id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-018301 | Functional Equations: Regularity Assumptions — Why Needed | 10 | 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 $f(1)=281$, 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": "The two methods are consistent and must coincide. Final answer: $\\boxed{281x}$.\nBoth methods force linearity and use $f(1)=281$ to identify the slope. The extra datum $f(\\frac{-7}{4})=\\frac{-1967}{4}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=281x$.",... | [
{
"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 $f(1)=281$ fixes the function to $f(x)=281x$. (Here the result is $\boxed{281x}$.) |
math-018302 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | Checkpoint: 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 $f(1)=193$, 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{193x}$.\nBoth methods force linearity and use $f(1)=193$ to identify the slope. The extra datum $f(\\frac{-17}{4})=\\frac{-3281}{4}$ 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... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=193$ fixes the function to $f(x)=193x$. (Here the result is $\boxed{193x}$.) |
math-018303 | Functional Equations: Additivity — Extension from Q to R | 10 | 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 $f(1)=335$, and that $f\!\left(\frac{5}{4}\right)=\frac{1675... | [
{
"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: $\\boxed{335x}$.\nBoth methods force linearity and use $f(1)=335$ to identify the slope. The extra datum $f(\\frac{5}{4})=\\frac{1675}{4}$ 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 $f(1)=335$ fixes the function to $f(x)=335x$. |
math-018304 | Functional Equations: Additive Maps — Density Argument | 10 | Explain each transformation: 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 $f(1)=121$, ... | [
{
"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: $\\boxed{121x}$.\nBoth methods force linearity and use $f(1)=121$ to identify the slope. The extra datum $f(-2)=-242$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=121x$.",
"robustness_analysis": "If... | [
{
"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 $f(x)=121x$. (Here the result is $\boxed{121x}$.) |
math-018305 | Functional Equations: Additive Maps — Density Argument | 10 | Solve and sanity-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 $f(1)=114$, and that $f\!\left(\frac{-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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{114x}$.\nBoth methods force linearity and use $f(1)=114$ to identify the slope. The extra datum $f(\\frac{-5}{4})=\\frac{-285}{2}$ 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 $f(1)=114$ fixes the function to $f(x)=114x$. |
math-018306 | Functional Equations: Additive Maps — Density Argument | 10 | Make each step logically reversible (or explain if not): 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 continuo... | [
{
"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: $\\boxed{700x}$.\nBoth methods force linearity and use $f(1)=700$ to identify the slope. The extra datum $f(\\frac{-23}{4})=-4025$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=700x$.",
"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 $f(1)=700$ fixes the function to $f(x)=700x$. (Here the result is $\boxed{700x}$.) |
math-018307 | Functional Equations: Additive Maps — Density Argument | 10 | 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 $f(1)=-311$, 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: $\\boxed{-311x}$.\nBoth methods force linearity and use $f(1)=-311$ to identify the slope. The extra datum $f(\\frac{-3}{2})=\\frac{933}{2}$ 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... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-311$ fixes the function to $f(x)=-311x$. (Here the result is $\boxed{-311x}$.) |
math-018308 | Functional Equations: Cauchy — Continuity Implies Linearity | 10 | 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 $f(1)=-281$, 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: $\\boxed{-281x}$.\nBoth methods force linearity and use $f(1)=-281$ to identify the slope. The extra datum $f(-3)=843$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-281x$.",
"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... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-281$ fixes the function to $f(x)=-281x$. |
math-018309 | Functional Equations: Regularity Assumptions — Why Needed | 10 | Start by stating any domain restrictions: 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 $f(1)=-247$, 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: $\\boxed{-247x}$.\nBoth methods force linearity and use $f(1)=-247$ to identify the slope. The extra datum $f(\\frac{4}{5})=\\frac{-988}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-247x$.",
"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 $f(1)=-247$ fixes the function to $f(x)=-247x$. |
math-018310 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | Carefully track domains: 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 $f(1)=667$, 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: $\\boxed{667x}$.\nBoth methods force linearity and use $f(1)=667$ to identify the slope. The extra datum $f(\\frac{-11}{5})=\\frac{-7337}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=667x$.",
"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 $f(1)=667$ fixes the function to $f(x)=667x$. (Here the result is $\boxed{667x}$.) |
math-018311 | Functional Equations: Additivity — Extension from Q to R | 10 | Compute the requested quantity: 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 $f(1)=526$, 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{526x}$.\nBoth methods force linearity and use $f(1)=526$ to identify the slope. The extra datum $f(\\frac{21}{5})=\\frac{11046}{5}$ 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 $f(1)=526$ fixes the function to $f(x)=526x$. (Here the result is $\boxed{526x}$.) |
math-018312 | Functional Equations: Additivity — Extension from Q to R | 10 | 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 $f(1)=-297$, 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: $\\boxed{-297x}$.\nBoth methods force linearity and use $f(1)=-297$ to identify the slope. The extra datum $f(\\frac{23}{6})=\\frac{-2277}{2}$ 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... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-297$ fixes the function to $f(x)=-297x$. |
math-018313 | Functional Equations: Cauchy — Continuity Implies Linearity | 10 | Make each step logically reversible (or explain if not): 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": "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: $\\boxed{-148x}$.\nBoth methods force linearity and use $f(1)=-148$ to identify the slope. The extra datum $f(\\frac{-23}{6})=\\frac{1702}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-148x$.",
... | [
{
"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 $f(1)=-148$ fixes the function to $f(x)=-148x$. (Here the result is $\boxed{-148x}$.) |
math-018314 | Functional Equations: Regularity Assumptions — Why Needed | 10 | Be explicit about assumptions: 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 $f(1)=-156$, 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: $\\boxed{-156x}$.\nBoth methods force linearity and use $f(1)=-156$ to identify the slope. The extra datum $f(\\frac{-3}{7})=\\frac{468}{7}$ 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 $f(1)=-156$ fixes the function to $f(x)=-156x$. |
math-018315 | Functional Equations: Additive Maps — Density Argument | 10 | Explain each transformation: 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 $f(1)=-77$, ... | [
{
"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: $\\boxed{-77x}$.\nBoth methods force linearity and use $f(1)=-77$ to identify the slope. The extra datum $f(\\frac{3}{4})=\\frac{-231}{4}$ 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... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-77$ fixes the function to $f(x)=-77x$. |
math-018316 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | 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 $f(1)=178$, 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: $\\boxed{178x}$.\nBoth methods force linearity and use $f(1)=178$ to identify the slope. The extra datum $f(\\frac{13}{12})=\\frac{1157}{6}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=178x$.",... | [
{
"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 $f(1)=178$ fixes the function to $f(x)=178x$. |
math-018317 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | Solve and sanity-check: 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 $f(1)=106$, 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: $\\boxed{106x}$.\nBoth methods force linearity and use $f(1)=106$ to identify the slope. The extra datum $f(4)=424$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=106x$.",
"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... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=106$ fixes the function to $f(x)=106x$. (Here the result is $\boxed{106x}$.) |
math-018318 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | 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 $f(1)=605$, 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": "Both approaches agree after simplification. Final answer: $\\boxed{605x}$.\nBoth methods force linearity and use $f(1)=605$ to identify the slope. The extra datum $f(\\frac{-17}{3})=\\frac{-10285}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=605x$.",
"... | [
{
"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 $f(1)=605$ fixes the function to $f(x)=605x$. |
math-018319 | Functional Equations: Cauchy — Continuity Implies Linearity | 10 | Answer with a short justification: 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 $f(1)=127$, 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{127x}$.\nBoth methods force linearity and use $f(1)=127$ to identify the slope. The extra datum $f(\\frac{-21}{10})=\\frac{-2667}{10}$ 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 $f(1)=127$ fixes the function to $f(x)=127x$. (Here the result is $\boxed{127x}$.) |
math-018320 | Functional Equations: Additivity — Extension from Q to R | 10 | Start by stating any domain restrictions: 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 $f(1)=-47$, 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": "The two methods are consistent and must coincide. Final answer: $\\boxed{-47x}$.\nBoth methods force linearity and use $f(1)=-47$ to identify the slope. The extra datum $f(\\frac{-11}{6})=\\frac{517}{6}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-47x$.",
... | [
{
"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 $f(1)=-47$ fixes the function to $f(x)=-47x$. |
math-018321 | Functional Equations: Additive Maps — Density Argument | 10 | Provide both a computational and a conceptual explanation: 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 $f(1)=664$... | [
{
"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: $\\boxed{664x}$.\nBoth methods force linearity and use $f(1)=664$ to identify the slope. The extra datum $f(\\frac{1}{4})=166$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=664x$.",
"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 $f(1)=664$ fixes the function to $f(x)=664x$. (Here the result is $\boxed{664x}$.) |
math-018322 | Functional Equations: Additive Maps — Density Argument | 10 | Compute the requested quantity: 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 $f(1)=7$, 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": "The two methods are consistent and must coincide. Final answer: $\\boxed{7x}$.\nBoth methods force linearity and use $f(1)=7$ to identify the slope. The extra datum $f(3)=21$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=7x$.",
"robustness_analysis": "Gener... | [
{
"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 $f(1)=7$ fixes the function to $f(x)=7x$. (Here the result is $\boxed{7x}$.) |
math-018323 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | 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 $f(1)=-679$, 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-679x}$.\nBoth methods force linearity and use $f(1)=-679$ to identify the slope. The extra datum $f(\\frac{15}{4})=\\frac{-10185}{4}$ 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 $f(1)=-679$ fixes the function to $f(x)=-679x$. |
math-018324 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | Complete the analysis: 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 $f(1)=-306$, 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: $\\boxed{-306x}$.\nBoth methods force linearity and use $f(1)=-306$ to identify the slope. The extra datum $f(\\frac{-8}{9})=272$ 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 $f(1)=-306$ fixes the function to $f(x)=-306x$. (Here the result is $\boxed{-306x}$.) |
math-018325 | Functional Equations: Additive Maps — Density Argument | 10 | 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": "Both approaches agree after simplification. Final answer: $\\boxed{-259x}$.\nBoth methods force linearity and use $f(1)=-259$ to identify the slope. The extra datum $f(\\frac{-5}{12})=\\frac{1295}{12}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-259x$.",
... | [
{
"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 $f(1)=-259$ fixes the function to $f(x)=-259x$. (Here the result is $\boxed{-259x}$.) |
math-018326 | Functional Equations: Additive Maps — Density Argument | 10 | Explain each transformation: 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 $f(1)=-220$,... | [
{
"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: $\\boxed{-220x}$.\nBoth methods force linearity and use $f(1)=-220$ to identify the slope. The extra datum $f(\\frac{-2}{11})=40$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-2... | [
{
"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 $f(1)=-220$ fixes the function to $f(x)=-220x$. |
math-018327 | Functional Equations: Cauchy — Continuity Implies Linearity | 10 | Solve and include a self-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)=-44... | [
{
"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: $\\boxed{-446x}$.\nBoth methods force linearity and use $f(1)=-446$ to identify the slope. The extra datum $f(\\frac{3}{2})=-669$ 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... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-446$ fixes the function to $f(x)=-446x$. |
math-018328 | Functional Equations: Additivity — Extension from Q to R | 10 | Checkpoint: 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 $f(1)=611$, 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: $\\boxed{611x}$.\nBoth methods force linearity and use $f(1)=611$ to identify the slope. The extra datum $f(\\frac{16}{7})=\\frac{9776}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=611x$.",
"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 $f(1)=611$ fixes the function to $f(x)=611x$. |
math-018329 | Functional Equations: Additive Maps — Density Argument | 10 | Give a theorem-based 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 $f(1)=349$, 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: $\\boxed{349x}$.\nBoth methods force linearity and use $f(1)=349$ to identify the slope. The extra datum $f(\\frac{13}{4})=\\frac{4537}{4}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=349x$.",
... | [
{
"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 $f(1)=349$ fixes the function to $f(x)=349x$. |
math-018330 | Functional Equations: Regularity Assumptions — Why Needed | 10 | Solve with verification: 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 $f(1)=602$, 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{602x}$.\nBoth methods force linearity and use $f(1)=602$ to identify the slope. The extra datum $f(-1)=-602$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=602x$.",
"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... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=602$ fixes the function to $f(x)=602x$. (Here the result is $\boxed{602x}$.) |
math-018331 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | 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 $f(1)=230$, 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: $\\boxed{230x}$.\nBoth methods force linearity and use $f(1)=230$ to identify the slope. The extra datum $f(\\frac{4}{11})=\\frac{920}{11}$ 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... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=230$ fixes the function to $f(x)=230x$. |
math-018332 | Functional Equations: Additive Maps — Density Argument | 10 | 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 $f(1)=-208$, 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: $\\boxed{-208x}$.\nBoth methods force linearity and use $f(1)=-208$ to identify the slope. The extra datum $f(\\frac{11}{9})=\\frac{-2288}{9}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-208x$... | [
{
"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 $f(1)=-208$ fixes the function to $f(x)=-208x$. (Here the result is $\boxed{-208x}$.) |
math-018333 | Functional Equations: Additivity — Extension from Q to R | 10 | 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 $f(1)=630$, 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{630x}$.\nBoth methods force linearity and use $f(1)=630$ to identify the slope. The extra datum $f(\\frac{-8}{9})=-560$ 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 $f(1)=630$ fixes the function to $f(x)=630x$. |
math-018334 | Functional Equations: Regularity Assumptions — Why Needed | 10 | Warm-up: 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 $f(1)=690$, and that $f\!\left(\frac{-18}{2}\right)=-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": "The two methods are consistent and must coincide. Final answer: $\\boxed{690x}$.\nBoth methods force linearity and use $f(1)=690$ to identify the slope. The extra datum $f(-9)=-6210$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=690x$.",
"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 $f(1)=690$ fixes the function to $f(x)=690x$. |
math-018335 | Functional Equations: Cauchy — Continuity Implies Linearity | 10 | 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 $f(1)=427$, 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: $\\boxed{427x}$.\nBoth methods force linearity and use $f(1)=427$ to identify the slope. The extra datum $f(\\frac{6}{7})=366$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=427x$.",
"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... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=427$ fixes the function to $f(x)=427x$. (Here the result is $\boxed{427x}$.) |
math-018336 | Functional Equations: Additivity — Extension from Q to R | 10 | Write the solution set clearly: 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 $f(1)=342$, 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": "Both approaches agree after simplification. Final answer: $\\boxed{342x}$.\nBoth methods force linearity and use $f(1)=342$ to identify the slope. The extra datum $f(\\frac{-5}{8})=\\frac{-855}{4}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=342x$.",
"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... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=342$ fixes the function to $f(x)=342x$. |
math-018337 | Functional Equations: Additive Maps — Density Argument | 10 | 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 $f(1)=-535$, 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: $\\boxed{-535x}$.\nBoth methods force linearity and use $f(1)=-535$ to identify the slope. The extra datum $f(-2)=1070$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-53... | [
{
"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 $f(1)=-535$ fixes the function to $f(x)=-535x$. (Here the result is $\boxed{-535x}$.) |
math-018338 | Functional Equations: Additivity — Extension from Q to R | 10 | 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 $f(1)=289$, 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: $\\boxed{289x}$.\nBoth methods force linearity and use $f(1)=289$ to identify the slope. The extra datum $f(\\frac{-21}{11})=\\frac{-6069}{11}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=289x$.",
... | [
{
"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 $f(1)=289$ fixes the function to $f(x)=289x$. |
math-018339 | Functional Equations: Regularity Assumptions — Why Needed | 10 | Provide both a computational and a conceptual explanation: 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 $f(1)=490$... | [
{
"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: $\\boxed{490x}$.\nBoth methods force linearity and use $f(1)=490$ to identify the slope. The extra datum $f(\\frac{11}{3})=\\frac{5390}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=490x$.",
... | [
{
"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 $f(1)=490$ fixes the function to $f(x)=490x$. |
math-018340 | Functional Equations: Cauchy — Continuity Implies Linearity | 10 | 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 $f(1)=429$, 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: $\\boxed{429x}$.\nBoth methods force linearity and use $f(1)=429$ to identify the slope. The extra datum $f(\\frac{7}{2})=\\frac{3003}{2}$ 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... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=429$ fixes the function to $f(x)=429x$. (Here the result is $\boxed{429x}$.) |
math-018341 | Functional Equations: Regularity Assumptions — Why Needed | 10 | 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 $f(1)=282$, 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: $\\boxed{282x}$.\nBoth methods force linearity and use $f(1)=282$ to identify the slope. The extra datum $f(\\frac{-22}{9})=\\frac{-2068}{3}$ 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... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=282$ fixes the function to $f(x)=282x$. |
math-018342 | Functional Equations: Additive Maps — Density Argument | 10 | Find the exact value: 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 $f(1)=453$, and that $f\!\left(\frac{25}{9}\rig... | [
{
"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: $\\boxed{453x}$.\nBoth methods force linearity and use $f(1)=453$ to identify the slope. The extra datum $f(\\frac{25}{9})=\\frac{3775}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=453x$.",
... | [
{
"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 $f(1)=453$ fixes the function to $f(x)=453x$. (Here the result is $\boxed{453x}$.) |
math-018343 | Functional Equations: Cauchy — Continuity Implies Linearity | 10 | 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 $f(1)=-793$, 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: $\\boxed{-793x}$.\nBoth methods force linearity and use $f(1)=-793$ to identify the slope. The extra datum $f(\\frac{21}{2})=\\frac{-16653}{2}$ 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 $f(1)=-793$ fixes the function to $f(x)=-793x$. |
math-018344 | Functional Equations: Additivity — Extension from Q to R | 10 | 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 $f(1)=-700$, 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: $\\boxed{-700x}$.\nBoth methods force linearity and use $f(1)=-700$ to identify the slope. The extra datum $f(\\frac{-23}{12})=\\frac{4025}{3}$ 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 $f(1)=-700$ fixes the function to $f(x)=-700x$. (Here the result is $\boxed{-700x}$.) |
math-018345 | Functional Equations: Additivity — Extension from Q to R | 10 | 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 $f(1)=-789$, and that $f\!\left(\frac{3... | [
{
"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: $\\boxed{-789x}$.\nBoth methods force linearity and use $f(1)=-789$ to identify the slope. The extra datum $f(1)=-789$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-789x$.",
"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... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-789$ fixes the function to $f(x)=-789x$. (Here the result is $\boxed{-789x}$.) |
math-018346 | Functional Equations: Additivity — Extension from Q to R | 10 | Find the exact value: 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 $f(1)=-275$, and that $f\!\left(\frac{-7}{10}\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": "The two methods are consistent and must coincide. Final answer: $\\boxed{-275x}$.\nBoth methods force linearity and use $f(1)=-275$ to identify the slope. The extra datum $f(\\frac{-7}{10})=\\frac{385}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-275x$.... | [
{
"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 $f(1)=-275$ fixes the function to $f(x)=-275x$. (Here the result is $\boxed{-275x}$.) |
math-018347 | Functional Equations: Cauchy — Continuity Implies Linearity | 10 | 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 $f(1)=785$, 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": "Both approaches agree after simplification. Final answer: $\\boxed{785x}$.\nBoth methods force linearity and use $f(1)=785$ to identify the slope. The extra datum $f(\\frac{20}{7})=\\frac{15700}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=785x$.",
"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 $f(1)=785$ fixes the function to $f(x)=785x$. |
math-018348 | Functional Equations: Additive Maps — Density Argument | 10 | Indicate where a theorem is used: 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 $f(1)=555$, 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{555x}$.\nBoth methods force linearity and use $f(1)=555$ to identify the slope. The extra datum $f(\\frac{25}{2})=\\frac{13875}{2}$ 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 $f(1)=555$ fixes the function to $f(x)=555x$. (Here the result is $\boxed{555x}$.) |
math-018349 | Functional Equations: Additivity — Extension from Q to R | 10 | Show all reasoning: 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 $f(1)=575$, 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": "Both approaches agree after simplification. Final answer: $\\boxed{575x}$.\nBoth methods force linearity and use $f(1)=575$ to identify the slope. The extra datum $f(\\frac{-3}{2})=\\frac{-1725}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=575x$.",
"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 $f(1)=575$ fixes the function to $f(x)=575x$. |
math-018350 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | Answer with a short justification: 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 $f(1)=206$, 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: $\\boxed{206x}$.\nBoth methods force linearity and use $f(1)=206$ to identify the slope. The extra datum $f(\\frac{1}{5})=\\frac{206}{5}$ 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... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=206$ fixes the function to $f(x)=206x$. |
math-018351 | Functional Equations: Additivity — Extension from Q to R | 10 | 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 $f(1)=280$, 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: $\\boxed{280x}$.\nBoth methods force linearity and use $f(1)=280$ to identify the slope. The extra datum $f(\\frac{9}{5})=504$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=280x$... | [
{
"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 $f(1)=280$ fixes the function to $f(x)=280x$. |
math-018352 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | Give a theorem-based solution: 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)=-460... | [
{
"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: $\\boxed{-460x}$.\nBoth methods force linearity and use $f(1)=-460$ to identify the slope. The extra datum $f(\\frac{-15}{11})=\\frac{6900}{11}$ 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... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-460$ fixes the function to $f(x)=-460x$. (Here the result is $\boxed{-460x}$.) |
math-018353 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | 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 $f(1)=420$, 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: $\\boxed{420x}$.\nBoth methods force linearity and use $f(1)=420$ to identify the slope. The extra datum $f(\\frac{-3}{2})=-630$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=420... | [
{
"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 $f(1)=420$ fixes the function to $f(x)=420x$. (Here the result is $\boxed{420x}$.) |
math-018354 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | 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 $f(1)=605$, 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{605x}$.\nBoth methods force linearity and use $f(1)=605$ to identify the slope. The extra datum $f(\\frac{-1}{10})=\\frac{-121}{2}$ 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 $f(1)=605$ fixes the function to $f(x)=605x$. |
math-018355 | Functional Equations: Cauchy — Continuity Implies Linearity | 10 | 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 $f(1)=-450$, 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: $\\boxed{-450x}$.\nBoth methods force linearity and use $f(1)=-450$ to identify the slope. The extra datum $f(\\frac{23}{2})=-5175$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-450x$.",
"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... | Core principle: 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-018356 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | 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 $f(1)=-153$, 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": "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 $f(2)=-306$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-153x$.",
"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... | Remember: 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 $f(x)=-153x$. |
math-018357 | Functional Equations: Additivity — Extension from Q to R | 10 | Give a fully justified solution: 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)=60... | [
{
"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: $\\boxed{602x}$.\nBoth methods force linearity and use $f(1)=602$ to identify the slope. The extra datum $f(\\frac{4}{3})=\\frac{2408}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=602x$.",
"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 $f(1)=602$ fixes the function to $f(x)=602x$. |
math-018358 | Functional Equations: Regularity Assumptions — Why Needed | 10 | Warm-up: 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 $f(1)=487$, and that $f\!\left(\frac{-10}{10}\right)=-... | [
{
"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: $\\boxed{487x}$.\nBoth methods force linearity and use $f(1)=487$ to identify the slope. The extra datum $f(-1)=-487$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=487x$.",
"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... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=487$ fixes the function to $f(x)=487x$. |
math-018359 | Functional Equations: Cauchy — Continuity Implies Linearity | 10 | Provide both a computational and a conceptual explanation: 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 $0$, 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{493x}$.\nBoth methods force linearity and use $f(1)=493$ to identify the slope. The extra datum $f(\\frac{-1}{2})=\\frac{-493}{2}$ 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 $f(1)=493$ fixes the function to $f(x)=493x$. |
math-018360 | Functional Equations: Cauchy — Continuity Implies Linearity | 10 | 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 $f(1)=565$, and that $f\!\left(\frac{11}{3}\right)=\frac{621... | [
{
"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: $\\boxed{565x}$.\nBoth methods force linearity and use $f(1)=565$ to identify the slope. The extra datum $f(\\frac{11}{3})=\\frac{6215}{3}$ 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... | Takeaway: 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-018361 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | Start by stating any domain restrictions: 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 $f(1)=59$, 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: $\\boxed{59x}$.\nBoth methods force linearity and use $f(1)=59$ to identify the slope. The extra datum $f(\\frac{1}{2})=\\frac{59}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(... | [
{
"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 $f(1)=59$ fixes the function to $f(x)=59x$. (Here the result is $\boxed{59x}$.) |
math-018362 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | Indicate where a theorem is used: 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)=3... | [
{
"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: $\\boxed{389x}$.\nBoth methods force linearity and use $f(1)=389$ to identify the slope. The extra datum $f(\\frac{5}{8})=\\frac{1945}{8}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=389x$.",
... | [
{
"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 $f(1)=389$ fixes the function to $f(x)=389x$. (Here the result is $\boxed{389x}$.) |
math-018363 | Functional Equations: Additive Maps — Density Argument | 10 | 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 $f(1)=-127$, 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: $\\boxed{-127x}$.\nBoth methods force linearity and use $f(1)=-127$ to identify the slope. The extra datum $f(\\frac{11}{12})=\\frac{-1397}{12}$ 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... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-127$ fixes the function to $f(x)=-127x$. |
math-018364 | Functional Equations: Additive Maps — Density Argument | 10 | 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 $f(1)=-384$, 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: $\\boxed{-384x}$.\nBoth methods force linearity and use $f(1)=-384$ to identify the slope. The extra datum $f(\\frac{15}{7})=\\frac{-5760}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-384x$... | [
{
"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 $f(1)=-384$ fixes the function to $f(x)=-384x$. (Here the result is $\boxed{-384x}$.) |
math-018365 | Functional Equations: Cauchy — Continuity Implies Linearity | 10 | Challenge: 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 $f(1)=-547$, and that $f\!\left(\frac{25}{11}\right)=\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: $\\boxed{-547x}$.\nBoth methods force linearity and use $f(1)=-547$ to identify the slope. The extra datum $f(\\frac{25}{11})=\\frac{-13675}{11}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-54... | [
{
"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 $f(1)=-547$ fixes the function to $f(x)=-547x$. |
math-018366 | Functional Equations: Cauchy — Continuity Implies Linearity | 10 | Explain what is being counted/optimized: 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 $f(1)=651$, 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: $\\boxed{651x}$.\nBoth methods force linearity and use $f(1)=651$ to identify the slope. The extra datum $f(\\frac{-12}{11})=\\frac{-7812}{11}$ 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... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=651$ fixes the function to $f(x)=651x$. (Here the result is $\boxed{651x}$.) |
math-018367 | Functional Equations: Cauchy — Continuity Implies Linearity | 10 | 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 $f(1)=325$, 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: $\\boxed{325x}$.\nBoth methods force linearity and use $f(1)=325$ to identify the slope. The extra datum $f(2)=650$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=325x$.",
"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... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=325$ fixes the function to $f(x)=325x$. |
math-018368 | Functional Equations: Additivity — Extension from Q to R | 10 | Solve and include a self-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)=700... | [
{
"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: $\\boxed{700x}$.\nBoth methods force linearity and use $f(1)=700$ to identify the slope. The extra datum $f(\\frac{-22}{7})=-2200$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=700x$.",
"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... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=700$ fixes the function to $f(x)=700x$. (Here the result is $\boxed{700x}$.) |
math-018369 | Functional Equations: Regularity Assumptions — Why Needed | 10 | Start by stating any domain restrictions: 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... | [
{
"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: $\\boxed{-493x}$.\nBoth methods force linearity and use $f(1)=-493$ to identify the slope. The extra datum $f(\\frac{20}{7})=\\frac{-9860}{7}$ 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... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-493$ fixes the function to $f(x)=-493x$. |
math-018370 | Functional Equations: Additivity — Extension from Q to R | 10 | 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 $f(1)=88$, and that $f\!\left(\frac{-5}{9}\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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{88x}$.\nBoth methods force linearity and use $f(1)=88$ to identify the slope. The extra datum $f(\\frac{-5}{9})=\\frac{-440}{9}$ 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... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=88$ fixes the function to $f(x)=88x$. |
math-018371 | Functional Equations: Additive Maps — Density Argument | 10 | 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 $f(1)=120$, 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": "The two methods are consistent and must coincide. Final answer: $\\boxed{120x}$.\nBoth methods force linearity and use $f(1)=120$ to identify the slope. The extra datum $f(\\frac{8}{3})=320$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=120x$.",
"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... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=120$ fixes the function to $f(x)=120x$. (Here the result is $\boxed{120x}$.) |
math-018372 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | Solve (and briefly cross-validate): 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 $f(1)=-769$, 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: $\\boxed{-769x}$.\nBoth methods force linearity and use $f(1)=-769$ to identify the slope. The extra datum $f(\\frac{7}{3})=\\frac{-5383}{3}$ 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 $f(1)=-769$ fixes the function to $f(x)=-769x$. (Here the result is $\boxed{-769x}$.) |
math-018373 | Functional Equations: Regularity Assumptions — Why Needed | 10 | Start by stating any domain restrictions: 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... | [
{
"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: $\\boxed{-271x}$.\nBoth methods force linearity and use $f(1)=-271$ to identify the slope. The extra datum $f(\\frac{14}{5})=\\frac{-3794}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-271x$... | [
{
"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 $f(1)=-271$ fixes the function to $f(x)=-271x$. |
math-018374 | Functional Equations: Cauchy — Continuity Implies Linearity | 10 | 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 $f(1)=642$, 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{642x}$.\nBoth methods force linearity and use $f(1)=642$ to identify the slope. The extra datum $f(-2)=-1284$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=642x... | [
{
"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 $f(1)=642$ fixes the function to $f(x)=642x$. |
math-018375 | Functional Equations: Regularity Assumptions — Why Needed | 10 | Provide both a computational and a conceptual explanation: 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)=-334... | [
{
"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: $\\boxed{-334x}$.\nBoth methods force linearity and use $f(1)=-334$ to identify the slope. The extra datum $f(-1)=334$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-334... | [
{
"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 $f(1)=-334$ fixes the function to $f(x)=-334x$. (Here the result is $\boxed{-334x}$.) |
math-018376 | Functional Equations: Additive Maps — Density Argument | 10 | Give a theorem-based solution: 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 $f(1)=251$... | [
{
"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: $\\boxed{251x}$.\nBoth methods force linearity and use $f(1)=251$ to identify the slope. The extra datum $f(\\frac{5}{12})=\\frac{1255}{12}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=251x$.",
"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... | 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 $f(x)=251x$. (Here the result is $\boxed{251x}$.) |
math-018377 | Functional Equations: Cauchy — Continuity Implies Linearity | 10 | 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)=-438... | [
{
"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: $\\boxed{-438x}$.\nBoth methods force linearity and use $f(1)=-438$ to identify the slope. The extra datum $f(\\frac{-19}{4})=\\frac{4161}{2}$ 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... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-438$ fixes the function to $f(x)=-438x$. |
math-018378 | Functional Equations: Additive Maps — Density Argument | 10 | Complete the analysis: 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 $f(1)=516$, and that $f\!\left(\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: $\\boxed{516x}$.\nBoth methods force linearity and use $f(1)=516$ to identify the slope. The extra datum $f(\\frac{-3}{4})=-387$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=516... | [
{
"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 $f(1)=516$ fixes the function to $f(x)=516x$. |
math-018379 | Functional Equations: Cauchy — Continuity Implies Linearity | 10 | Determine the requested value: 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 $f(1)=-198$, 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: $\\boxed{-198x}$.\nBoth methods force linearity and use $f(1)=-198$ to identify the slope. The extra datum $f(\\frac{-17}{5})=\\frac{3366}{5}$ 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 $f(1)=-198$ fixes the function to $f(x)=-198x$. |
math-018380 | Functional Equations: Additivity — Extension from Q to R | 10 | Derive the result step-by-step: 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 $f(1)=758$, 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: $\\boxed{758x}$.\nBoth methods force linearity and use $f(1)=758$ to identify the slope. The extra datum $f(\\frac{22}{7})=\\frac{16676}{7}$ 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... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=758$ fixes the function to $f(x)=758x$. |
math-018381 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | 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 $f(1)=-67$, and that $f\!\left(\frac{-9}{6}\right)=\frac{201... | [
{
"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: $\\boxed{-67x}$.\nBoth methods force linearity and use $f(1)=-67$ to identify the slope. The extra datum $f(\\frac{-3}{2})=\\frac{201}{2}$ 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 $f(1)=-67$ fixes the function to $f(x)=-67x$. (Here the result is $\boxed{-67x}$.) |
math-018382 | Functional Equations: Regularity Assumptions — Why Needed | 10 | 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 $f(1)=182$, 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: $\\boxed{182x}$.\nBoth methods force linearity and use $f(1)=182$ to identify the slope. The extra datum $f(7)=1274$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=182x$.... | [
{
"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 $f(1)=182$ fixes the function to $f(x)=182x$. |
math-018383 | Functional Equations: Additivity — Extension from Q to R | 10 | Solve and sanity-check: 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 $f(1)=-129$, 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: $\\boxed{-129x}$.\nBoth methods force linearity and use $f(1)=-129$ to identify the slope. The extra datum $f(\\frac{13}{3})=-559$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-129x$.",
"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... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-129$ fixes the function to $f(x)=-129x$. (Here the result is $\boxed{-129x}$.) |
math-018384 | Functional Equations: Cauchy — Continuity Implies Linearity | 10 | Solve (and briefly cross-validate): 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 $f(1)=459$, 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": "Both approaches agree after simplification. Final answer: $\\boxed{459x}$.\nBoth methods force linearity and use $f(1)=459$ to identify the slope. The extra datum $f(\\frac{-1}{4})=\\frac{-459}{4}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=459x$.",
"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 $f(1)=459$ fixes the function to $f(x)=459x$. (Here the result is $\boxed{459x}$.) |
math-018385 | Functional Equations: Regularity Assumptions — Why Needed | 10 | Complete the analysis: 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 $f(1)=377$, and that $f\!\left(\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: $\\boxed{377x}$.\nBoth methods force linearity and use $f(1)=377$ to identify the slope. The extra datum $f(2)=754$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=377x$.",
"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... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=377$ fixes the function to $f(x)=377x$. (Here the result is $\boxed{377x}$.) |
math-018386 | Functional Equations: Additive Maps — Density Argument | 10 | Explain each transformation: 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 $f(1)=-658$, 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: $\\boxed{-658x}$.\nBoth methods force linearity and use $f(1)=-658$ to identify the slope. The extra datum $f(2)=-1316$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-658x$.",
"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... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-658$ fixes the function to $f(x)=-658x$. |
math-018387 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | Derive the result step-by-step: 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 $f(1)=306$, 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{306x}$.\nBoth methods force linearity and use $f(1)=306$ to identify the slope. The extra datum $f(\\frac{1}{6})=51$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=306x$.... | [
{
"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 $f(1)=306$ fixes the function to $f(x)=306x$. |
math-018388 | Functional Equations: Additivity — Extension from Q to R | 10 | Solve (and briefly cross-validate): 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 $f(1)=163$, 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": "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 $f(-2)=-326$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=163x$.",
"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... | Remember: 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-018389 | Functional Equations: Regularity Assumptions — Why Needed | 10 | Warm-up: 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 $f(1)=577$, and that $f\!\left(\frac{20}{12}\ri... | [
{
"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: $\\boxed{577x}$.\nBoth methods force linearity and use $f(1)=577$ to identify the slope. The extra datum $f(\\frac{5}{3})=\\frac{2885}{3}$ 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... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=577$ fixes the function to $f(x)=577x$. |
math-018390 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | Track quantifiers 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 $f(1)=551$, 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": "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 $f(-2)=-1102$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=551x$.",
"robustness_analysis": "G... | [
{
"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-018391 | Functional Equations: Regularity Assumptions — Why Needed | 10 | 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 $f(1)=-494$, 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": "The two methods are consistent and must coincide. Final answer: $\\boxed{-494x}$.\nBoth methods force linearity and use $f(1)=-494$ to identify the slope. The extra datum $f(\\frac{1}{2})=-247$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-494x$.",
"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 $f(1)=-494$ fixes the function to $f(x)=-494x$. (Here the result is $\boxed{-494x}$.) |
math-018392 | Functional Equations: Regularity Assumptions — Why Needed | 10 | 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 $f(1)=348$, 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: $\\boxed{348x}$.\nBoth methods force linearity and use $f(1)=348$ to identify the slope. The extra datum $f(\\frac{-3}{4})=-261$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=348x$.",
"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... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=348$ fixes the function to $f(x)=348x$. (Here the result is $\boxed{348x}$.) |
math-018393 | Functional Equations: Cauchy — Continuity Implies Linearity | 10 | 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 $f(1)=305$, 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: $\\boxed{305x}$.\nBoth methods force linearity and use $f(1)=305$ to identify the slope. The extra datum $f(\\frac{-25}{4})=\\frac{-7625}{4}$ 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... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=305$ fixes the function to $f(x)=305x$. (Here the result is $\boxed{305x}$.) |
math-018394 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | Challenge: 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 $f(1)=-375$, and that $f\!\left(\frac{20}{6}\right)=-1250$... | [
{
"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: $\\boxed{-375x}$.\nBoth methods force linearity and use $f(1)=-375$ to identify the slope. The extra datum $f(\\frac{10}{3})=-1250$ 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... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-375$ fixes the function to $f(x)=-375x$. (Here the result is $\boxed{-375x}$.) |
math-018395 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | Make each step logically reversible (or explain if not): 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": "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: $\\boxed{726x}$.\nBoth methods force linearity and use $f(1)=726$ to identify the slope. The extra datum $f(\\frac{-10}{7})=\\frac{-7260}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=726x$."... | [
{
"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 $f(1)=726$ fixes the function to $f(x)=726x$. (Here the result is $\boxed{726x}$.) |
math-018396 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-483x}$.\nBoth methods force linearity and use $f(1)=-483$ to identify the slope. The extra datum $f(\\frac{-24}{7})=1656$ 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 $f(1)=-483$ fixes the function to $f(x)=-483x$. (Here the result is $\boxed{-483x}$.) |
math-018397 | Functional Equations: Additivity — Extension from Q to R | 10 | Solve with verification: 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 $f(1)=-735$, 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: $\\boxed{-735x}$.\nBoth methods force linearity and use $f(1)=-735$ to identify the slope. The extra datum $f(\\frac{10}{9})=\\frac{-2450}{3}$ 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 $f(1)=-735$ fixes the function to $f(x)=-735x$. (Here the result is $\boxed{-735x}$.) |
math-018398 | Functional Equations: Additivity — Extension from Q to R | 10 | Task: 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 $f(1)=413$, 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{413x}$.\nBoth methods force linearity and use $f(1)=413$ to identify the slope. The extra datum $f(\\frac{-24}{11})=\\frac{-9912}{11}$ 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... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=413$ fixes the function to $f(x)=413x$. |
math-018399 | Functional Equations: Additivity — Extension from Q to R | 10 | Complete the analysis: 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 $f(1)=5$, and that $f\!\left(\frac{-15}{8}\rig... | [
{
"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: $\\boxed{5x}$.\nBoth methods force linearity and use $f(1)=5$ to identify the slope. The extra datum $f(\\frac{-15}{8})=\\frac{-75}{8}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=5x$.",
"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... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=5$ fixes the function to $f(x)=5x$. |
math-018400 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | Compute the requested quantity: 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 $f(1)=770$, 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: $\\boxed{770x}$.\nBoth methods force linearity and use $f(1)=770$ to identify the slope. The extra datum $f(\\frac{1}{5})=154$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=770x$... | [
{
"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 $f(1)=770$ fixes the function to $f(x)=770x$. (Here the result is $\boxed{770x}$.) |
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