id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-016001 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Work carefully and justify each inference: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-56$, and that... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-56x}$.\nBoth methods force linearity and use $f(1)=-56$ to identify the slope. The extra datum $f(\\frac{-12}{7})=96$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-56x... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-56$ fixes the function to $f(x)=-56x$. (Here the result is $\boxed{-56x}$.) |
math-016002 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Problem: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-641$, and that $f\!\left(\frac{-1}{8}\right)=\f... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-641x}$.\nBoth methods force linearity and use $f(1)=-641$ to identify the slope. The extra datum $f(\\frac{-1}{8})=\\frac{641}{8}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclu... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-641$ fixes the function to $f(x)=-641x$. (Here the result is $\boxed{-641x}$.) |
math-016003 | Functional Equations: Additive Maps — Density Argument | 9 | Work carefully and justify each inference: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=558$, a... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{558x}$.\nBoth methods force linearity and use $f(1)=558$ to identify the slope. The extra datum $f(\\frac{15}{7})=\\frac{8370}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=558x$.",
"rob... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=558$ fixes the function to $f(x)=558x$. |
math-016004 | Functional Equations: Additive Maps — Density Argument | 9 | Find the exact value: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=105$, and tha... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{105x}$.\nBoth methods force linearity and use $f(1)=105$ to identify the slope. The extra datum $f(\\frac{-19}{12})=\\frac{-665}{4}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both concl... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=105$ fixes the function to $f(x)=105x$. (Here the result is $\boxed{105x}$.) |
math-016005 | Functional Equations: Additivity — Extension from Q to R | 9 | Prompt: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=327$, and that $f\!\left(\f... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{327x}$.\nBoth methods force linearity and use $f(1)=327$ to identify the slope. The extra datum $f(\\frac{-3}{4})=\\frac{-981}{4}$ is consistent with $f(x)=kx$ and serves as a built-in check. Bot... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=327$ fixes the function to $f(x)=327x$. |
math-016006 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Show all reasoning: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-713$, and that $f\!\left(\frac{-25}{... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-713x}$.\nBoth methods force linearity and use $f(1)=-713$ to identify the slope. The extra datum $f(\\frac{-5}{2})=\\frac{3565}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-713x$.... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-713$ fixes the function to $f(x)=-713x$. |
math-016007 | Functional Equations: Additive Maps — Density Argument | 9 | Give a fully justified solution: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-3$, and that $f\!\left(... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-3x}$.\nBoth methods force linearity and use $f(1)=-3$ to identify the slope. The extra datum $f(\\frac{-17}{4})=\\frac{51}{4}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both c... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-3$ fixes the function to $f(x)=-3x$. |
math-016008 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Provide both a computational and a conceptual explanation: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is contin... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{762x}$.\nBoth methods force linearity and use $f(1)=762$ to identify the slope. The extra datum $f(\\frac{-15}{2})=-5715$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=762x$.",
"robust... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=762$ fixes the function to $f(x)=762x$. |
math-016009 | Functional Equations: Additivity — Extension from Q to R | 9 | Track units/moduli carefully: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=548$, and that $f\!\left(\frac{-2... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{548x}$.\nBoth methods force linearity and use $f(1)=548$ to identify the slope. The extra datum $f(\\frac{-21}{8})=\\frac{-2877}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=548x$.",
"r... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=548$ fixes the function to $f(x)=548x$. (Here the result is $\boxed{548x}$.) |
math-016010 | Functional Equations: Additive Maps — Density Argument | 9 | Answer using clear logical steps: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=121$, and that $f\!\lef... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{121x}$.\nBoth methods force linearity and use $f(1)=121$ to identify the slope. The extra datum $f(\\frac{1}{3})=\\frac{121}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both ... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=121$ fixes the function to $f(x)=121x$. |
math-016011 | Functional Equations: Additive Maps — Density Argument | 9 | Give a fully justified solution: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=167$, and that $f\!\left(\frac... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{167x}$.\nBoth methods force linearity and use $f(1)=167$ to identify the slope. The extra datum $f(\\frac{4}{5})=\\frac{668}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both ... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=167$ fixes the function to $f(x)=167x$. (Here the result is $\boxed{167x}$.) |
math-016012 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Explain what is being counted/optimized: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-576$, and that $f\!\l... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-576x}$.\nBoth methods force linearity and use $f(1)=-576$ to identify the slope. The extra datum $f(\\frac{-5}{3})=960$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-576$ fixes the function to $f(x)=-576x$. (Here the result is $\boxed{-576x}$.) |
math-016013 | Functional Equations: Additivity — Extension from Q to R | 9 | Compute the requested quantity: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=209... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{209x}$.\nBoth methods force linearity and use $f(1)=209$ to identify the slope. The extra datum $f(\\frac{-1}{4})=\\frac{-209}{4}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=209x$.",
"rob... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=209$ fixes the function to $f(x)=209x$. |
math-016014 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Do not skip justification steps: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-583$, and that $f\!\left(\fra... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-583x}$.\nBoth methods force linearity and use $f(1)=-583$ to identify the slope. The extra datum $f(\\frac{1}{2})=\\frac{-583}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-583x$."... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-583$ fixes the function to $f(x)=-583x$. (Here the result is $\boxed{-583x}$.) |
math-016015 | Functional Equations: Additivity — Extension from Q to R | 9 | Exercise: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=625$, and that $f\!\left(\frac{3}{5}\rig... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{625x}$.\nBoth methods force linearity and use $f(1)=625$ to identify the slope. The extra datum $f(\\frac{3}{5})=375$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=625x$.",
"robustness_analy... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=625$ fixes the function to $f(x)=625x$. (Here the result is $\boxed{625x}$.) |
math-016016 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Provide a rigorous solution: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=27$, and that $f\!\left(\frac{5}{6... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{27x}$.\nBoth methods force linearity and use $f(1)=27$ to identify the slope. The extra datum $f(\\frac{5}{6})=\\frac{45}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both con... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=27$ fixes the function to $f(x)=27x$. |
math-016017 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Explain why your operations are valid: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-356$, and ... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-356x}$.\nBoth methods force linearity and use $f(1)=-356$ to identify the slope. The extra datum $f(2)=-712$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-356x$.",
"robustness_analys... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-356$ fixes the function to $f(x)=-356x$. (Here the result is $\boxed{-356x}$.) |
math-016018 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Keep the final answer in boxed form: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-425x}$.\nBoth methods force linearity and use $f(1)=-425$ to identify the slope. The extra datum $f(\\frac{-13}{7})=\\frac{5525}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-425x$... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-425$ fixes the function to $f(x)=-425x$. (Here the result is $\boxed{-425x}$.) |
math-016019 | Functional Equations: Additive Maps — Density Argument | 9 | Question: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-294$, and that $f\!\left... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-294x}$.\nBoth methods force linearity and use $f(1)=-294$ to identify the slope. The extra datum $f(\\frac{-24}{5})=\\frac{7056}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conc... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-294$ fixes the function to $f(x)=-294x$. (Here the result is $\boxed{-294x}$.) |
math-016020 | Functional Equations: Additive Maps — Density Argument | 9 | Proceed methodically: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=239$, and that $f\!\left(\fr... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{239x}$.\nBoth methods force linearity and use $f(1)=239$ to identify the slope. The extra datum $f(4)=956$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=239x$.",
"robustness_analysis": "Gene... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=239$ fixes the function to $f(x)=239x$. (Here the result is $\boxed{239x}$.) |
math-016021 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Try to avoid pattern-matching; explain why: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=58$, and that... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{58x}$.\nBoth methods force linearity and use $f(1)=58$ to identify the slope. The extra datum $f(\\frac{-5}{2})=-145$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=58x$.",
"robustness_... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=58$ fixes the function to $f(x)=58x$. (Here the result is $\boxed{58x}$.) |
math-016022 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Write the solution set clearly: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=240$, and that $f\!\left(... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{240x}$.\nBoth methods force linearity and use $f(1)=240$ to identify the slope. The extra datum $f(\\frac{-5}{9})=\\frac{-400}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclud... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=240$ fixes the function to $f(x)=240x$. |
math-016023 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Explain why your operations are valid: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-179x}$.\nBoth methods force linearity and use $f(1)=-179$ to identify the slope. The extra datum $f(\\frac{-1}{6})=\\frac{179}{6}$ is consistent with $f(x)=kx$ and serves as a built-in check. Bo... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-179$ fixes the function to $f(x)=-179x$. |
math-016024 | Functional Equations: Additive Maps — Density Argument | 9 | Try to avoid pattern-matching; explain why: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-535$, and that $f\... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-535x}$.\nBoth methods force linearity and use $f(1)=-535$ to identify the slope. The extra datum $f(\\frac{5}{11})=\\frac{-2675}{11}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-535x... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-535$ fixes the function to $f(x)=-535x$. |
math-016025 | Functional Equations: Additivity — Extension from Q to R | 9 | Work this out carefully: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=664$, and that $f\!\left(\frac{-6}{10}... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{664x}$.\nBoth methods force linearity and use $f(1)=664$ to identify the slope. The extra datum $f(\\frac{-3}{5})=\\frac{-1992}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Bo... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=664$ fixes the function to $f(x)=664x$. |
math-016026 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Explain each transformation: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-99$, and that $f\!\left(\fr... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-99x}$.\nBoth methods force linearity and use $f(1)=-99$ to identify the slope. The extra datum $f(\\frac{-19}{5})=\\frac{1881}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-99x$.",... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-99$ fixes the function to $f(x)=-99x$. (Here the result is $\boxed{-99x}$.) |
math-016027 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Do not skip justification steps: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=51... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{515x}$.\nBoth methods force linearity and use $f(1)=515$ to identify the slope. The extra datum $f(\\frac{-7}{9})=\\frac{-3605}{9}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=515x$.",... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=515$ fixes the function to $f(x)=515x$. (Here the result is $\boxed{515x}$.) |
math-016028 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Indicate where a theorem is used: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=450$, and that $f\!\left(\fra... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{450x}$.\nBoth methods force linearity and use $f(1)=450$ to identify the slope. The extra datum $f(\\frac{-23}{8})=\\frac{-5175}{4}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=450x$.",
"r... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=450$ fixes the function to $f(x)=450x$. (Here the result is $\boxed{450x}$.) |
math-016029 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Compute the requested quantity: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-554$, and that $f\!\left... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-554x}$.\nBoth methods force linearity and use $f(1)=-554$ to identify the slope. The extra datum $f(\\frac{9}{7})=\\frac{-4986}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both concl... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-554$ fixes the function to $f(x)=-554x$. |
math-016030 | Functional Equations: Additive Maps — Density Argument | 9 | Track units/moduli carefully: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=38$, and that $f\!\left(\frac{21}... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{38x}$.\nBoth methods force linearity and use $f(1)=38$ to identify the slope. The extra datum $f(\\frac{7}{2})=133$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=38x$.",... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=38$ fixes the function to $f(x)=38x$. |
math-016031 | Functional Equations: Additivity — Extension from Q to R | 9 | Answer using clear logical steps: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-163$, and that ... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-163x}$.\nBoth methods force linearity and use $f(1)=-163$ to identify the slope. The extra datum $f(\\frac{24}{5})=\\frac{-3912}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-163x$... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-163$ fixes the function to $f(x)=-163x$. (Here the result is $\boxed{-163x}$.) |
math-016032 | Functional Equations: Additive Maps — Density Argument | 9 | Solve and then verify: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-191$, and that $f\!\left(\frac{-8}{11}\... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-191x}$.\nBoth methods force linearity and use $f(1)=-191$ to identify the slope. The extra datum $f(\\frac{-8}{11})=\\frac{1528}{11}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-191x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-191$ fixes the function to $f(x)=-191x$. (Here the result is $\boxed{-191x}$.) |
math-016033 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Explain why your operations are valid: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{586x}$.\nBoth methods force linearity and use $f(1)=586$ to identify the slope. The extra datum $f(\\frac{-15}{2})=-4395$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=586x$.",
"robustness_a... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=586$ fixes the function to $f(x)=586x$. (Here the result is $\boxed{586x}$.) |
math-016034 | Functional Equations: Additive Maps — Density Argument | 9 | Solve and sanity-check: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=452$, and t... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{452x}$.\nBoth methods force linearity and use $f(1)=452$ to identify the slope. The extra datum $f(\\frac{1}{5})=\\frac{452}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude ... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=452$ fixes the function to $f(x)=452x$. (Here the result is $\boxed{452x}$.) |
math-016035 | Functional Equations: Additivity — Extension from Q to R | 9 | Give a theorem-based solution: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=490$, and that $f\!\left(\... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{490x}$.\nBoth methods force linearity and use $f(1)=490$ to identify the slope. The extra datum $f(2)=980$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=490x$.",
"robu... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=490$ fixes the function to $f(x)=490x$. (Here the result is $\boxed{490x}$.) |
math-016036 | Functional Equations: Additivity — Extension from Q to R | 9 | Be explicit about assumptions: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=13$, and that $f\!\... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{13x}$.\nBoth methods force linearity and use $f(1)=13$ to identify the slope. The extra datum $f(\\frac{-7}{9})=\\frac{-91}{9}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=13x$.",
"r... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=13$ fixes the function to $f(x)=13x$. (Here the result is $\boxed{13x}$.) |
math-016037 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Problem: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-10$, and that $f\!\left(\... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-10x}$.\nBoth methods force linearity and use $f(1)=-10$ to identify the slope. The extra datum $f(\\frac{-8}{11})=\\frac{80}{11}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-10x$.",
"rob... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-10$ fixes the function to $f(x)=-10x$. (Here the result is $\boxed{-10x}$.) |
math-016038 | Functional Equations: Additive Maps — Density Argument | 9 | Give reasoning, not just computation: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=799$, and th... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{799x}$.\nBoth methods force linearity and use $f(1)=799$ to identify the slope. The extra datum $f(\\frac{-5}{6})=\\frac{-3995}{6}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=799x$.",... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=799$ fixes the function to $f(x)=799x$. (Here the result is $\boxed{799x}$.) |
math-016039 | Functional Equations: Additivity — Extension from Q to R | 9 | Question: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=581$, and that $f\!\left(\frac{-15}{4}\r... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{581x}$.\nBoth methods force linearity and use $f(1)=581$ to identify the slope. The extra datum $f(\\frac{-15}{4})=\\frac{-8715}{4}$ is consistent with $f(x)=kx$ and serves as a built-in check. B... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=581$ fixes the function to $f(x)=581x$. (Here the result is $\boxed{581x}$.) |
math-016040 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Solve (and briefly cross-validate): Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{733x}$.\nBoth methods force linearity and use $f(1)=733$ to identify the slope. The extra datum $f(\\frac{-2}{3})=\\frac{-1466}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=733x$.",
"ro... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=733$ fixes the function to $f(x)=733x$. (Here the result is $\boxed{733x}$.) |
math-016041 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Work this out carefully: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=331$, and that $f\!\left(\frac{-... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{331x}$.\nBoth methods force linearity and use $f(1)=331$ to identify the slope. The extra datum $f(\\frac{-12}{5})=\\frac{-3972}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both concl... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=331$ fixes the function to $f(x)=331x$. |
math-016042 | Functional Equations: Additive Maps — Density Argument | 9 | Derive the result step-by-step: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=125... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{125x}$.\nBoth methods force linearity and use $f(1)=125$ to identify the slope. The extra datum $f(\\frac{-14}{11})=\\frac{-1750}{11}$ is consistent with $f(x)=kx$ and serves as a built-in check.... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=125$ fixes the function to $f(x)=125x$. (Here the result is $\boxed{125x}$.) |
math-016043 | Functional Equations: Additivity — Extension from Q to R | 9 | Work this out carefully: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-474$, and... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-474x}$.\nBoth methods force linearity and use $f(1)=-474$ to identify the slope. The extra datum $f(\\frac{-23}{4})=\\frac{5451}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conc... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-474$ fixes the function to $f(x)=-474x$. |
math-016044 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Indicate where a theorem is used: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=671$, and that $f\!\left(\fra... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{671x}$.\nBoth methods force linearity and use $f(1)=671$ to identify the slope. The extra datum $f(\\frac{-23}{2})=\\frac{-15433}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=671x$.",
"... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=671$ fixes the function to $f(x)=671x$. (Here the result is $\boxed{671x}$.) |
math-016045 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Show all reasoning: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-203$, and that $f\!\left(\frac{-2}{3}\righ... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-203x}$.\nBoth methods force linearity and use $f(1)=-203$ to identify the slope. The extra datum $f(\\frac{-2}{3})=\\frac{406}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Bo... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-203$ fixes the function to $f(x)=-203x$. (Here the result is $\boxed{-203x}$.) |
math-016046 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Track units/moduli carefully: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-571$, and that $f\!\left(\... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-571x}$.\nBoth methods force linearity and use $f(1)=-571$ to identify the slope. The extra datum $f(\\frac{6}{5})=\\frac{-3426}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both concl... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-571$ fixes the function to $f(x)=-571x$. |
math-016047 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Proceed methodically: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-710$, and th... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-710x}$.\nBoth methods force linearity and use $f(1)=-710$ to identify the slope. The extra datum $f(\\frac{9}{5})=-1278$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-710x$.",
"robustness_... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-710$ fixes the function to $f(x)=-710x$. (Here the result is $\boxed{-710x}$.) |
math-016048 | Functional Equations: Additivity — Extension from Q to R | 9 | Proceed methodically: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=94$, and that... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{94x}$.\nBoth methods force linearity and use $f(1)=94$ to identify the slope. The extra datum $f(\\frac{-7}{4})=\\frac{-329}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=94x$.",
"... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=94$ fixes the function to $f(x)=94x$. |
math-016049 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Work carefully and justify each inference: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-551$, and that $f\!... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-551x}$.\nBoth methods force linearity and use $f(1)=-551$ to identify the slope. The extra datum $f(\\frac{7}{5})=\\frac{-3857}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-551x$.",
"... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-551$ fixes the function to $f(x)=-551x$. (Here the result is $\boxed{-551x}$.) |
math-016050 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Solve and then verify: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=562$, and that $f\!\left(\frac{24}... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{562x}$.\nBoth methods force linearity and use $f(1)=562$ to identify the slope. The extra datum $f(\\frac{24}{5})=\\frac{13488}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclu... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=562$ fixes the function to $f(x)=562x$. |
math-016051 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Solve and then verify: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=572$, and that $f\!\left(\frac{10}... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{572x}$.\nBoth methods force linearity and use $f(1)=572$ to identify the slope. The extra datum $f(\\frac{5}{4})=715$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=572x$.",
"robustness_analy... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=572$ fixes the function to $f(x)=572x$. (Here the result is $\boxed{572x}$.) |
math-016052 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Solve and justify each step: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=558$, and that $f\!\left(\fr... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{558x}$.\nBoth methods force linearity and use $f(1)=558$ to identify the slope. The extra datum $f(\\frac{11}{4})=\\frac{3069}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=558x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=558$ fixes the function to $f(x)=558x$. (Here the result is $\boxed{558x}$.) |
math-016053 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Question: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=247$, and that $f\!\left(\frac{13}{8}\right)=\f... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{247x}$.\nBoth methods force linearity and use $f(1)=247$ to identify the slope. The extra datum $f(\\frac{13}{8})=\\frac{3211}{8}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclud... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=247$ fixes the function to $f(x)=247x$. |
math-016054 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Give reasoning, not just computation: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=3$, and that $f\!\l... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{3x}$.\nBoth methods force linearity and use $f(1)=3$ to identify the slope. The extra datum $f(-2)=-6$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=3x$.",
"robustness_analysis": "If the pro... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=3$ fixes the function to $f(x)=3x$. |
math-016055 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Keep the final answer in boxed form: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-400$, and that $f\!... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-400x}$.\nBoth methods force linearity and use $f(1)=-400$ to identify the slope. The extra datum $f(\\frac{-21}{10})=840$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-400x$.",
"robu... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-400$ fixes the function to $f(x)=-400x$. |
math-016056 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Do not skip justification steps: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-153$, and that $f\!\lef... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-153x}$.\nBoth methods force linearity and use $f(1)=-153$ to identify the slope. The extra datum $f(\\frac{-21}{4})=\\frac{3213}{4}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-153x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-153$ fixes the function to $f(x)=-153x$. (Here the result is $\boxed{-153x}$.) |
math-016057 | Functional Equations: Additivity — Extension from Q to R | 9 | Task: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=73$, and that $f\!\left(\frac{-17}{6}\right)=\frac{... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{73x}$.\nBoth methods force linearity and use $f(1)=73$ to identify the slope. The extra datum $f(\\frac{-17}{6})=\\frac{-1241}{6}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclud... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=73$ fixes the function to $f(x)=73x$. |
math-016058 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Give a fully justified solution: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=544$, and that $f... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{544x}$.\nBoth methods force linearity and use $f(1)=544$ to identify the slope. The extra datum $f(\\frac{7}{3})=\\frac{3808}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=544$ fixes the function to $f(x)=544x$. (Here the result is $\boxed{544x}$.) |
math-016059 | Functional Equations: Additive Maps — Density Argument | 9 | Track units/moduli carefully: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-118$, and that $f\!... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-118x}$.\nBoth methods force linearity and use $f(1)=-118$ to identify the slope. The extra datum $f(\\frac{4}{9})=\\frac{-472}{9}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclu... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-118$ fixes the function to $f(x)=-118x$. (Here the result is $\boxed{-118x}$.) |
math-016060 | Functional Equations: Additivity — Extension from Q to R | 9 | State any required conditions first: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=642$, and tha... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{642x}$.\nBoth methods force linearity and use $f(1)=642$ to identify the slope. The extra datum $f(-5)=-3210$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=642x$.",
"r... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=642$ fixes the function to $f(x)=642x$. |
math-016061 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Work this out carefully: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-247$, and that $f\!\left(\frac{... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-247x}$.\nBoth methods force linearity and use $f(1)=-247$ to identify the slope. The extra datum $f(\\frac{-19}{9})=\\frac{4693}{9}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-247x$... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-247$ fixes the function to $f(x)=-247x$. (Here the result is $\boxed{-247x}$.) |
math-016062 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Provide both a computational and a conceptual explanation: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-251x}$.\nBoth methods force linearity and use $f(1)=-251$ to identify the slope. The extra datum $f(\\frac{2}{3})=\\frac{-502}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-251x$."... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-251$ fixes the function to $f(x)=-251x$. |
math-016063 | Functional Equations: Additivity — Extension from Q to R | 9 | Task: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-492$, and that $f\!\left(\frac{-23}{8}\righ... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-492x}$.\nBoth methods force linearity and use $f(1)=-492$ to identify the slope. The extra datum $f(\\frac{-23}{8})=\\frac{2829}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-492x$... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-492$ fixes the function to $f(x)=-492x$. |
math-016064 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Where appropriate, name the theorem you use: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-521$... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-521x}$.\nBoth methods force linearity and use $f(1)=-521$ to identify the slope. The extra datum $f(\\frac{-17}{12})=\\frac{8857}{12}$ is consistent with $f(x)=kx$ and serves as a built-in check... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-521$ fixes the function to $f(x)=-521x$. (Here the result is $\boxed{-521x}$.) |
math-016065 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Track units/moduli carefully: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=762$, and that $f\!\left(\f... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{762x}$.\nBoth methods force linearity and use $f(1)=762$ to identify the slope. The extra datum $f(\\frac{-5}{6})=-635$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=762... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=762$ fixes the function to $f(x)=762x$. (Here the result is $\boxed{762x}$.) |
math-016066 | Functional Equations: Additive Maps — Density Argument | 9 | Give an answer and a quick verification: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-313$, an... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-313x}$.\nBoth methods force linearity and use $f(1)=-313$ to identify the slope. The extra datum $f(\\frac{-17}{9})=\\frac{5321}{9}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conc... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-313$ fixes the function to $f(x)=-313x$. (Here the result is $\boxed{-313x}$.) |
math-016067 | Functional Equations: Additive Maps — Density Argument | 9 | Keep the final answer in boxed form: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{660x}$.\nBoth methods force linearity and use $f(1)=660$ to identify the slope. The extra datum $f(8)=5280$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=660x$.",
"rob... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=660$ fixes the function to $f(x)=660x$. (Here the result is $\boxed{660x}$.) |
math-016068 | Functional Equations: Additive Maps — Density Argument | 9 | Solve and justify each step: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-177$,... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-177x}$.\nBoth methods force linearity and use $f(1)=-177$ to identify the slope. The extra datum $f(\\frac{4}{3})=-236$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-177$ fixes the function to $f(x)=-177x$. |
math-016069 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Checkpoint: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=565$, and that $f\!\left(\frac{-1}{12}\right)... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{565x}$.\nBoth methods force linearity and use $f(1)=565$ to identify the slope. The extra datum $f(\\frac{-1}{12})=\\frac{-565}{12}$ is consistent with $f(x)=kx$ and serves as a built-in check. B... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=565$ fixes the function to $f(x)=565x$. (Here the result is $\boxed{565x}$.) |
math-016070 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Explain what is being counted/optimized: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=78$, and that $f\!\lef... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{78x}$.\nBoth methods force linearity and use $f(1)=78$ to identify the slope. The extra datum $f(\\frac{-16}{7})=\\frac{-1248}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=78x$.",
"robu... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=78$ fixes the function to $f(x)=78x$. (Here the result is $\boxed{78x}$.) |
math-016071 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Do not skip justification steps: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=177$, and that $f... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{177x}$.\nBoth methods force linearity and use $f(1)=177$ to identify the slope. The extra datum $f(\\frac{10}{3})=590$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=177x$.",
"robustnes... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=177$ fixes the function to $f(x)=177x$. (Here the result is $\boxed{177x}$.) |
math-016072 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Prompt: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=364$, and that $f\!\left(\frac{-8}{8}\righ... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{364x}$.\nBoth methods force linearity and use $f(1)=364$ to identify the slope. The extra datum $f(-1)=-364$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=364x$.",
"robustness_analysis... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=364$ fixes the function to $f(x)=364x$. (Here the result is $\boxed{364x}$.) |
math-016073 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Where appropriate, name the theorem you use: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-350$, and t... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-350x}$.\nBoth methods force linearity and use $f(1)=-350$ to identify the slope. The extra datum $f(-1)=350$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-350x$.",
"robustness_analys... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-350$ fixes the function to $f(x)=-350x$. |
math-016074 | Functional Equations: Additive Maps — Density Argument | 9 | Show all reasoning: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-576$, and that... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-576x}$.\nBoth methods force linearity and use $f(1)=-576$ to identify the slope. The extra datum $f(\\frac{5}{3})=-960$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-5... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-576$ fixes the function to $f(x)=-576x$. (Here the result is $\boxed{-576x}$.) |
math-016075 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Solve and include a self-check: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=612$, and that $f\!\left(... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{612x}$.\nBoth methods force linearity and use $f(1)=612$ to identify the slope. The extra datum $f(-2)=-1224$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=612x$.",
"robustness_analysi... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=612$ fixes the function to $f(x)=612x$. (Here the result is $\boxed{612x}$.) |
math-016076 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Do not skip justification steps: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=486$, and that $f\!\left... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{486x}$.\nBoth methods force linearity and use $f(1)=486$ to identify the slope. The extra datum $f(\\frac{-7}{6})=-567$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=486x$.",
"robustness_ana... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=486$ fixes the function to $f(x)=486x$. |
math-016077 | Functional Equations: Additivity — Extension from Q to R | 9 | Track units/moduli carefully: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-647$, and that $f\!\left(\frac{7... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-647x}$.\nBoth methods force linearity and use $f(1)=-647$ to identify the slope. The extra datum $f(\\frac{7}{2})=\\frac{-4529}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. B... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-647$ fixes the function to $f(x)=-647x$. (Here the result is $\boxed{-647x}$.) |
math-016078 | Functional Equations: Additive Maps — Density Argument | 9 | Compute the requested quantity: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=677$, and that $f\... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{677x}$.\nBoth methods force linearity and use $f(1)=677$ to identify the slope. The extra datum $f(\\frac{-22}{7})=\\frac{-14894}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. ... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=677$ fixes the function to $f(x)=677x$. |
math-016079 | Functional Equations: Additive Maps — Density Argument | 9 | Solve with verification: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=385$, and that $f\!\left(\frac{2}{11}\... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{385x}$.\nBoth methods force linearity and use $f(1)=385$ to identify the slope. The extra datum $f(\\frac{2}{11})=70$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=385x$.",
"robustness_analy... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=385$ fixes the function to $f(x)=385x$. (Here the result is $\boxed{385x}$.) |
math-016080 | Functional Equations: Additive Maps — Density Argument | 9 | Prompt: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-528$, and that $f\!\left(\frac{2}{8}\right)=-132... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-528x}$.\nBoth methods force linearity and use $f(1)=-528$ to identify the slope. The extra datum $f(\\frac{1}{4})=-132$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-528x$.",
"robust... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-528$ fixes the function to $f(x)=-528x$. |
math-016081 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Use two approaches if possible: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-46... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-469x}$.\nBoth methods force linearity and use $f(1)=-469$ to identify the slope. The extra datum $f(\\frac{-10}{9})=\\frac{4690}{9}$ is consistent with $f(x)=kx$ and serves as a built-in check. ... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-469$ fixes the function to $f(x)=-469x$. (Here the result is $\boxed{-469x}$.) |
math-016082 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Find the exact value: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=456$, and tha... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{456x}$.\nBoth methods force linearity and use $f(1)=456$ to identify the slope. The extra datum $f(\\frac{-7}{2})=-1596$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=45... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=456$ fixes the function to $f(x)=456x$. (Here the result is $\boxed{456x}$.) |
math-016083 | Functional Equations: Additive Maps — Density Argument | 9 | Keep the final answer in boxed form: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{800x}$.\nBoth methods force linearity and use $f(1)=800$ to identify the slope. The extra datum $f(\\frac{-5}{3})=\\frac{-4000}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=800x$.",
"ro... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=800$ fixes the function to $f(x)=800x$. |
math-016084 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Make each step logically reversible (or explain if not): Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-321x}$.\nBoth methods force linearity and use $f(1)=-321$ to identify the slope. The extra datum $f(\\frac{5}{3})=-535$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-321$ fixes the function to $f(x)=-321x$. (Here the result is $\boxed{-321x}$.) |
math-016085 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Solve with verification: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=90$, and that $f\!\left(\... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{90x}$.\nBoth methods force linearity and use $f(1)=90$ to identify the slope. The extra datum $f(\\frac{3}{7})=\\frac{270}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both co... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=90$ fixes the function to $f(x)=90x$. (Here the result is $\boxed{90x}$.) |
math-016086 | Functional Equations: Additivity — Extension from Q to R | 9 | Keep the final answer in boxed form: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-681$, and that $f\!\left(... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-681x}$.\nBoth methods force linearity and use $f(1)=-681$ to identify the slope. The extra datum $f(2)=-1362$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-68... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-681$ fixes the function to $f(x)=-681x$. |
math-016087 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Work this out carefully: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=697$, and that $f\!\left(... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{697x}$.\nBoth methods force linearity and use $f(1)=697$ to identify the slope. The extra datum $f(\\frac{-23}{6})=\\frac{-16031}{6}$ is consistent with $f(x)=kx$ and serves as a built-in check. ... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=697$ fixes the function to $f(x)=697x$. |
math-016088 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Do not skip justification steps: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-7... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-790x}$.\nBoth methods force linearity and use $f(1)=-790$ to identify the slope. The extra datum $f(-6)=4740$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-790x$.",
"robustness_analysis": ... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-790$ fixes the function to $f(x)=-790x$. (Here the result is $\boxed{-790x}$.) |
math-016089 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Warm-up: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=618$, and that $f\!\left(\frac{10}{10}\right)=618$.
(... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{618x}$.\nBoth methods force linearity and use $f(1)=618$ to identify the slope. The extra datum $f(1)=618$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=618x$.",
"robu... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=618$ fixes the function to $f(x)=618x$. |
math-016090 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Explain why your operations are valid: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{431x}$.\nBoth methods force linearity and use $f(1)=431$ to identify the slope. The extra datum $f(\\frac{3}{10})=\\frac{1293}{10}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=431x$.",... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=431$ fixes the function to $f(x)=431x$. |
math-016091 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Prompt: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-727$, and that $f\!\left(\frac{10}{8}\right)=\frac{-36... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-727x}$.\nBoth methods force linearity and use $f(1)=-727$ to identify the slope. The extra datum $f(\\frac{5}{4})=\\frac{-3635}{4}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-727x$.... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-727$ fixes the function to $f(x)=-727x$. (Here the result is $\boxed{-727x}$.) |
math-016092 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Give a fully justified solution: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-326$, and that $f\!\lef... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-326x}$.\nBoth methods force linearity and use $f(1)=-326$ to identify the slope. The extra datum $f(\\frac{5}{9})=\\frac{-1630}{9}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both concl... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-326$ fixes the function to $f(x)=-326x$. (Here the result is $\boxed{-326x}$.) |
math-016093 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Be explicit about assumptions: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=743$... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{743x}$.\nBoth methods force linearity and use $f(1)=743$ to identify the slope. The extra datum $f(\\frac{1}{2})=\\frac{743}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=743x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=743$ fixes the function to $f(x)=743x$. (Here the result is $\boxed{743x}$.) |
math-016094 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Solve and sanity-check: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-688$, and ... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-688x}$.\nBoth methods force linearity and use $f(1)=-688$ to identify the slope. The extra datum $f(\\frac{-17}{8})=1462$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-688$ fixes the function to $f(x)=-688x$. |
math-016095 | Functional Equations: Additivity — Extension from Q to R | 9 | Explain why your operations are valid: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=467$, and that $f\... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{467x}$.\nBoth methods force linearity and use $f(1)=467$ to identify the slope. The extra datum $f(\\frac{-24}{7})=\\frac{-11208}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=467x$.",
"... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=467$ fixes the function to $f(x)=467x$. (Here the result is $\boxed{467x}$.) |
math-016096 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Answer with a short justification: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-318x}$.\nBoth methods force linearity and use $f(1)=-318$ to identify the slope. The extra datum $f(6)=-1908$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-318x$.",
"robustness_analy... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-318$ fixes the function to $f(x)=-318x$. (Here the result is $\boxed{-318x}$.) |
math-016097 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Give a fully justified solution: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-70$, and that $f... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-70x}$.\nBoth methods force linearity and use $f(1)=-70$ to identify the slope. The extra datum $f(\\frac{-11}{6})=\\frac{385}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-70x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-70$ fixes the function to $f(x)=-70x$. |
math-016098 | Functional Equations: Regularity Assumptions — Why Needed | 9 | State any required conditions first: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-501$, and th... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-501x}$.\nBoth methods force linearity and use $f(1)=-501$ to identify the slope. The extra datum $f(\\frac{4}{5})=\\frac{-2004}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-501x$.",
"... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-501$ fixes the function to $f(x)=-501x$. (Here the result is $\boxed{-501x}$.) |
math-016099 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Explain what is being counted/optimized: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=90$, and ... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{90x}$.\nBoth methods force linearity and use $f(1)=90$ to identify the slope. The extra datum $f(11)=990$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=90x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=90$ fixes the function to $f(x)=90x$. |
math-016100 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Use two approaches if possible: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=277$, and that $f\... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{277x}$.\nBoth methods force linearity and use $f(1)=277$ to identify the slope. The extra datum $f(\\frac{-8}{3})=\\frac{-2216}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=277x$.",
"ro... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=277$ fixes the function to $f(x)=277x$. |
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