id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-016201 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Keep the final answer in boxed form: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=135$, 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: $\\boxed{135x}$.\nBoth methods force linearity and use $f(1)=135$ to identify the slope. The extra datum $f(\\frac{-3}{10})=\\frac{-81}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=135x$.",
"rob... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=135$ fixes the function to $f(x)=135x$. |
math-016202 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Problem: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=64$, and that $f\!\left(\frac{8}{2}\right)=256$.... | [
{
"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 $f(4)=256$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=64x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=64$ fixes the function to $f(x)=64x$. (Here the result is $\boxed{64x}$.) |
math-016203 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | 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)=247... | [
{
"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: $\\boxed{247x}$.\nBoth methods force linearity and use $f(1)=247$ to identify the slope. The extra datum $f(\\frac{-24}{5})=\\frac{-5928}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=247x$."... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=247$ fixes the function to $f(x)=247x$. (Here the result is $\boxed{247x}$.) |
math-016204 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | 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 $f(1)=578$, 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: $\\boxed{578x}$.\nBoth methods force linearity and use $f(1)=578$ to identify the slope. The extra datum $f(1)=578$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=578x$.",
"robu... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=578$ fixes the function to $f(x)=578x$. |
math-016205 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Warm-up: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=423$, and that $f\!\left(\frac{8}{10}\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": "The two methods are consistent and must coincide. Final answer: $\\boxed{423x}$.\nBoth methods force linearity and use $f(1)=423$ to identify the slope. The extra datum $f(\\frac{4}{5})=\\frac{1692}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=423x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=423$ fixes the function to $f(x)=423x$. (Here the result is $\boxed{423x}$.) |
math-016206 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Start by stating any domain restrictions: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that... | [
{
"method_name": "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: $\\boxed{269x}$.\nBoth methods force linearity and use $f(1)=269$ to identify the slope. The extra datum $f(\\frac{9}{7})=\\frac{2421}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=269$ fixes the function to $f(x)=269x$. |
math-016207 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Answer with a short justification: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-565$, 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: $\\boxed{-565x}$.\nBoth methods force linearity and use $f(1)=-565$ to identify the slope. The extra datum $f(\\frac{1}{6})=\\frac{-565}{6}$ 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... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-565$ fixes the function to $f(x)=-565x$. (Here the result is $\boxed{-565x}$.) |
math-016208 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | 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": "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: $\\boxed{142x}$.\nBoth methods force linearity and use $f(1)=142$ to identify the slope. The extra datum $f(\\frac{5}{11})=\\frac{710}{11}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=142x$.",
"rob... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=142$ fixes the function to $f(x)=142x$. (Here the result is $\boxed{142x}$.) |
math-016209 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | 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 $f(1)=-668$, 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: $\\boxed{-668x}$.\nBoth methods force linearity and use $f(1)=-668$ to identify the slope. The extra datum $f(\\frac{-1}{2})=334$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-668x$.",
"robustness_a... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-668$ fixes the function to $f(x)=-668x$. (Here the result is $\boxed{-668x}$.) |
math-016210 | Functional Equations: Additivity — Extension from Q to R | 9 | Give an answer and a quick verification: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=435$, 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: $\\boxed{435x}$.\nBoth methods force linearity and use $f(1)=435$ to identify the slope. The extra datum $f(\\frac{-24}{11})=\\frac{-10440}{11}$ 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... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=435$ fixes the function to $f(x)=435x$. (Here the result is $\boxed{435x}$.) |
math-016211 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Work carefully and justify each inference: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=353$, 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: $\\boxed{353x}$.\nBoth methods force linearity and use $f(1)=353$ to identify the slope. The extra datum $f(\\frac{-11}{2})=\\frac{-3883}{2}$ 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 $f(1)=353$ fixes the function to $f(x)=353x$. (Here the result is $\boxed{353x}$.) |
math-016212 | Functional Equations: Additivity — Extension from Q to R | 9 | Checkpoint: 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 $f(1)=224$, and that $f\!\left(\frac{16}{2}\right)=1792$.... | [
{
"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: $\\boxed{224x}$.\nBoth methods force linearity and use $f(1)=224$ to identify the slope. The extra datum $f(8)=1792$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=224x$.",
"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 $f(1)=224$ fixes the function to $f(x)=224x$. |
math-016213 | Functional Equations: Additive Maps — Density Argument | 9 | Where appropriate, name the theorem you use: 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 $0$, 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-692x}$.\nBoth methods force linearity and use $f(1)=-692$ to identify the slope. The extra datum $f(-4)=2768$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-69... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-692$ fixes the function to $f(x)=-692x$. |
math-016214 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Proceed methodically: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-181$, and that $f\!\left(\f... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-181x}$.\nBoth methods force linearity and use $f(1)=-181$ to identify the slope. The extra datum $f(\\frac{-7}{4})=\\frac{1267}{4}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both concl... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-181$ fixes the function to $f(x)=-181x$. (Here the result is $\boxed{-181x}$.) |
math-016215 | Functional Equations: Additive Maps — Density Argument | 9 | Answer with a short justification: 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 $f(1)=307$, and that $f\!\left(\fr... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{307x}$.\nBoth methods force linearity and use $f(1)=307$ to identify the slope. The extra datum $f(\\frac{21}{2})=\\frac{6447}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Bot... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=307$ fixes the function to $f(x)=307x$. (Here the result is $\boxed{307x}$.) |
math-016216 | Functional Equations: Additive Maps — Density Argument | 9 | 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 $f(1)=-165$, 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: $\\boxed{-165x}$.\nBoth methods force linearity and use $f(1)=-165$ to identify the slope. The extra datum $f(\\frac{8}{9})=\\frac{-440}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Bo... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-165$ fixes the function to $f(x)=-165x$. (Here the result is $\boxed{-165x}$.) |
math-016217 | Functional Equations: Additive Maps — Density Argument | 9 | Indicate where a theorem is used: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-200$, 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-200x}$.\nBoth methods force linearity and use $f(1)=-200$ to identify the slope. The extra datum $f(\\frac{-20}{7})=\\frac{4000}{7}$ 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... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-200$ fixes the function to $f(x)=-200x$. (Here the result is $\boxed{-200x}$.) |
math-016218 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Complete the analysis: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-660$, and that $f\!\left(\frac{2}{5}\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": "Both approaches agree after simplification. Final answer: $\\boxed{-660x}$.\nBoth methods force linearity and use $f(1)=-660$ to identify the slope. The extra datum $f(\\frac{2}{5})=-264$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-660x$.",
"robustness_a... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-660$ fixes the function to $f(x)=-660x$. |
math-016219 | Functional Equations: Additivity — Extension from Q to R | 9 | Write the solution set clearly: 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)=272... | [
{
"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: $\\boxed{272x}$.\nBoth methods force linearity and use $f(1)=272$ to identify the slope. The extra datum $f(\\frac{-17}{7})=\\frac{-4624}{7}$ 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 $f(1)=272$ fixes the function to $f(x)=272x$. (Here the result is $\boxed{272x}$.) |
math-016220 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Show all reasoning: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-601$, 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: $\\boxed{-601x}$.\nBoth methods force linearity and use $f(1)=-601$ to identify the slope. The extra datum $f(2)=-1202$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-601x$.",
"robustness_analy... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-601$ fixes the function to $f(x)=-601x$. |
math-016221 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Provide a rigorous solution: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=389$, and that $f\!\left(\fr... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{389x}$.\nBoth methods force linearity and use $f(1)=389$ to identify the slope. The extra datum $f(\\frac{2}{11})=\\frac{778}{11}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=389x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=389$ fixes the function to $f(x)=389x$. |
math-016222 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Explain why your operations are valid: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-43$, 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: $\\boxed{-43x}$.\nBoth methods force linearity and use $f(1)=-43$ to identify the slope. The extra datum $f(\\frac{-19}{12})=\\frac{817}{12}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-43x$.",
"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... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-43$ fixes the function to $f(x)=-43x$. |
math-016223 | Functional Equations: Additivity — Extension from Q to R | 9 | Answer using clear logical steps: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=678$, and that $f\!\lef... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{678x}$.\nBoth methods force linearity and use $f(1)=678$ to identify the slope. The extra datum $f(\\frac{11}{12})=\\frac{1243}{2}$ 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... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=678$ fixes the function to $f(x)=678x$. |
math-016224 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Try to avoid pattern-matching; explain why: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=414$, and tha... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{414x}$.\nBoth methods force linearity and use $f(1)=414$ to identify the slope. The extra datum $f(\\frac{11}{3})=1518$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=414x$.",
"robustne... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=414$ fixes the function to $f(x)=414x$. |
math-016225 | Functional Equations: Additive Maps — Density Argument | 9 | Carefully track domains: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-577$, 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: $\\boxed{-577x}$.\nBoth methods force linearity and use $f(1)=-577$ to identify the slope. The extra datum $f(\\frac{1}{2})=\\frac{-577}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-577x$.",
"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... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-577$ fixes the function to $f(x)=-577x$. (Here the result is $\boxed{-577x}$.) |
math-016226 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Keep the final answer in boxed form: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-135$, 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: $\\boxed{-135x}$.\nBoth methods force linearity and use $f(1)=-135$ to identify the slope. The extra datum $f(\\frac{3}{2})=\\frac{-405}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-135x$.",
"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... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-135$ fixes the function to $f(x)=-135x$. (Here the result is $\boxed{-135x}$.) |
math-016227 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Show all reasoning: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=141$, and that $f\!\left(\frac{2}{5}\right)... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{141x}$.\nBoth methods force linearity and use $f(1)=141$ to identify the slope. The extra datum $f(\\frac{2}{5})=\\frac{282}{5}$ 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 $f(1)=141$ fixes the function to $f(x)=141x$. (Here the result is $\boxed{141x}$.) |
math-016228 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Compute the requested quantity: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-380$, 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: $\\boxed{-380x}$.\nBoth methods force linearity and use $f(1)=-380$ to identify the slope. The extra datum $f(-2)=760$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-380x$.",
"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 $f(1)=-380$ fixes the function to $f(x)=-380x$. (Here the result is $\boxed{-380x}$.) |
math-016229 | Functional Equations: Additivity — Extension from Q to R | 9 | Carefully track domains: 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 $f(1)=465$, and that $f\!\left(\frac{2... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "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 $f(\\frac{5}{2})=\\frac{2325}{2}$ 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... | Key idea: 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-016230 | Functional Equations: Additivity — Extension from Q to R | 9 | Solve with verification: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=291$, 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: $\\boxed{291x}$.\nBoth methods force linearity and use $f(1)=291$ to identify the slope. The extra datum $f(\\frac{13}{5})=\\frac{3783}{5}$ 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 $f(1)=291$ fixes the function to $f(x)=291x$. (Here the result is $\boxed{291x}$.) |
math-016231 | Functional Equations: Additive Maps — Density Argument | 9 | Show all reasoning: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=261$, 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: $\\boxed{261x}$.\nBoth methods force linearity and use $f(1)=261$ to identify the slope. The extra datum $f(\\frac{-5}{2})=\\frac{-1305}{2}$ 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 $f(1)=261$ fixes the function to $f(x)=261x$. |
math-016232 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Work carefully and justify each inference: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-607$, and tha... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-607x}$.\nBoth methods force linearity and use $f(1)=-607$ to identify the slope. The extra datum $f(\\frac{-16}{11})=\\frac{9712}{11}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-607... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-607$ fixes the function to $f(x)=-607x$. (Here the result is $\boxed{-607x}$.) |
math-016233 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Problem: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-348$, and that $f\!\left(\frac{-4}{6}\right)=23... | [
{
"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: $\\boxed{-348x}$.\nBoth methods force linearity and use $f(1)=-348$ to identify the slope. The extra datum $f(\\frac{-2}{3})=232$ 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 $f(1)=-348$ fixes the function to $f(x)=-348x$. (Here the result is $\boxed{-348x}$.) |
math-016234 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Track units/moduli 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 $f(1)=27$, ... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{27x}$.\nBoth methods force linearity and use $f(1)=27$ to identify the slope. The extra datum $f(\\frac{24}{5})=\\frac{648}{5}$ 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 $f(1)=27$ fixes the function to $f(x)=27x$. (Here the result is $\boxed{27x}$.) |
math-016235 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Track quantifiers 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 $f(1)=-424$,... | [
{
"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: $\\boxed{-424x}$.\nBoth methods force linearity and use $f(1)=-424$ to identify the slope. The extra datum $f(\\frac{4}{3})=\\frac{-1696}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-424x$.... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-424$ fixes the function to $f(x)=-424x$. (Here the result is $\boxed{-424x}$.) |
math-016236 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Give a fully justified solution: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=90$, and that $f\!\left(\frac{... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{90x}$.\nBoth methods force linearity and use $f(1)=90$ to identify the slope. The extra datum $f(\\frac{5}{2})=225$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=90x$.",... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=90$ fixes the function to $f(x)=90x$. (Here the result is $\boxed{90x}$.) |
math-016237 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | 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 $f(1)=282$, 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: $\\boxed{282x}$.\nBoth methods force linearity and use $f(1)=282$ to identify the slope. The extra datum $f(\\frac{6}{7})=\\frac{1692}{7}$ 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... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=282$ fixes the function to $f(x)=282x$. |
math-016238 | Functional Equations: Additive Maps — Density Argument | 9 | 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 $f(1)=-57$, 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: $\\boxed{-57x}$.\nBoth methods force linearity and use $f(1)=-57$ to identify the slope. The extra datum $f(\\frac{13}{8})=\\frac{-741}{8}$ is consistent with $f(x)=kx$ and serves as a built-in check. Bot... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-57$ fixes the function to $f(x)=-57x$. (Here the result is $\boxed{-57x}$.) |
math-016239 | Functional Equations: Regularity Assumptions — Why Needed | 9 | 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": "The two methods are consistent and must coincide. Final answer: $\\boxed{-534x}$.\nBoth methods force linearity and use $f(1)=-534$ to identify the slope. The extra datum $f(\\frac{-5}{2})=1335$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-534x$.",
"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... | Takeaway: 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-016240 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Try to avoid pattern-matching; explain why: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=265$, and tha... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{265x}$.\nBoth methods force linearity and use $f(1)=265$ to identify the slope. The extra datum $f(\\frac{11}{7})=\\frac{2915}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=265x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=265$ fixes the function to $f(x)=265x$. |
math-016241 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Give a fully justified solution: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=260$, 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: $\\boxed{260x}$.\nBoth methods force linearity and use $f(1)=260$ to identify the slope. The extra datum $f(\\frac{9}{11})=\\frac{2340}{11}$ 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 $f(1)=260$ fixes the function to $f(x)=260x$. (Here the result is $\boxed{260x}$.) |
math-016242 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Where appropriate, name the theorem you use: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=714$,... | [
{
"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: $\\boxed{714x}$.\nBoth methods force linearity and use $f(1)=714$ to identify the slope. The extra datum $f(\\frac{19}{3})=4522$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=714x$.",
"robustne... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=714$ fixes the function to $f(x)=714x$. |
math-016243 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Solve and sanity-check: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-28$, 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: $\\boxed{-28x}$.\nBoth methods force linearity and use $f(1)=-28$ to identify the slope. The extra datum $f(\\frac{1}{2})=-14$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-28$ fixes the function to $f(x)=-28x$. (Here the result is $\boxed{-28x}$.) |
math-016244 | Functional Equations: Additivity — Extension from Q to R | 9 | Work carefully and justify each inference: 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 $0$, tha... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{82x}$.\nBoth methods force linearity and use $f(1)=82$ to identify the slope. The extra datum $f(\\frac{8}{11})=\\frac{656}{11}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=82x$.",
"robust... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=82$ fixes the function to $f(x)=82x$. |
math-016245 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Give an answer and a quick verification: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=545$, 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: $\\boxed{545x}$.\nBoth methods force linearity and use $f(1)=545$ to identify the slope. The extra datum $f(\\frac{-11}{10})=\\frac{-1199}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=545x$.... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=545$ fixes the function to $f(x)=545x$. (Here the result is $\boxed{545x}$.) |
math-016246 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Solve (and briefly cross-validate): Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{751x}$.\nBoth methods force linearity and use $f(1)=751$ to identify the slope. The extra datum $f(-3)=-2253$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=751x$.",
"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 $f(1)=751$ fixes the function to $f(x)=751x$. (Here the result is $\boxed{751x}$.) |
math-016247 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Use two approaches if possible: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-77... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-779x}$.\nBoth methods force linearity and use $f(1)=-779$ to identify the slope. The extra datum $f(\\frac{23}{8})=\\frac{-17917}{8}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-779x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-779$ fixes the function to $f(x)=-779x$. |
math-016248 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | 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 $f(1)=158$, 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: $\\boxed{158x}$.\nBoth methods force linearity and use $f(1)=158$ to identify the slope. The extra datum $f(\\frac{-1}{12})=\\frac{-79}{6}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=158x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=158$ fixes the function to $f(x)=158x$. (Here the result is $\boxed{158x}$.) |
math-016249 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Answer using clear logical steps: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-15$, 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: $\\boxed{-15x}$.\nBoth methods force linearity and use $f(1)=-15$ to identify the slope. The extra datum $f(8)=-120$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-15x$.",
"robustness_analysis": "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 $f(1)=-15$ fixes the function to $f(x)=-15x$. (Here the result is $\boxed{-15x}$.) |
math-016250 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Answer using clear logical steps: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-761$, 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": "Both approaches agree after simplification. Final answer: $\\boxed{-761x}$.\nBoth methods force linearity and use $f(1)=-761$ to identify the slope. The extra datum $f(9)=-6849$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-761x$.",
"robustness_analysis": ... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-761$ fixes the function to $f(x)=-761x$. |
math-016251 | Functional Equations: Additivity — Extension from Q to R | 9 | Write the solution set clearly: 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)=-19... | [
{
"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: $\\boxed{-190x}$.\nBoth methods force linearity and use $f(1)=-190$ to identify the slope. The extra datum $f(\\frac{-3}{11})=\\frac{570}{11}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-190x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-190$ fixes the function to $f(x)=-190x$. |
math-016252 | Functional Equations: Additivity — Extension from Q to R | 9 | 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 $f(1)=213$, 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: $\\boxed{213x}$.\nBoth methods force linearity and use $f(1)=213$ to identify the slope. The extra datum $f(\\frac{-13}{4})=\\frac{-2769}{4}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=213x$.",
"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 $f(1)=213$ fixes the function to $f(x)=213x$. (Here the result is $\boxed{213x}$.) |
math-016253 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Solve with verification: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-151$, 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": "Both approaches agree after simplification. Final answer: $\\boxed{-151x}$.\nBoth methods force linearity and use $f(1)=-151$ to identify the slope. The extra datum $f(6)=-906$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-151x$.",
"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 $f(1)=-151$ fixes the function to $f(x)=-151x$. |
math-016254 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Track units/moduli carefully: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-745$, 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: $\\boxed{-745x}$.\nBoth methods force linearity and use $f(1)=-745$ to identify the slope. The extra datum $f(-1)=745$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-745... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-745$ fixes the function to $f(x)=-745x$. |
math-016255 | Functional Equations: Additive Maps — Density Argument | 9 | 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)=-33... | [
{
"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: $\\boxed{-334x}$.\nBoth methods force linearity and use $f(1)=-334$ to identify the slope. The extra datum $f(\\frac{-25}{4})=\\frac{4175}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-334x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-334$ fixes the function to $f(x)=-334x$. (Here the result is $\boxed{-334x}$.) |
math-016256 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Indicate where a theorem is used: 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 $f(1)=-508$, 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: $\\boxed{-508x}$.\nBoth methods force linearity and use $f(1)=-508$ to identify the slope. The extra datum $f(\\frac{-9}{4})=1143$ 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... | 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-016257 | Functional Equations: Regularity Assumptions — Why Needed | 9 | 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 $f(1)=-426$, and tha... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-426x}$.\nBoth methods force linearity and use $f(1)=-426$ to identify the slope. The extra datum $f(\\frac{-23}{11})=\\frac{9798}{11}$ 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... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-426$ fixes the function to $f(x)=-426x$. |
math-016258 | Functional Equations: Regularity Assumptions — Why Needed | 9 | 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 $f(1)=726$, and that $f\!\left(\frac{-20... | [
{
"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: $\\boxed{726x}$.\nBoth methods force linearity and use $f(1)=726$ to identify the slope. The extra datum $f(\\frac{-10}{3})=-2420$ 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... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=726$ fixes the function to $f(x)=726x$. (Here the result is $\boxed{726x}$.) |
math-016259 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Be explicit about assumptions: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-664$, and that $f\!\left(\frac{... | [
{
"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: $\\boxed{-664x}$.\nBoth methods force linearity and use $f(1)=-664$ to identify the slope. The extra datum $f(\\frac{8}{9})=\\frac{-5312}{9}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both concl... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-664$ fixes the function to $f(x)=-664x$. (Here the result is $\boxed{-664x}$.) |
math-016260 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | 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 $f(1)=59$, and that $f\!\lef... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{59x}$.\nBoth methods force linearity and use $f(1)=59$ to identify the slope. The extra datum $f(\\frac{-7}{10})=\\frac{-413}{10}$ is consistent with $f(x)=kx$ and serves as a built-in check. Bot... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=59$ fixes the function to $f(x)=59x$. (Here the result is $\boxed{59x}$.) |
math-016261 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Provide both a computational and a conceptual explanation: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is contin... | [
{
"method_name": "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: $\\boxed{110x}$.\nBoth methods force linearity and use $f(1)=110$ to identify the slope. The extra datum $f(-4)=-440$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=110x$... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=110$ fixes the function to $f(x)=110x$. (Here the result is $\boxed{110x}$.) |
math-016262 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Warm-up: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-465$, and that $f\!\left(\frac{-1}{9}\right)=\frac{15... | [
{
"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: $\\boxed{-465x}$.\nBoth methods force linearity and use $f(1)=-465$ to identify the slope. The extra datum $f(\\frac{-1}{9})=\\frac{155}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Bo... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: 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 $f(x)=-465x$. |
math-016263 | Functional Equations: Additive Maps — Density Argument | 9 | Explain why your operations are valid: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{628x}$.\nBoth methods force linearity and use $f(1)=628$ to identify the slope. The extra datum $f(\\frac{9}{10})=\\frac{2826}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=628x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=628$ fixes the function to $f(x)=628x$. |
math-016264 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Compute the requested quantity: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-547$, 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: $\\boxed{-547x}$.\nBoth methods force linearity and use $f(1)=-547$ to identify the slope. The extra datum $f(\\frac{23}{2})=\\frac{-12581}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both con... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-547$ fixes the function to $f(x)=-547x$. (Here the result is $\boxed{-547x}$.) |
math-016265 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Solve and justify each step: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-636$, 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: $\\boxed{-636x}$.\nBoth methods force linearity and use $f(1)=-636$ to identify the slope. The extra datum $f(-1)=636$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-636x$.",
"... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-636$ fixes the function to $f(x)=-636x$. |
math-016266 | Functional Equations: Additivity — Extension from Q to R | 9 | Track units/moduli carefully: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-487$, and that $f\!\left(\frac{1... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-487x}$.\nBoth methods force linearity and use $f(1)=-487$ to identify the slope. The extra datum $f(2)=-974$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-487x$.",
"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... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-487$ fixes the function to $f(x)=-487x$. (Here the result is $\boxed{-487x}$.) |
math-016267 | Functional Equations: Additive Maps — Density Argument | 9 | Derive the result step-by-step: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-10... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-106x}$.\nBoth methods force linearity and use $f(1)=-106$ to identify the slope. The extra datum $f(\\frac{-15}{7})=\\frac{1590}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conc... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-106$ fixes the function to $f(x)=-106x$. (Here the result is $\boxed{-106x}$.) |
math-016268 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Exercise: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=292$, and that $f\!\left(\frac{12}{7}\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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{292x}$.\nBoth methods force linearity and use $f(1)=292$ to identify the slope. The extra datum $f(\\frac{12}{7})=\\frac{3504}{7}$ 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 $f(1)=292$ fixes the function to $f(x)=292x$. |
math-016269 | Functional Equations: Additive Maps — Density Argument | 9 | 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 $f(1)=368$, and that $f\!\left(\frac{-7}{12}\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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{368x}$.\nBoth methods force linearity and use $f(1)=368$ to identify the slope. The extra datum $f(\\frac{-7}{12})=\\frac{-644}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Bo... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=368$ fixes the function to $f(x)=368x$. (Here the result is $\boxed{368x}$.) |
math-016270 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Complete the analysis: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-31$, and that $f\!\left(\frac{-7}{2}\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: $\\boxed{-31x}$.\nBoth methods force linearity and use $f(1)=-31$ to identify the slope. The extra datum $f(\\frac{-7}{2})=\\frac{217}{2}$ 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... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-31$ fixes the function to $f(x)=-31x$. (Here the result is $\boxed{-31x}$.) |
math-016271 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Track quantifiers carefully: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=442$, and that $f\!\left(\fr... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{442x}$.\nBoth methods force linearity and use $f(1)=442$ to identify the slope. The extra datum $f(\\frac{-23}{5})=\\frac{-10166}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conc... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=442$ fixes the function to $f(x)=442x$. |
math-016272 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Provide a rigorous solution: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=65$, a... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{65x}$.\nBoth methods force linearity and use $f(1)=65$ to identify the slope. The extra datum $f(\\frac{8}{9})=\\frac{520}{9}$ 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... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=65$ fixes the function to $f(x)=65x$. (Here the result is $\boxed{65x}$.) |
math-016273 | Functional Equations: Additivity — Extension from Q to R | 9 | 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 $f(1)=-686$, 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: $\\boxed{-686x}$.\nBoth methods force linearity and use $f(1)=-686$ to identify the slope. The extra datum $f(\\frac{7}{5})=\\frac{-4802}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-686x$.",
"... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-686$ fixes the function to $f(x)=-686x$. |
math-016274 | Functional Equations: Additive Maps — Density Argument | 9 | Question: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=741$, and that $f\!\left(\frac{20}{10}\right)=1... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{741x}$.\nBoth methods force linearity and use $f(1)=741$ to identify the slope. The extra datum $f(2)=1482$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=741x$.",
"rob... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=741$ fixes the function to $f(x)=741x$. (Here the result is $\boxed{741x}$.) |
math-016275 | Functional Equations: Additivity — Extension from Q to R | 9 | 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 $f(1)=-3$, 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-3x}$.\nBoth methods force linearity and use $f(1)=-3$ to identify the slope. The extra datum $f(-1)=3$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-3x$.",
"robustne... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-3$ fixes the function to $f(x)=-3x$. |
math-016276 | Functional Equations: Additivity — Extension from Q to R | 9 | Provide both a computational and a conceptual explanation: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1... | [
{
"method_name": "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: $\\boxed{94x}$.\nBoth methods force linearity and use $f(1)=94$ to identify the slope. The extra datum $f(\\frac{-12}{5})=\\frac{-1128}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Bot... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=94$ fixes the function to $f(x)=94x$. |
math-016277 | Functional Equations: Additive Maps — Density Argument | 9 | Show all reasoning: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=392$, and that $f\!\left(\frac{18}{6}\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": "Both approaches agree after simplification. Final answer: $\\boxed{392x}$.\nBoth methods force linearity and use $f(1)=392$ to identify the slope. The extra datum $f(3)=1176$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=392x$.",
"robustness_analysis": "If ... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=392$ fixes the function to $f(x)=392x$. (Here the result is $\boxed{392x}$.) |
math-016278 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Provide a rigorous solution: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=327$, and that $f\!\left(\fr... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{327x}$.\nBoth methods force linearity and use $f(1)=327$ to identify the slope. The extra datum $f(\\frac{-4}{5})=\\frac{-1308}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=327x$.",
"ro... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=327$ fixes the function to $f(x)=327x$. |
math-016279 | Functional Equations: Additive Maps — Density Argument | 9 | 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 $f(1)=-240$, 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: $\\boxed{-240x}$.\nBoth methods force linearity and use $f(1)=-240$ to identify the slope. The extra datum $f(\\frac{13}{4})=-780$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-240x$.",
"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 $f(1)=-240$ fixes the function to $f(x)=-240x$. (Here the result is $\boxed{-240x}$.) |
math-016280 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Prompt: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-302$, 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: $\\boxed{-302x}$.\nBoth methods force linearity and use $f(1)=-302$ to identify the slope. The extra datum $f(\\frac{-5}{8})=\\frac{755}{4}$ 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 $f(1)=-302$ fixes the function to $f(x)=-302x$. (Here the result is $\boxed{-302x}$.) |
math-016281 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Answer using clear logical steps: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=650$, and that $f\!\lef... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{650x}$.\nBoth methods force linearity and use $f(1)=650$ to identify the slope. The extra datum $f(\\frac{-24}{7})=\\frac{-15600}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. ... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=650$ fixes the function to $f(x)=650x$. (Here the result is $\boxed{650x}$.) |
math-016282 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Exercise: 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 $f\!\left(\frac{-5}{3}\right)=\... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-731x}$.\nBoth methods force linearity and use $f(1)=-731$ to identify the slope. The extra datum $f(\\frac{-5}{3})=\\frac{3655}{3}$ 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 $f(1)=-731$ fixes the function to $f(x)=-731x$. |
math-016283 | Functional Equations: Additivity — Extension from Q to R | 9 | Solve and include a self-check: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=569$, 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: $\\boxed{569x}$.\nBoth methods force linearity and use $f(1)=569$ to identify the slope. The extra datum $f(\\frac{23}{2})=\\frac{13087}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Bo... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=569$ fixes the function to $f(x)=569x$. (Here the result is $\boxed{569x}$.) |
math-016284 | Functional Equations: Additivity — Extension from Q to R | 9 | 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 $f(1)=-537$, and that $f\!\left(\frac{... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-537x}$.\nBoth methods force linearity and use $f(1)=-537$ to identify the slope. The extra datum $f(\\frac{3}{4})=\\frac{-1611}{4}$ 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 $f(1)=-537$ fixes the function to $f(x)=-537x$. (Here the result is $\boxed{-537x}$.) |
math-016285 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Track quantifiers carefully: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=633$, and that $f\!\left(\fr... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{633x}$.\nBoth methods force linearity and use $f(1)=633$ to identify the slope. The extra datum $f(\\frac{-2}{3})=-422$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=633x$.",
"robustness_ana... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=633$ fixes the function to $f(x)=633x$. (Here the result is $\boxed{633x}$.) |
math-016286 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | 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 $f(1)=-489$, and that $f\!\left(\frac... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-489x}$.\nBoth methods force linearity and use $f(1)=-489$ to identify the slope. The extra datum $f(\\frac{9}{11})=\\frac{-4401}{11}$ is consistent with $f(x)=kx$ and serves as a built-in check.... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-489$ fixes the function to $f(x)=-489x$. |
math-016287 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Solve and justify each step: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-535$,... | [
{
"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: $\\boxed{-535x}$.\nBoth methods force linearity and use $f(1)=-535$ to identify the slope. The extra datum $f(\\frac{-12}{5})=1284$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-535x$.",
"robustness... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-535$ fixes the function to $f(x)=-535x$. (Here the result is $\boxed{-535x}$.) |
math-016288 | Functional Equations: Additive Maps — Density Argument | 9 | Question: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=307$, 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: $\\boxed{307x}$.\nBoth methods force linearity and use $f(1)=307$ to identify the slope. The extra datum $f(\\frac{11}{3})=\\frac{3377}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=307x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=307$ fixes the function to $f(x)=307x$. (Here the result is $\boxed{307x}$.) |
math-016289 | Functional Equations: Additivity — Extension from Q to R | 9 | Explain what is being counted/optimized: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=544$, 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: $\\boxed{544x}$.\nBoth methods force linearity and use $f(1)=544$ to identify the slope. The extra datum $f(\\frac{-11}{6})=\\frac{-2992}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both concl... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=544$ fixes the function to $f(x)=544x$. |
math-016290 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Do not skip justification steps: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-480$, 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": "The two methods are consistent and must coincide. Final answer: $\\boxed{-480x}$.\nBoth methods force linearity and use $f(1)=-480$ to identify the slope. The extra datum $f(\\frac{1}{2})=-240$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-480x$.",
"robust... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-480$ fixes the function to $f(x)=-480x$. (Here the result is $\boxed{-480x}$.) |
math-016291 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Solve with verification: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-401$, and that $f\!\left(\frac{20}{3}... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-401x}$.\nBoth methods force linearity and use $f(1)=-401$ to identify the slope. The extra datum $f(\\frac{20}{3})=\\frac{-8020}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-401x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(x)=-401x$. |
math-016292 | Functional Equations: Additivity — Extension from Q to R | 9 | Give a theorem-based solution: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=659$... | [
{
"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: $\\boxed{659x}$.\nBoth methods force linearity and use $f(1)=659$ to identify the slope. The extra datum $f(\\frac{20}{7})=\\frac{13180}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. Bo... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=659$ fixes the function to $f(x)=659x$. |
math-016293 | Functional Equations: Additive Maps — Density Argument | 9 | 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 $f(1)=245$, and that $f\!\left(\frac{-11}{3}\right)=\frac{-2695... | [
{
"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: $\\boxed{245x}$.\nBoth methods force linearity and use $f(1)=245$ to identify the slope. The extra datum $f(\\frac{-11}{3})=\\frac{-2695}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=245x$.",
"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... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=245$ fixes the function to $f(x)=245x$. |
math-016294 | Functional Equations: Additive Maps — Density Argument | 9 | Give a fully justified solution: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=523$, 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: $\\boxed{523x}$.\nBoth methods force linearity and use $f(1)=523$ to identify the slope. The extra datum $f(\\frac{-11}{8})=\\frac{-5753}{8}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=523x$."... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=523$ fixes the function to $f(x)=523x$. |
math-016295 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Be explicit about assumptions: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=382$, 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{382x}$.\nBoth methods force linearity and use $f(1)=382$ to identify the slope. The extra datum $f(1)=382$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=382x$.",
"robu... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=382$ fixes the function to $f(x)=382x$. |
math-016296 | Functional Equations: Additivity — Extension from Q to R | 9 | Proceed methodically: 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 $f(1)=-488$, and that $f\!\left(\frac{-4}{5}\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: $\\boxed{-488x}$.\nBoth methods force linearity and use $f(1)=-488$ to identify the slope. The extra datum $f(\\frac{-4}{5})=\\frac{1952}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both concl... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: 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 $f(x)=-488x$. |
math-016297 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | 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 $f(1)=-109$, and that $f\!\left(\frac{-1... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-109x}$.\nBoth methods force linearity and use $f(1)=-109$ to identify the slope. The extra datum $f(\\frac{-17}{2})=\\frac{1853}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-109x$... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-109$ fixes the function to $f(x)=-109x$. (Here the result is $\boxed{-109x}$.) |
math-016298 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Prompt: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=548$, and that $f\!\left(\frac{23}{3}\righ... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{548x}$.\nBoth methods force linearity and use $f(1)=548$ to identify the slope. The extra datum $f(\\frac{23}{3})=\\frac{12604}{3}$ 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... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=548$ fixes the function to $f(x)=548x$. |
math-016299 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Write the solution set clearly: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-762$, 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: $\\boxed{-762x}$.\nBoth methods force linearity and use $f(1)=-762$ to identify the slope. The extra datum $f(\\frac{10}{7})=\\frac{-7620}{7}$ 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 $f(1)=-762$ fixes the function to $f(x)=-762x$. |
math-016300 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Answer with a short justification: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=... | [
{
"method_name": "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: $\\boxed{159x}$.\nBoth methods force linearity and use $f(1)=159$ to identify the slope. The extra datum $f(\\frac{17}{7})=\\frac{2703}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=159x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=159$ fixes the function to $f(x)=159x$. (Here the result is $\boxed{159x}$.) |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.