id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-018201 | 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 $f(1)=136$, and that $f\!\left(\frac{1}{9}\righ... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{136x}$.\nBoth methods force linearity and use $f(1)=136$ to identify the slope. The extra datum $f(\\frac{1}{9})=\\frac{136}{9}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=136x$.",
... | [
{
"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 $f(1)=136$ fixes the function to $f(x)=136x$. (Here the result is $\boxed{136x}$.) |
math-018202 | Functional Equations: Additive Maps — Density Argument | 10 | Find the exact 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 $f(1)=-104$, and that $f\!\left(\frac{-23... | [
{
"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: $\\boxed{-104x}$.\nBoth methods force linearity and use $f(1)=-104$ to identify the slope. The extra datum $f(\\frac{-23}{3})=\\frac{2392}{3}$ 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... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-104$ fixes the function to $f(x)=-104x$. |
math-018203 | Functional Equations: Additive Maps — Density Argument | 10 | Task: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=641$, and that $f\!\left(\frac{21}{5}\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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{641x}$.\nBoth methods force linearity and use $f(1)=641$ to identify the slope. The extra datum $f(\\frac{21}{5})=\\frac{13461}{5}$ 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... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=641$ fixes the function to $f(x)=641x$. (Here the result is $\boxed{641x}$.) |
math-018204 | Real Analysis: Additive Functions — Pathologies Avoided | 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 $f(1)=534$, 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{534x}$.\nBoth methods force linearity and use $f(1)=534$ to identify the slope. The extra datum $f(\\frac{4}{11})=\\frac{2136}{11}$ 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 $f(1)=534$ fixes the function to $f(x)=534x$. (Here the result is $\boxed{534x}$.) |
math-018205 | Functional Equations: Additive Maps — Density Argument | 10 | Make each step logically reversible (or explain if not): 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 $f(1)=-386$,... | [
{
"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: $\\boxed{-386x}$.\nBoth methods force linearity and use $f(1)=-386$ to identify the slope. The extra datum $f(\\frac{1}{10})=\\frac{-193}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-386x$.",
"... | [
{
"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 $f(1)=-386$ fixes the function to $f(x)=-386x$. (Here the result is $\boxed{-386x}$.) |
math-018206 | Functional Equations: Additivity — Extension from Q to R | 10 | Track units/moduli carefully: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-789$, 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: $\\boxed{-789x}$.\nBoth methods force linearity and use $f(1)=-789$ to identify the slope. The extra datum $f(\\frac{-21}{10})=\\frac{16569}{10}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both c... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-789$ fixes the function to $f(x)=-789x$. |
math-018207 | Functional Equations: Cauchy — Continuity Implies Linearity | 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 $f(1)=229$, 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{229x}$.\nBoth methods force linearity and use $f(1)=229$ to identify the slope. The extra datum $f(\\frac{3}{5})=\\frac{687}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both ... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=229$ fixes the function to $f(x)=229x$. (Here the result is $\boxed{229x}$.) |
math-018208 | Functional Equations: Additivity — Extension from Q to R | 10 | Explain what is being counted/optimized: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-206$, and that $f\!\l... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-206x}$.\nBoth methods force linearity and use $f(1)=-206$ to identify the slope. The extra datum $f(\\frac{8}{5})=\\frac{-1648}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-206x$.",
"... | [
{
"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 $f(1)=-206$ fixes the function to $f(x)=-206x$. (Here the result is $\boxed{-206x}$.) |
math-018209 | Functional Equations: Cauchy — Continuity Implies Linearity | 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 $f(1)=430$, and that $f\!... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{430x}$.\nBoth methods force linearity and use $f(1)=430$ to identify the slope. The extra datum $f(-4)=-1720$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=430x$.",
"robustness_analysis": "S... | [
{
"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 $f(1)=430$ fixes the function to $f(x)=430x$. (Here the result is $\boxed{430x}$.) |
math-018210 | Functional Equations: Regularity Assumptions — Why Needed | 10 | Prompt: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-425$, and that $f\!\left(\frac{21}{11}\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: $\\boxed{-425x}$.\nBoth methods force linearity and use $f(1)=-425$ to identify the slope. The extra datum $f(\\frac{21}{11})=\\frac{-8925}{11}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both co... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-425$ fixes the function to $f(x)=-425x$. (Here the result is $\boxed{-425x}$.) |
math-018211 | Functional Equations: Cauchy — Continuity Implies Linearity | 10 | Complete the analysis: 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 $f(1)=-742$, and that $f\!\left(\frac{19... | [
{
"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: $\\boxed{-742x}$.\nBoth methods force linearity and use $f(1)=-742$ to identify the slope. The extra datum $f(\\frac{19}{11})=\\frac{-14098}{11}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both c... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-742$ fixes the function to $f(x)=-742x$. (Here the result is $\boxed{-742x}$.) |
math-018212 | Functional Equations: Regularity Assumptions — Why Needed | 10 | Problem: 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 $f(1)=-702$, and that $f\!\left(\frac{-21}{3}\right)=4914$.
... | [
{
"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: $\\boxed{-702x}$.\nBoth methods force linearity and use $f(1)=-702$ to identify the slope. The extra datum $f(-7)=4914$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-702x$.",
... | [
{
"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 $f(1)=-702$ fixes the function to $f(x)=-702x$. (Here the result is $\boxed{-702x}$.) |
math-018213 | Functional Equations: Additivity — Extension from Q to R | 10 | Solve and then verify: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=774$, and that $f\!\left(\frac{25}{11}\r... | [
{
"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: $\\boxed{774x}$.\nBoth methods force linearity and use $f(1)=774$ to identify the slope. The extra datum $f(\\frac{25}{11})=\\frac{19350}{11}$ 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 $f(1)=774$ fixes the function to $f(x)=774x$. (Here the result is $\boxed{774x}$.) |
math-018214 | Functional Equations: Additive Maps — Density Argument | 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 $f(1)=-662$, 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: $\\boxed{-662x}$.\nBoth methods force linearity and use $f(1)=-662$ to identify the slope. The extra datum $f(\\frac{-6}{5})=\\frac{3972}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. B... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-662$ fixes the function to $f(x)=-662x$. (Here the result is $\boxed{-662x}$.) |
math-018215 | 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 $f(1)=-445$, 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: $\\boxed{-445x}$.\nBoth methods force linearity and use $f(1)=-445$ to identify the slope. The extra datum $f(\\frac{-12}{5})=1068$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-445x$.",
"robustness... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-445$ fixes the function to $f(x)=-445x$. (Here the result is $\boxed{-445x}$.) |
math-018216 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | Solve and include a self-check: 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 $f(1)=-414$, 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: $\\boxed{-414x}$.\nBoth methods force linearity and use $f(1)=-414$ to identify the slope. The extra datum $f(\\frac{-6}{5})=\\frac{2484}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-414x$.",
"... | [
{
"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 $f(1)=-414$ fixes the function to $f(x)=-414x$. (Here the result is $\boxed{-414x}$.) |
math-018217 | Functional Equations: Additivity — Extension from Q to R | 10 | Problem: 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 $f(1)=93$, and that $f\!\left(\frac{-5}{12}\rig... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{93x}$.\nBoth methods force linearity and use $f(1)=93$ to identify the slope. The extra datum $f(\\frac{-5}{12})=\\frac{-155}{4}$ 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... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=93$ fixes the function to $f(x)=93x$. (Here the result is $\boxed{93x}$.) |
math-018218 | Real Analysis: Additive Functions — Pathologies Avoided | 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 $f(1)=322$, and that $f\!\left(\fra... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{322x}$.\nBoth methods force linearity and use $f(1)=322$ to identify the slope. The extra datum $f(\\frac{-25}{3})=\\frac{-8050}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=322x$."... | [
{
"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 $f(1)=322$ fixes the function to $f(x)=322x$. (Here the result is $\boxed{322x}$.) |
math-018219 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | Try to avoid pattern-matching; explain why: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-608$, 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: $\\boxed{-608x}$.\nBoth methods force linearity and use $f(1)=-608$ to identify the slope. The extra datum $f(\\frac{5}{2})=-1520$ 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 $f(1)=-608$ fixes the function to $f(x)=-608x$. |
math-018220 | Functional Equations: Additive Maps — Density Argument | 10 | Derive the result step-by-step: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=322$, 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: $\\boxed{322x}$.\nBoth methods force linearity and use $f(1)=322$ to identify the slope. The extra datum $f(\\frac{8}{5})=\\frac{2576}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=322x$.",
"robu... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=322$ fixes the function to $f(x)=322x$. |
math-018221 | Functional Equations: Additivity — Extension from Q to R | 10 | Write the solution set clearly: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-561$, 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: $\\boxed{-561x}$.\nBoth methods force linearity and use $f(1)=-561$ to identify the slope. The extra datum $f(\\frac{8}{9})=\\frac{-1496}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-561x$.",
"... | [
{
"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 $f(1)=-561$ fixes the function to $f(x)=-561x$. |
math-018222 | Functional Equations: Additivity — Extension from Q to R | 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 $f(1)=30$, 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: $\\boxed{30x}$.\nBoth methods force linearity and use $f(1)=30$ to identify the slope. The extra datum $f(\\frac{-22}{7})=\\frac{-660}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=30x$.",
"robus... | [
{
"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 $f(1)=30$ fixes the function to $f(x)=30x$. (Here the result is $\boxed{30x}$.) |
math-018223 | Real Analysis: Additive Functions — Pathologies Avoided | 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 $f(1)=389$, 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: $\\boxed{389x}$.\nBoth methods force linearity and use $f(1)=389$ to identify the slope. The extra datum $f(\\frac{14}{3})=\\frac{5446}{3}$ 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... | Remember: 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-018224 | Functional Equations: Additive Maps — Density Argument | 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 $f(1)=-425$, and that $f\!\le... | [
{
"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: $\\boxed{-425x}$.\nBoth methods force linearity and use $f(1)=-425$ to identify the slope. The extra datum $f(\\frac{22}{7})=\\frac{-9350}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. ... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-425$ fixes the function to $f(x)=-425x$. |
math-018225 | Functional Equations: Additive Maps — Density Argument | 10 | Problem: 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 $f(1)=771$, and that $f\!\left(\frac{10}{10}\right)=771$.
(... | [
{
"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: $\\boxed{771x}$.\nBoth methods force linearity and use $f(1)=771$ to identify the slope. The extra datum $f(1)=771$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=771x$.",
"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 $f(1)=771$ fixes the function to $f(x)=771x$. |
math-018226 | Functional Equations: Additive Maps — Density Argument | 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: $\\boxed{21x}$.\nBoth methods force linearity and use $f(1)=21$ to identify the slope. The extra datum $f(\\frac{-15}{4})=\\frac{-315}{4}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=21x$.",
"robus... | [
{
"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 $f(1)=21$ fixes the function to $f(x)=21x$. (Here the result is $\boxed{21x}$.) |
math-018227 | Functional Equations: Regularity Assumptions — Why Needed | 10 | Exercise: 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 $f(1)=408$, and that $f\!\left(\frac{12}{10}\right)=\frac{2... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{408x}$.\nBoth methods force linearity and use $f(1)=408$ to identify the slope. The extra datum $f(\\frac{6}{5})=\\frac{2448}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=408$ fixes the function to $f(x)=408x$. (Here the result is $\boxed{408x}$.) |
math-018228 | Functional Equations: Additive Maps — Density Argument | 10 | Provide a rigorous 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 $f(1)=525$, 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{525x}$.\nBoth methods force linearity and use $f(1)=525$ to identify the slope. The extra datum $f(\\frac{3}{5})=315$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=525x$... | [
{
"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 $f(1)=525$ fixes the function to $f(x)=525x$. |
math-018229 | Functional Equations: Regularity Assumptions — Why Needed | 10 | Solve (and briefly cross-validate): 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 $f(1)=256$, 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: $\\boxed{256x}$.\nBoth methods force linearity and use $f(1)=256$ to identify the slope. The extra datum $f(\\frac{19}{6})=\\frac{2432}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclud... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=256$ fixes the function to $f(x)=256x$. (Here the result is $\boxed{256x}$.) |
math-018230 | Functional Equations: Additivity — Extension from Q to R | 10 | Carefully track domains: 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 $f(1)=207$, and that $f\!\left(\frac{7}{8}\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": "Both approaches agree after simplification. Final answer: $\\boxed{207x}$.\nBoth methods force linearity and use $f(1)=207$ to identify the slope. The extra datum $f(\\frac{7}{8})=\\frac{1449}{8}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=207x$.",
"robu... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=207$ fixes the function to $f(x)=207x$. |
math-018231 | Real Analysis: Additive Functions — Pathologies Avoided | 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 $f(1)=-152$, and that $f\!\left(\frac{23}{11}\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: $\\boxed{-152x}$.\nBoth methods force linearity and use $f(1)=-152$ to identify the slope. The extra datum $f(\\frac{23}{11})=\\frac{-3496}{11}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-152... | [
{
"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 $f(1)=-152$ fixes the function to $f(x)=-152x$. (Here the result is $\boxed{-152x}$.) |
math-018232 | Functional Equations: Additive Maps — Density Argument | 10 | Problem: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-602$, and that $f\!\left(\frac{14}{10}\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": "The two methods are consistent and must coincide. Final answer: $\\boxed{-602x}$.\nBoth methods force linearity and use $f(1)=-602$ to identify the slope. The extra datum $f(\\frac{7}{5})=\\frac{-4214}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-602x$.... | [
{
"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 $f(1)=-602$ fixes the function to $f(x)=-602x$. (Here the result is $\boxed{-602x}$.) |
math-018233 | Real Analysis: Additive Functions — Pathologies Avoided | 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 $f(1)=651$, and that $f\!\left(\fra... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{651x}$.\nBoth methods force linearity and use $f(1)=651$ to identify the slope. The extra datum $f(\\frac{1}{2})=\\frac{651}{2}$ 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 $f(1)=651$ fixes the function to $f(x)=651x$. |
math-018234 | Functional Equations: Additivity — Extension from Q to R | 10 | Do not skip justification steps: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=134$, 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: $\\boxed{134x}$.\nBoth methods force linearity and use $f(1)=134$ to identify the slope. The extra datum $f(\\frac{21}{5})=\\frac{2814}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=134x$.",
"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 $f(1)=134$ fixes the function to $f(x)=134x$. (Here the result is $\boxed{134x}$.) |
math-018235 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | Solve and then verify: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-456$, and 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-456x}$.\nBoth methods force linearity and use $f(1)=-456$ to identify the slope. The extra datum $f(1)=-456$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-456... | [
{
"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 $f(1)=-456$ fixes the function to $f(x)=-456x$. (Here the result is $\boxed{-456x}$.) |
math-018236 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | Solve and include a self-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 $f(1)=-382$, 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: $\\boxed{-382x}$.\nBoth methods force linearity and use $f(1)=-382$ to identify the slope. The extra datum $f(\\frac{13}{9})=\\frac{-4966}{9}$ is consistent with $f(x)=kx$ and serves as a built-in check. ... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-382$ fixes the function to $f(x)=-382x$. (Here the result is $\boxed{-382x}$.) |
math-018237 | Functional Equations: Additivity — Extension from Q to R | 10 | Be explicit about assumptions: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=397$, 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: $\\boxed{397x}$.\nBoth methods force linearity and use $f(1)=397$ to identify the slope. The extra datum $f(-3)=-1191$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=397x$.",
"robustness_analysis": "I... | [
{
"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 $f(1)=397$ fixes the function to $f(x)=397x$. (Here the result is $\boxed{397x}$.) |
math-018238 | Functional Equations: Additivity — Extension from Q to R | 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 $f(1)=364$, and that $f\!\left(\frac{18... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{364x}$.\nBoth methods force linearity and use $f(1)=364$ to identify the slope. The extra datum $f(9)=3276$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=364x$.",
"robustness_analysis"... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=364$ fixes the function to $f(x)=364x$. |
math-018239 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | Keep the final answer in boxed form: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=385$, 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: $\\boxed{385x}$.\nBoth methods force linearity and use $f(1)=385$ to identify the slope. The extra datum $f(\\frac{2}{5})=154$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=385x$.",
"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 $f(1)=385$ fixes the function to $f(x)=385x$. (Here the result is $\boxed{385x}$.) |
math-018240 | Functional Equations: Additivity — Extension from Q to R | 10 | Work this out carefully: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=226$, 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{226x}$.\nBoth methods force linearity and use $f(1)=226$ to identify the slope. The extra datum $f(\\frac{3}{8})=\\frac{339}{4}$ 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... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=226$ fixes the function to $f(x)=226x$. (Here the result is $\boxed{226x}$.) |
math-018241 | Functional Equations: Additive Maps — Density Argument | 10 | Give a theorem-based 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 $f(1)=393$, 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: $\\boxed{393x}$.\nBoth methods force linearity and use $f(1)=393$ to identify the slope. The extra datum $f(\\frac{-17}{2})=\\frac{-6681}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. B... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=393$ fixes the function to $f(x)=393x$. (Here the result is $\boxed{393x}$.) |
math-018242 | Functional Equations: Additive Maps — Density Argument | 10 | Exercise: 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 $f(1)=55$, 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: $\\boxed{55x}$.\nBoth methods force linearity and use $f(1)=55$ to identify the slope. The extra datum $f(2)=110$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=55x$.",
"robustness_analysis": "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 $f(1)=55$ fixes the function to $f(x)=55x$. (Here the result is $\boxed{55x}$.) |
math-018243 | Functional Equations: Additive Maps — Density Argument | 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 $f(1)=332$, 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: $\\boxed{332x}$.\nBoth methods force linearity and use $f(1)=332$ to identify the slope. The extra datum $f(\\frac{-8}{9})=\\frac{-2656}{9}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclu... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=332$ fixes the function to $f(x)=332x$. (Here the result is $\boxed{332x}$.) |
math-018244 | Functional Equations: Cauchy — Continuity Implies Linearity | 10 | Provide both a computational and a conceptual explanation: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1... | [
{
"method_name": "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: $\\boxed{376x}$.\nBoth methods force linearity and use $f(1)=376$ to identify the slope. The extra datum $f(\\frac{-15}{4})=-1410$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=3... | [
{
"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 $f(1)=376$ fixes the function to $f(x)=376x$. (Here the result is $\boxed{376x}$.) |
math-018245 | 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 $f(1)=452$, ... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{452x}$.\nBoth methods force linearity and use $f(1)=452$ to identify the slope. The extra datum $f(\\frac{-8}{3})=\\frac{-3616}{3}$ 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 $f(1)=452$ fixes the function to $f(x)=452x$. |
math-018246 | Functional Equations: Regularity Assumptions — Why Needed | 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)=795... | [
{
"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: $\\boxed{795x}$.\nBoth methods force linearity and use $f(1)=795$ to identify the slope. The extra datum $f(\\frac{-16}{11})=\\frac{-12720}{11}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=795x$.",
... | [
{
"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 $f(1)=795$ fixes the function to $f(x)=795x$. |
math-018247 | Functional Equations: Additivity — Extension from Q to R | 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 $f(1)=-692$, 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: $\\boxed{-692x}$.\nBoth methods force linearity and use $f(1)=-692$ to identify the slope. The extra datum $f(\\frac{-10}{11})=\\frac{6920}{11}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both co... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-692$ fixes the function to $f(x)=-692x$. |
math-018248 | Functional Equations: Additivity — Extension from Q to R | 10 | Prompt: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-275$, 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: $\\boxed{-275x}$.\nBoth methods force linearity and use $f(1)=-275$ to identify the slope. The extra datum $f(1)=-275$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-275x$.",
"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 $f(1)=-275$ fixes the function to $f(x)=-275x$. (Here the result is $\boxed{-275x}$.) |
math-018249 | Functional Equations: Regularity Assumptions — Why Needed | 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 $f(1)=-618$, 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": "Both approaches agree after simplification. Final answer: $\\boxed{-618x}$.\nBoth methods force linearity and use $f(1)=-618$ to identify the slope. The extra datum $f(\\frac{5}{3})=-1030$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-618x$.",
"robustness_... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-618$ fixes the function to $f(x)=-618x$. (Here the result is $\boxed{-618x}$.) |
math-018250 | Functional Equations: Additive Maps — Density Argument | 10 | Keep the final answer in boxed form: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=164$, 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: $\\boxed{164x}$.\nBoth methods force linearity and use $f(1)=164$ to identify the slope. The extra datum $f(\\frac{11}{4})=451$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=164x$.",
"robustnes... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=164$ fixes the function to $f(x)=164x$. (Here the result is $\boxed{164x}$.) |
math-018251 | Functional Equations: Additivity — Extension from Q to R | 10 | State any required conditions first: 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 $f(1)=539$, 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: $\\boxed{539x}$.\nBoth methods force linearity and use $f(1)=539$ to identify the slope. The extra datum $f(-1)=-539$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=539x$.",
"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 $f(1)=539$ fixes the function to $f(x)=539x$. (Here the result is $\boxed{539x}$.) |
math-018252 | Functional Equations: Cauchy — Continuity Implies Linearity | 10 | Question: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-9$, and that $f\!\left(\frac{-17}{4}\ri... | [
{
"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: $\\boxed{-9x}$.\nBoth methods force linearity and use $f(1)=-9$ to identify the slope. The extra datum $f(\\frac{-17}{4})=\\frac{153}{4}$ 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... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-9$ fixes the function to $f(x)=-9x$. |
math-018253 | Functional Equations: Additivity — Extension from Q to R | 10 | Solve and justify each step: 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 $f(1)=465$, and that $f\!\left(\frac{5}{... | [
{
"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: $\\boxed{465x}$.\nBoth methods force linearity and use $f(1)=465$ to identify the slope. The extra datum $f(\\frac{5}{8})=\\frac{2325}{8}$ 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... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=465$ fixes the function to $f(x)=465x$. (Here the result is $\boxed{465x}$.) |
math-018254 | Functional Equations: Regularity Assumptions — Why Needed | 10 | Start by stating any domain restrictions: 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 $f(1)=731$, 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: $\\boxed{731x}$.\nBoth methods force linearity and use $f(1)=731$ to identify the slope. The extra datum $f(\\frac{14}{9})=\\frac{10234}{9}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclu... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=731$ fixes the function to $f(x)=731x$. |
math-018255 | Functional Equations: Regularity Assumptions — Why Needed | 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 $f(1)=-609$, and t... | [
{
"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: $\\boxed{-609x}$.\nBoth methods force linearity and use $f(1)=-609$ to identify the slope. The extra datum $f(\\frac{-7}{6})=\\frac{1421}{2}$ 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 $f(1)=-609$ fixes the function to $f(x)=-609x$. (Here the result is $\boxed{-609x}$.) |
math-018256 | 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 $f(1)=789$, 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: $\\boxed{789x}$.\nBoth methods force linearity and use $f(1)=789$ to identify the slope. The extra datum $f(\\frac{23}{8})=\\frac{18147}{8}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=789x$.",
"ro... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=789$ fixes the function to $f(x)=789x$. (Here the result is $\boxed{789x}$.) |
math-018257 | Functional Equations: Additive Maps — Density Argument | 10 | Complete the analysis: 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 $f(1)=86$, 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{86x}$.\nBoth methods force linearity and use $f(1)=86$ to identify the slope. The extra datum $f(\\frac{4}{3})=\\frac{344}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both co... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=86$ fixes the function to $f(x)=86x$. (Here the result is $\boxed{86x}$.) |
math-018258 | Real Analysis: Additive Functions — Pathologies Avoided | 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)=-790... | [
{
"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: $\\boxed{-790x}$.\nBoth methods force linearity and use $f(1)=-790$ to identify the slope. The extra datum $f(\\frac{2}{3})=\\frac{-1580}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. B... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-790$ fixes the function to $f(x)=-790x$. |
math-018259 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | Answer with a short justification: 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 $f(1)=443$, and that $f\!\le... | [
{
"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: $\\boxed{443x}$.\nBoth methods force linearity and use $f(1)=443$ to identify the slope. The extra datum $f(\\frac{-23}{7})=\\frac{-10189}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=443x$.",
"... | [
{
"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 $f(1)=443$ fixes the function to $f(x)=443x$. |
math-018260 | Functional Equations: Additive Maps — Density Argument | 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 $f(1)=42$, and that $f\!\le... | [
{
"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: $\\boxed{42x}$.\nBoth methods force linearity and use $f(1)=42$ to identify the slope. The extra datum $f(\\frac{-1}{10})=\\frac{-21}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=42x$.",
"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... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=42$ fixes the function to $f(x)=42x$. |
math-018261 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | State any required conditions first: 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: $\\boxed{735x}$.\nBoth methods force linearity and use $f(1)=735$ to identify the slope. The extra datum $f(\\frac{-15}{8})=\\frac{-11025}{8}$ 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-018262 | Real Analysis: Additive Functions — Pathologies Avoided | 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 $f(1)=511$, 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: $\\boxed{511x}$.\nBoth methods force linearity and use $f(1)=511$ to identify the slope. The extra datum $f(\\frac{-11}{5})=\\frac{-5621}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. B... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=511$ fixes the function to $f(x)=511x$. |
math-018263 | Functional Equations: Cauchy — Continuity Implies Linearity | 10 | Solve with verification: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=341$, and that $f\!\left(\frac{-3}{9}\... | [
{
"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: $\\boxed{341x}$.\nBoth methods force linearity and use $f(1)=341$ to identify the slope. The extra datum $f(\\frac{-1}{3})=\\frac{-341}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclud... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=341$ fixes the function to $f(x)=341x$. |
math-018264 | Functional Equations: Additive Maps — Density Argument | 10 | Task: 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 $f(1)=-478$, and that $f\!\left(\frac{-22}{4}\right)=2629$.
(a... | [
{
"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: $\\boxed{-478x}$.\nBoth methods force linearity and use $f(1)=-478$ to identify the slope. The extra datum $f(\\frac{-11}{2})=2629$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-478x$.",
"robu... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-478$ fixes the function to $f(x)=-478x$. |
math-018265 | Functional Equations: Cauchy — Continuity Implies Linearity | 10 | Explain each transformation: 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 $f(1)=508$, and that $f\!\left(\frac{-21... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{508x}$.\nBoth methods force linearity and use $f(1)=508$ to identify the slope. The extra datum $f(\\frac{-21}{2})=-5334$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=5... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=508$ fixes the function to $f(x)=508x$. (Here the result is $\boxed{508x}$.) |
math-018266 | Functional Equations: Cauchy — Continuity Implies Linearity | 10 | Solve and sanity-check: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-470$, 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: $\\boxed{-470x}$.\nBoth methods force linearity and use $f(1)=-470$ to identify the slope. The extra datum $f(\\frac{-11}{9})=\\frac{5170}{9}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conc... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-470$ fixes the function to $f(x)=-470x$. (Here the result is $\boxed{-470x}$.) |
math-018267 | 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 $f(1)=-4$, 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: $\\boxed{-4x}$.\nBoth methods force linearity and use $f(1)=-4$ to identify the slope. The extra datum $f(\\frac{-24}{7})=\\frac{96}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both c... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-4$ fixes the function to $f(x)=-4x$. (Here the result is $\boxed{-4x}$.) |
math-018268 | Functional Equations: Regularity Assumptions — Why Needed | 10 | Work carefully and justify each inference: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=138$, and that... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{138x}$.\nBoth methods force linearity and use $f(1)=138$ to identify the slope. The extra datum $f(\\frac{-5}{6})=-115$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=138... | [
{
"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 $f(1)=138$ fixes the function to $f(x)=138x$. |
math-018269 | Functional Equations: Additive Maps — Density Argument | 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 $f(1)=66$, 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: $\\boxed{66x}$.\nBoth methods force linearity and use $f(1)=66$ to identify the slope. The extra datum $f(-1)=-66$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=66x$.",
... | [
{
"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 $f(1)=66$ fixes the function to $f(x)=66x$. (Here the result is $\boxed{66x}$.) |
math-018270 | Functional Equations: Additivity — Extension from Q to R | 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)=-... | [
{
"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: $\\boxed{-784x}$.\nBoth methods force linearity and use $f(1)=-784$ to identify the slope. The extra datum $f(\\frac{8}{3})=\\frac{-6272}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-784x$.... | [
{
"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 $f(1)=-784$ fixes the function to $f(x)=-784x$. |
math-018271 | 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 $f(1)=-138$, 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: $\\boxed{-138x}$.\nBoth methods force linearity and use $f(1)=-138$ to identify the slope. The extra datum $f(\\frac{6}{7})=\\frac{-828}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-138x$."... | [
{
"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 $f(1)=-138$ fixes the function to $f(x)=-138x$. |
math-018272 | Functional Equations: Cauchy — Continuity Implies Linearity | 10 | Determine the requested value: 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 $f(1)=-557$, 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: $\\boxed{-557x}$.\nBoth methods force linearity and use $f(1)=-557$ to identify the slope. The extra datum $f(\\frac{18}{11})=\\frac{-10026}{11}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-557x$.",... | [
{
"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 $f(1)=-557$ fixes the function to $f(x)=-557x$. (Here the result is $\boxed{-557x}$.) |
math-018273 | Functional Equations: Additivity — Extension from Q to R | 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 $f(1)=-383$, 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: $\\boxed{-383x}$.\nBoth methods force linearity and use $f(1)=-383$ to identify the slope. The extra datum $f(\\frac{-15}{4})=\\frac{5745}{4}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conc... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-383$ fixes the function to $f(x)=-383x$. |
math-018274 | Functional Equations: Regularity Assumptions — Why Needed | 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 $f(1)=-655$, and that $f\!\left(\fra... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-655x}$.\nBoth methods force linearity and use $f(1)=-655$ to identify the slope. The extra datum $f(\\frac{-5}{3})=\\frac{3275}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. B... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-655$ fixes the function to $f(x)=-655x$. |
math-018275 | Functional Equations: Additivity — Extension from Q to R | 10 | Task: 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 $f(1)=389$, and that $f\!\left(\frac{17}{11}\right)=\frac{6613}... | [
{
"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: $\\boxed{389x}$.\nBoth methods force linearity and use $f(1)=389$ to identify the slope. The extra datum $f(\\frac{17}{11})=\\frac{6613}{11}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=389x$.",
"r... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=389$ fixes the function to $f(x)=389x$. (Here the result is $\boxed{389x}$.) |
math-018276 | Functional Equations: Additive Maps — Density Argument | 10 | Compute the requested quantity: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=34$... | [
{
"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: $\\boxed{34x}$.\nBoth methods force linearity and use $f(1)=34$ to identify the slope. The extra datum $f(-8)=-272$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=34x$.",
"robus... | [
{
"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 $f(1)=34$ fixes the function to $f(x)=34x$. (Here the result is $\boxed{34x}$.) |
math-018277 | 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 $f(1)=-401$, 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-401x}$.\nBoth methods force linearity and use $f(1)=-401$ to identify the slope. The extra datum $f(\\frac{3}{4})=\\frac{-1203}{4}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both concl... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-401$ fixes the function to $f(x)=-401x$. (Here the result is $\boxed{-401x}$.) |
math-018278 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | Keep the final answer in boxed form: 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 $f(1)=-221$, 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": "The two methods are consistent and must coincide. Final answer: $\\boxed{-221x}$.\nBoth methods force linearity and use $f(1)=-221$ to identify the slope. The extra datum $f(\\frac{-2}{11})=\\frac{442}{11}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-221x$... | [
{
"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 $f(1)=-221$ fixes the function to $f(x)=-221x$. (Here the result is $\boxed{-221x}$.) |
math-018279 | Functional Equations: Regularity Assumptions — Why Needed | 10 | Task: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=570$, and that $f\!\left(\frac{-16}{9}\right)=\frac... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{570x}$.\nBoth methods force linearity and use $f(1)=570$ to identify the slope. The extra datum $f(\\frac{-16}{9})=\\frac{-3040}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=570x$.",
"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 $f(1)=570$ fixes the function to $f(x)=570x$. (Here the result is $\boxed{570x}$.) |
math-018280 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | Carefully track domains: 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 $f(1)=-282$, 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: $\\boxed{-282x}$.\nBoth methods force linearity and use $f(1)=-282$ to identify the slope. The extra datum $f(\\frac{2}{11})=\\frac{-564}{11}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conc... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-282$ fixes the function to $f(x)=-282x$. (Here the result is $\boxed{-282x}$.) |
math-018281 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | Answer using clear logical steps: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-419$, and that ... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-419x}$.\nBoth methods force linearity and use $f(1)=-419$ to identify the slope. The extra datum $f(\\frac{-23}{8})=\\frac{9637}{8}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-419x$... | [
{
"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 $f(1)=-419$ fixes the function to $f(x)=-419x$. |
math-018282 | Functional Equations: Cauchy — Continuity Implies Linearity | 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 $f(1)=-393$, 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: $\\boxed{-393x}$.\nBoth methods force linearity and use $f(1)=-393$ to identify the slope. The extra datum $f(\\frac{-13}{8})=\\frac{5109}{8}$ 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 $f(1)=-393$ fixes the function to $f(x)=-393x$. (Here the result is $\boxed{-393x}$.) |
math-018283 | Functional Equations: Additive Maps — Density Argument | 10 | Make each step logically reversible (or explain if not): 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 $f(1)=192$, ... | [
{
"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: $\\boxed{192x}$.\nBoth methods force linearity and use $f(1)=192$ to identify the slope. The extra datum $f(\\frac{-15}{4})=-720$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=192x$.",
"robustn... | [
{
"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 $f(1)=192$ fixes the function to $f(x)=192x$. (Here the result is $\boxed{192x}$.) |
math-018284 | Functional Equations: Additive Maps — Density Argument | 10 | State any required conditions first: 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 $f(1)=-125$, 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: $\\boxed{-125x}$.\nBoth methods force linearity and use $f(1)=-125$ to identify the slope. The extra datum $f(\\frac{5}{4})=\\frac{-625}{4}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-125x$."... | [
{
"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 $f(1)=-125$ fixes the function to $f(x)=-125x$. |
math-018285 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | Problem: 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 $f(1)=309$, and that $f\!\left(\frac{6}{7}\righ... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{309x}$.\nBoth methods force linearity and use $f(1)=309$ to identify the slope. The extra datum $f(\\frac{6}{7})=\\frac{1854}{7}$ 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... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=309$ fixes the function to $f(x)=309x$. (Here the result is $\boxed{309x}$.) |
math-018286 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | Determine the requested 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)=-735... | [
{
"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: $\\boxed{-735x}$.\nBoth methods force linearity and use $f(1)=-735$ to identify the slope. The extra datum $f(3)=-2205$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-735x$.",
... | [
{
"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 $f(1)=-735$ fixes the function to $f(x)=-735x$. |
math-018287 | Functional Equations: Additive Maps — Density Argument | 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 $f(1)=-488$, and t... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-488x}$.\nBoth methods force linearity and use $f(1)=-488$ to identify the slope. The extra datum $f(\\frac{4}{3})=\\frac{-1952}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-488x$.... | [
{
"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 $f(1)=-488$ fixes the function to $f(x)=-488x$. (Here the result is $\boxed{-488x}$.) |
math-018288 | 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 $f(1)=373$, 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: $\\boxed{373x}$.\nBoth methods force linearity and use $f(1)=373$ to identify the slope. The extra datum $f(\\frac{1}{3})=\\frac{373}{3}$ 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... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=373$ fixes the function to $f(x)=373x$. |
math-018289 | Functional Equations: Additivity — Extension from Q to R | 10 | Solve and then verify: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=189$, and that $f\!\left(\frac{10}{4}\ri... | [
{
"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: $\\boxed{189x}$.\nBoth methods force linearity and use $f(1)=189$ to identify the slope. The extra datum $f(\\frac{5}{2})=\\frac{945}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=189x$.",
"robus... | [
{
"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 $f(1)=189$ fixes the function to $f(x)=189x$. (Here the result is $\boxed{189x}$.) |
math-018290 | 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 $f(1)=-361$, and that $f\!\left(\frac{-16}{2}\right)=2888$.
... | [
{
"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: $\\boxed{-361x}$.\nBoth methods force linearity and use $f(1)=-361$ to identify the slope. The extra datum $f(-8)=2888$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-361x$.",
... | [
{
"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 $f(1)=-361$ fixes the function to $f(x)=-361x$. |
math-018291 | Functional Equations: Additive Maps — Density Argument | 10 | Solve and then verify: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-64$, and that $f\!\left(\frac{-2}{10}\r... | [
{
"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: $\\boxed{-64x}$.\nBoth methods force linearity and use $f(1)=-64$ to identify the slope. The extra datum $f(\\frac{-1}{5})=\\frac{64}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both ... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-64$ fixes the function to $f(x)=-64x$. (Here the result is $\boxed{-64x}$.) |
math-018292 | Functional Equations: Additivity — Extension from Q to R | 10 | Find the exact value: 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 $f(1)=362$, 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{362x}$.\nBoth methods force linearity and use $f(1)=362$ to identify the slope. The extra datum $f(\\frac{10}{11})=\\frac{3620}{11}$ 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 $f(1)=362$ fixes the function to $f(x)=362x$. |
math-018293 | Functional Equations: Additivity — Extension from Q to R | 10 | Use two approaches if possible: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-762$, and that $f\!\left... | [
{
"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: $\\boxed{-762x}$.\nBoth methods force linearity and use $f(1)=-762$ to identify the slope. The extra datum $f(\\frac{1}{4})=\\frac{-381}{2}$ 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 $f(1)=-762$ fixes the function to $f(x)=-762x$. |
math-018294 | Functional Equations: Regularity Assumptions — Why Needed | 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 $f(1)=341$, 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: $\\boxed{341x}$.\nBoth methods force linearity and use $f(1)=341$ to identify the slope. The extra datum $f(\\frac{-14}{11})=-434$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=3... | [
{
"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 $f(1)=341$ fixes the function to $f(x)=341x$. (Here the result is $\boxed{341x}$.) |
math-018295 | Functional Equations: Additivity — Extension from Q to R | 10 | Compute the requested quantity: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=386... | [
{
"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: $\\boxed{386x}$.\nBoth methods force linearity and use $f(1)=386$ to identify the slope. The extra datum $f(\\frac{-17}{10})=\\frac{-3281}{5}$ 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 $f(1)=386$ fixes the function to $f(x)=386x$. (Here the result is $\boxed{386x}$.) |
math-018296 | Functional Equations: Additivity — Extension from Q to R | 10 | Explain what is being counted/optimized: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-747$, 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-747x}$.\nBoth methods force linearity and use $f(1)=-747$ to identify the slope. The extra datum $f(\\frac{17}{9})=-1411$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-747$ fixes the function to $f(x)=-747x$. (Here the result is $\boxed{-747x}$.) |
math-018297 | Functional Equations: Regularity Assumptions — Why Needed | 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 $f(1)=323$, 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: $\\boxed{323x}$.\nBoth methods force linearity and use $f(1)=323$ to identify the slope. The extra datum $f(\\frac{11}{10})=\\frac{3553}{10}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=323x$."... | [
{
"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 $f(1)=323$ fixes the function to $f(x)=323x$. |
math-018298 | Real Analysis: Additive Functions — Pathologies Avoided | 10 | Derive the result step-by-step: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-55... | [
{
"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: $\\boxed{-557x}$.\nBoth methods force linearity and use $f(1)=-557$ to identify the slope. The extra datum $f(\\frac{8}{3})=\\frac{-4456}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. B... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-557$ fixes the function to $f(x)=-557x$. (Here the result is $\boxed{-557x}$.) |
math-018299 | Real Analysis: Additive Functions — Pathologies Avoided | 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 $f(1)=276$, 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: $\\boxed{276x}$.\nBoth methods force linearity and use $f(1)=276$ to identify the slope. The extra datum $f(\\frac{-11}{4})=-759$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=276x$.",
"robustness_an... | [
{
"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 $f(1)=276$ fixes the function to $f(x)=276x$. (Here the result is $\boxed{276x}$.) |
math-018300 | 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 $f(1)=-36$, and that $f\!\left(\frac... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-36x}$.\nBoth methods force linearity and use $f(1)=-36$ to identify the slope. The extra datum $f(\\frac{-8}{5})=\\frac{288}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-36x$.",
... | [
{
"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 $f(1)=-36$ fixes the function to $f(x)=-36x$. |
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