuous on $(0,138)$? 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 ... | 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-018581 | Real Analysis: Uniform Continuity (Core) | 10 | Solve and sanity-check: Let $f:[-441,441]\to\mathbb{R}$ be $f(x)=(7)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-441,441]$.
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 $x,y\\in[-441,441]$, $|f(x)-f(y)|=|7|\\,|x^2-y^2|=|7|\\,|x-y||x+y|$.",
"Step 2: Bound $|x... | {
"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 ... | 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-018582 | Real Analysis: Uniform Continuity (Variant A) | 10 | Determine the requested value: Let $f:(0,429)\to\mathbb{R}$ be $f(x)=\ln(27x)$. Is $f$ uniformly continuous on $(0,429)$? 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 $x_n=\\frac{429}{n}$ and $y_n=\\frac{429}{2n}$.",
"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": "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-018583 | Real Analysis: Uniform Continuity (Variant C) | 10 | Give an answer and a quick verification: Let $f:(0,1454)\to\mathbb{R}$ be $f(x)=\ln(7x)$. Is $f$ uniformly continuous on $(0,1454)$? 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-018584 | Real Analysis: Uniform Continuity (Core) | 10 | Write the solution set clearly: Let $f:(0,978)\to\mathbb{R}$ be $f(x)=\frac{1}{(13)x}$. Is $f$ uniformly continuous on $(0,978)$? 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-018585 | Real Analysis: Uniform Continuity (Core) | 10 | Question: Let $f:(0,1207)\to\mathbb{R}$ be $f(x)=\frac{1}{(-23)x}$. Is $f$ uniformly continuous on $(0,1207)$? 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-018586 | Real Analysis: Uniform Continuity (Variant A) | 10 | Solve (and briefly cross-validate): Let $f:(0,1069)\to\mathbb{R}$ be $f(x)=\frac{1}{(26)x}$. Is $f$ uniformly continuous on $(0,1069)$? 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 $x_n=\\frac{1069}{n}$ and $y_n=\\frac{1069}{2n}$ in $(0,1069)$.",
"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-018587 | Real Analysis: Uniform Continuity (Variant B) | 10 | Start by stating any domain restrictions: Let $f:(0,553)\to\mathbb{R}$ be $f(x)=\ln(50x)$. Is $f$ uniformly continuous on $(0,553)$? 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. |
math-018588 | Real Analysis: Uniform Continuity (Core) | 10 | Provide a rigorous solution: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(25)x^2$ 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": "Both approaches agree after simplification. 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.",
"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 ... | 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-018589 | Real Analysis: Uniform Continuity (Variant B) | 10 | Exercise: Let $f:(0,824)\to\mathbb{R}$ be $f(x)=\sqrt{51x}$. Prove that $f$ is uniformly continuous on $(0,824)$.
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": "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. |
math-018590 | Real Analysis: Uniform Continuity (Core) | 10 | Problem: Let $f:(0,21)\to\mathbb{R}$ be $f(x)=\sqrt{7x}$. Prove that $f$ is uniformly continuous on $(0,21)$.
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": "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-018591 | Real Analysis: Uniform Continuity (Core) | 10 | Solve (and briefly cross-validate): Let $f:(0,820)\to\mathbb{R}$ be $f(x)=\ln(36x)$. Is $f$ uniformly continuous on $(0,820)$? 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 ... | 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-018592 | Real Analysis: Uniform Continuity (Core) | 10 | Give reasoning, not just computation: Let $f:(0,1498)\to\mathbb{R}$ be $f(x)=\ln(18x)$. Is $f$ uniformly continuous on $(0,1498)$? 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 $x_n=\\frac{1498}{n}$ and $y_n=\\frac{1498}{2n}$.",
"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 ... | 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-018593 | Real Analysis: Uniform Continuity (Variant A) | 10 | Be explicit about assumptions: Let $f:(0,345)\to\mathbb{R}$ be $f(x)=\sqrt{16x}$. Prove that $f$ is uniformly continuous on $(0,345)$.
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 $g:[0,345]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,345]$.",
"Step 2: The interval $[0,3... | {
"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. (Here the result is $\boxed{\text{Yes}$.) |
math-018594 | Real Analysis: Uniform Continuity (Variant B) | 10 | Explain what is being counted/optimized: Let $f:[-1888,1888]\to\mathbb{R}$ be $f(x)=(-20)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1888,1888]$.
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 $[-1888,1888]$ 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": "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-018595 | Real Analysis: Uniform Continuity (Variant C) | 10 | Warm-up: Let $f:(0,1521)\to\mathbb{R}$ be $f(x)=\frac{1}{(10)x}$. Is $f$ uniformly continuous on $(0,1521)$? 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 $x_n=\\frac{1521}{n}$ and $y_n=\\frac{1521}{2n}$ in $(0,1521)$.",
"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 ... | 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-018596 | Real Analysis: Uniform Continuity (Variant A) | 10 | Answer using clear logical steps: Let $f:(0,976)\to\mathbb{R}$ be $f(x)=\frac{1}{(28)x}$. Is $f$ uniformly continuous on $(0,976)$? 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. |
math-018597 | Real Analysis: Uniform Continuity (Core) | 10 | Use two approaches if possible: Let $f:(0,1516)\to\mathbb{R}$ be $f(x)=\frac{1}{(3)x}$. Is $f$ uniformly continuous on $(0,1516)$? 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 $x_n=\\frac{1516}{n}$ and $y_n=\\frac{1516}{2n}$ in $(0,1516)$.",
"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 ... | 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-018598 | Real Analysis: Uniform Continuity (Variant B) | 10 | Provide both a computational and a conceptual explanation: Let $f:(0,967)\to\mathbb{R}$ be $f(x)=\frac{1}{(-30)x}$. Is $f$ uniformly continuous on $(0,967)$? 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 $x_n=\\frac{967}{n}$ and $y_n=\\frac{967}{2n}$ in $(0,967)$.",
"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. (Here the result is $\boxed{\text{No}$.) |
math-018599 | Real Analysis: Uniform Continuity (Variant B) | 10 | Try to avoid pattern-matching; explain why: Let $f:(0,501)\to\mathbb{R}$ be $f(x)=\sqrt{41x}$. Prove that $f$ is uniformly continuous on $(0,501)$.
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 $g:[0,501]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,501]$.",
"Step 2: The interval $[0,5... | {
"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. |
math-018600 | Real Analysis: Uniform Continuity (Core) | 10 | Use two approaches if possible: Let $f:(0,437)\to\mathbb{R}$ be $f(x)=\frac{1}{(19)x}$. Is $f$ uniformly continuous on $(0,437)$? 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 $x_n=\\frac{437}{n}$ and $y_n=\\frac{437}{2n}$ in $(0,437)$.",
"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 $\boxed{\text{No}$.) |
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