id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-016101 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Prompt: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=719$, and that $f\!\left(\frac{-14}{6}\right)=\fr... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{719x}$.\nBoth methods force linearity and use $f(1)=719$ to identify the slope. The extra datum $f(\\frac{-7}{3})=\\frac{-5033}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=719x$.",
"ro... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=719$ fixes the function to $f(x)=719x$. (Here the result is $\boxed{719x}$.) |
math-016102 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Problem: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-182$, and that $f\!\left(\frac{25}{8}\right)=\frac{-2... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-182x}$.\nBoth methods force linearity and use $f(1)=-182$ to identify the slope. The extra datum $f(\\frac{25}{8})=\\frac{-2275}{4}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-182x$... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-182$ fixes the function to $f(x)=-182x$. (Here the result is $\boxed{-182x}$.) |
math-016103 | Functional Equations: Additive Maps — Density Argument | 9 | Track units/moduli carefully: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-374$, and that $f\!... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-374x}$.\nBoth methods force linearity and use $f(1)=-374$ to identify the slope. The extra datum $f(\\frac{19}{7})=\\frac{-7106}{7}$ 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 $f(1)=-374$ fixes the function to $f(x)=-374x$. (Here the result is $\boxed{-374x}$.) |
math-016104 | Functional Equations: Additive Maps — Density Argument | 9 | Give reasoning, not just computation: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-75$, and that $f\!... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-75x}$.\nBoth methods force linearity and use $f(1)=-75$ to identify the slope. The extra datum $f(\\frac{-18}{7})=\\frac{1350}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-75x$.",... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-75$ fixes the function to $f(x)=-75x$. |
math-016105 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Give reasoning, not just computation: 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 $f(1)=-374$, and that $f\!\left... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-374x}$.\nBoth methods force linearity and use $f(1)=-374$ to identify the slope. The extra datum $f(\\frac{-14}{5})=\\frac{5236}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-374x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-374$ fixes the function to $f(x)=-374x$. (Here the result is $\boxed{-374x}$.) |
math-016106 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Challenge: 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 $f(1)=-604$, 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": "The two methods are consistent and must coincide. Final answer: $\\boxed{-604x}$.\nBoth methods force linearity and use $f(1)=-604$ to identify the slope. The extra datum $f(\\frac{-3}{8})=\\frac{453}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-604x$."... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-604$ fixes the function to $f(x)=-604x$. (Here the result is $\boxed{-604x}$.) |
math-016107 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Show all reasoning: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=603$, and that $f\!\left(\frac{-5}{3}... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{603x}$.\nBoth methods force linearity and use $f(1)=603$ to identify the slope. The extra datum $f(\\frac{-5}{3})=-1005$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=60... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=603$ fixes the function to $f(x)=603x$. |
math-016108 | 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 $f(1)=4$, and that $f\!\left(\frac{16... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{4x}$.\nBoth methods force linearity and use $f(1)=4$ to identify the slope. The extra datum $f(\\frac{16}{7})=\\frac{64}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conc... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=4$ fixes the function to $f(x)=4x$. |
math-016109 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Question: 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 $f(1)=-637$, and that $f\!\left(\frac{3}{7}\right)=-273$.
... | [
{
"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: $\\boxed{-637x}$.\nBoth methods force linearity and use $f(1)=-637$ to identify the slope. The extra datum $f(\\frac{3}{7})=-273$ 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 $f(1)=-637$ fixes the function to $f(x)=-637x$. (Here the result is $\boxed{-637x}$.) |
math-016110 | Real Analysis: Additive Functions — Pathologies Avoided | 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 $f(1)=-543$, 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: $\\boxed{-543x}$.\nBoth methods force linearity and use $f(1)=-543$ to identify the slope. The extra datum $f(8)=-4344$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-543x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-543$ fixes the function to $f(x)=-543x$. (Here the result is $\boxed{-543x}$.) |
math-016111 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Checkpoint: 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 $f(1)=-256$, and that $f\!\left(\frac{-24}{9... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-256x}$.\nBoth methods force linearity and use $f(1)=-256$ to identify the slope. The extra datum $f(\\frac{-8}{3})=\\frac{2048}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-256x$.... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-256$ fixes the function to $f(x)=-256x$. (Here the result is $\boxed{-256x}$.) |
math-016112 | Real Analysis: Additive Functions — Pathologies Avoided | 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 $f(1)=-369$, 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: $\\boxed{-369x}$.\nBoth methods force linearity and use $f(1)=-369$ to identify the slope. The extra datum $f(\\frac{-17}{8})=\\frac{6273}{8}$ is consistent with $f(x)=kx$ and serves as a built-in check. ... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-369$ fixes the function to $f(x)=-369x$. (Here the result is $\boxed{-369x}$.) |
math-016113 | Functional Equations: Regularity Assumptions — Why Needed | 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 $f(1)=-404$, 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: $\\boxed{-404x}$.\nBoth methods force linearity and use $f(1)=-404$ to identify the slope. The extra datum $f(\\frac{-23}{9})=\\frac{9292}{9}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conc... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-404$ fixes the function to $f(x)=-404x$. |
math-016114 | Functional Equations: Regularity Assumptions — Why Needed | 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 $f(1)=362$, and that $f\!\left(\frac{-12}{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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{362x}$.\nBoth methods force linearity and use $f(1)=362$ to identify the slope. The extra datum $f(-2)=-724$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=362x$.",
"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 $f(1)=362$ fixes the function to $f(x)=362x$. (Here the result is $\boxed{362x}$.) |
math-016115 | Functional Equations: Additive Maps — Density Argument | 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 $f(1)=-728$, 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": "Both approaches agree after simplification. Final answer: $\\boxed{-728x}$.\nBoth methods force linearity and use $f(1)=-728$ to identify the slope. The extra datum $f(\\frac{-18}{5})=\\frac{13104}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-728x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-728$ fixes the function to $f(x)=-728x$. (Here the result is $\boxed{-728x}$.) |
math-016116 | Functional Equations: Additivity — Extension from Q to R | 9 | Work this out carefully: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=300$, 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: $\\boxed{300x}$.\nBoth methods force linearity and use $f(1)=300$ to identify the slope. The extra datum $f(3)=900$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=300x$.",
"robu... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=300$ fixes the function to $f(x)=300x$. (Here the result is $\boxed{300x}$.) |
math-016117 | Functional Equations: Additivity — Extension from Q to R | 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 $f(1)=795$, and that $f\!\lef... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{795x}$.\nBoth methods force linearity and use $f(1)=795$ to identify the slope. The extra datum $f(-1)=-795$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=795x$.",
"ro... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=795$ fixes the function to $f(x)=795x$. (Here the result is $\boxed{795x}$.) |
math-016118 | Functional Equations: Additive Maps — Density Argument | 9 | Make each step logically reversible (or explain if not): Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-54x}$.\nBoth methods force linearity and use $f(1)=-54$ to identify the slope. The extra datum $f(\\frac{-17}{8})=\\frac{459}{4}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-54x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-54$ fixes the function to $f(x)=-54x$. (Here the result is $\boxed{-54x}$.) |
math-016119 | Functional Equations: Additive Maps — Density Argument | 9 | Solve (and briefly cross-validate): Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=96$, 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: $\\boxed{96x}$.\nBoth methods force linearity and use $f(1)=96$ to identify the slope. The extra datum $f(\\frac{23}{9})=\\frac{736}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=96x$.",
"robustn... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=96$ fixes the function to $f(x)=96x$. (Here the result is $\boxed{96x}$.) |
math-016120 | Functional Equations: Additivity — Extension from Q to R | 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 $f(1)=-381$, and that $f\!\left(\frac{-1}{5}... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-381x}$.\nBoth methods force linearity and use $f(1)=-381$ to identify the slope. The extra datum $f(\\frac{-1}{5})=\\frac{381}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclu... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-381$ fixes the function to $f(x)=-381x$. (Here the result is $\boxed{-381x}$.) |
math-016121 | Functional Equations: Additive Maps — Density Argument | 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 $f(1)=-91$, 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": "The two methods are consistent and must coincide. Final answer: $\\boxed{-91x}$.\nBoth methods force linearity and use $f(1)=-91$ to identify the slope. The extra datum $f(1)=-91$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-91x$.",
"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 $f(1)=-91$ fixes the function to $f(x)=-91x$. (Here the result is $\boxed{-91x}$.) |
math-016122 | Real Analysis: Additive Functions — Pathologies Avoided | 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 $f(1)=-620$, 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: $\\boxed{-620x}$.\nBoth methods force linearity and use $f(1)=-620$ to identify the slope. The extra datum $f(\\frac{23}{2})=-7130$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclu... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-620$ fixes the function to $f(x)=-620x$. (Here the result is $\boxed{-620x}$.) |
math-016123 | Functional Equations: Cauchy — Continuity Implies Linearity | 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-155x}$.\nBoth methods force linearity and use $f(1)=-155$ to identify the slope. The extra datum $f(-1)=155$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-155x$.",
"... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-155$ fixes the function to $f(x)=-155x$. (Here the result is $\boxed{-155x}$.) |
math-016124 | Functional Equations: Additive Maps — Density Argument | 9 | Write the solution set clearly: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=501$, 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: $\\boxed{501x}$.\nBoth methods force linearity and use $f(1)=501$ to identify the slope. The extra datum $f(-2)=-1002$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=501x$.",
"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 $f(1)=501$ fixes the function to $f(x)=501x$. (Here the result is $\boxed{501x}$.) |
math-016125 | 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 $f(1)=-467$, and that $f\... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-467x}$.\nBoth methods force linearity and use $f(1)=-467$ to identify the slope. The extra datum $f(-5)=2335$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-467x$.",
"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... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-467$ fixes the function to $f(x)=-467x$. |
math-016126 | Functional Equations: Additivity — Extension from Q to R | 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 $f(1)=161$, and that $f\!\left(... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{161x}$.\nBoth methods force linearity and use $f(1)=161$ to identify the slope. The extra datum $f(\\frac{-9}{8})=\\frac{-1449}{8}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclu... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=161$ fixes the function to $f(x)=161x$. (Here the result is $\boxed{161x}$.) |
math-016127 | Functional Equations: Additive Maps — Density Argument | 9 | Give reasoning, not just computation: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=468$, and that $f\!... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{468x}$.\nBoth methods force linearity and use $f(1)=468$ to identify the slope. The extra datum $f(\\frac{11}{9})=572$ 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 $f(1)=468$ fixes the function to $f(x)=468x$. (Here the result is $\boxed{468x}$.) |
math-016128 | Functional Equations: Additive Maps — Density Argument | 9 | Explain what is being counted/optimized: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=535$, 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: $\\boxed{535x}$.\nBoth methods force linearity and use $f(1)=535$ to identify the slope. The extra datum $f(\\frac{-11}{4})=\\frac{-5885}{4}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=535x$.",
"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 $f(1)=535$ fixes the function to $f(x)=535x$. (Here the result is $\boxed{535x}$.) |
math-016129 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Give reasoning, not just computation: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=800$, and th... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{800x}$.\nBoth methods force linearity and use $f(1)=800$ to identify the slope. The extra datum $f(\\frac{3}{4})=600$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=800x$.",
"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... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=800$ fixes the function to $f(x)=800x$. (Here the result is $\boxed{800x}$.) |
math-016130 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Solve and justify each 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 $f(1)=365$, and that $f\!\l... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{365x}$.\nBoth methods force linearity and use $f(1)=365$ to identify the slope. The extra datum $f(10)=3650$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=365x$.",
"robustness_analysis": "Ro... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=365$ fixes the function to $f(x)=365x$. |
math-016131 | Functional Equations: Additivity — Extension from Q to R | 9 | Keep the final answer in boxed form: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-772$, and th... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-772x}$.\nBoth methods force linearity and use $f(1)=-772$ to identify the slope. The extra datum $f(\\frac{1}{2})=-386$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-7... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-772$ fixes the function to $f(x)=-772x$. (Here the result is $\boxed{-772x}$.) |
math-016132 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Where appropriate, name the theorem you use: 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 $f(1)=-222$, and that $f... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-222x}$.\nBoth methods force linearity and use $f(1)=-222$ to identify the slope. The extra datum $f(\\frac{-17}{8})=\\frac{1887}{4}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-222x$... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-222$ fixes the function to $f(x)=-222x$. (Here the result is $\boxed{-222x}$.) |
math-016133 | 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{377x}$.\nBoth methods force linearity and use $f(1)=377$ to identify the slope. The extra datum $f(4)=1508$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=377x$.... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=377$ fixes the function to $f(x)=377x$. |
math-016134 | Functional Equations: Cauchy — Continuity Implies Linearity | 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 $f(1)=541$, 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: $\\boxed{541x}$.\nBoth methods force linearity and use $f(1)=541$ to identify the slope. The extra datum $f(\\frac{-22}{5})=\\frac{-11902}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=541x$.... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=541$ fixes the function to $f(x)=541x$. |
math-016135 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Checkpoint: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=799$, and that $f\!\left(\frac{18}{4}\... | [
{
"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: $\\boxed{799x}$.\nBoth methods force linearity and use $f(1)=799$ to identify the slope. The extra datum $f(\\frac{9}{2})=\\frac{7191}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=799x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=799$ fixes the function to $f(x)=799x$. |
math-016136 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Solve and justify each 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 $f(1)=21$, 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: $\\boxed{21x}$.\nBoth methods force linearity and use $f(1)=21$ to identify the slope. The extra datum $f(3)=63$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=21x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=21$ fixes the function to $f(x)=21x$. (Here the result is $\boxed{21x}$.) |
math-016137 | Functional Equations: Cauchy — Continuity Implies Linearity | 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 $f(1)=313$, and that $f\!\left(\frac{-20}{5}\right)=-1252$.
... | [
{
"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: $\\boxed{313x}$.\nBoth methods force linearity and use $f(1)=313$ to identify the slope. The extra datum $f(-4)=-1252$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=313x$.",
"robustness_analysis": "I... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=313$ fixes the function to $f(x)=313x$. (Here the result is $\boxed{313x}$.) |
math-016138 | Functional Equations: Additivity — Extension from Q to R | 9 | Prompt: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=225$, 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: $\\boxed{225x}$.\nBoth methods force linearity and use $f(1)=225$ to identify the slope. The extra datum $f(-1)=-225$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=225x$.",
"ro... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=225$ fixes the function to $f(x)=225x$. (Here the result is $\boxed{225x}$.) |
math-016139 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Find the exact value: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-707$, and that $f\!\left(\frac{18}... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-707x}$.\nBoth methods force linearity and use $f(1)=-707$ to identify the slope. The extra datum $f(6)=-4242$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-70... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-707$ fixes the function to $f(x)=-707x$. |
math-016140 | Functional Equations: Regularity Assumptions — Why Needed | 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 $f(1)=-109$, and that $f\!\... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-109x}$.\nBoth methods force linearity and use $f(1)=-109$ to identify the slope. The extra datum $f(\\frac{-4}{3})=\\frac{436}{3}$ 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-016141 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Keep the final answer in boxed form: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{502x}$.\nBoth methods force linearity and use $f(1)=502$ to identify the slope. The extra datum $f(\\frac{-17}{4})=\\frac{-4267}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. B... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=502$ fixes the function to $f(x)=502x$. |
math-016142 | Functional Equations: Cauchy — Continuity Implies Linearity | 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 $f(1)=-158$, 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: $\\boxed{-158x}$.\nBoth methods force linearity and use $f(1)=-158$ to identify the slope. The extra datum $f(\\frac{5}{2})=-395$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-158$ fixes the function to $f(x)=-158x$. (Here the result is $\boxed{-158x}$.) |
math-016143 | 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)=-24... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-245x}$.\nBoth methods force linearity and use $f(1)=-245$ to identify the slope. The extra datum $f(\\frac{-1}{4})=\\frac{245}{4}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-245x$."... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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-016144 | 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 $f(1)=-168$, and that $f\!\left(\frac{2}{6}\right)=-56... | [
{
"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: $\\boxed{-168x}$.\nBoth methods force linearity and use $f(1)=-168$ to identify the slope. The extra datum $f(\\frac{1}{3})=-56$ 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 $f(1)=-168$ fixes the function to $f(x)=-168x$. |
math-016145 | Functional Equations: Regularity Assumptions — Why Needed | 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 $f(1)=523$, and that $f\!\left(\frac{... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{523x}$.\nBoth methods force linearity and use $f(1)=523$ to identify the slope. The extra datum $f(\\frac{23}{5})=\\frac{12029}{5}$ 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-016146 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Explain what is being counted/optimized: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=661$, and... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{661x}$.\nBoth methods force linearity and use $f(1)=661$ to identify the slope. The extra datum $f(\\frac{7}{10})=\\frac{4627}{10}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=661x$.",... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=661$ fixes the function to $f(x)=661x$. (Here the result is $\boxed{661x}$.) |
math-016147 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Warm-up: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-365$, and that $f\!\left(\frac{-3}{8}\right)=\f... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-365x}$.\nBoth methods force linearity and use $f(1)=-365$ to identify the slope. The extra datum $f(\\frac{-3}{8})=\\frac{1095}{8}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-365x$.",
"... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-365$ fixes the function to $f(x)=-365x$. |
math-016148 | Functional Equations: Additive Maps — Density Argument | 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 $f(1)=-711$, 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: $\\boxed{-711x}$.\nBoth methods force linearity and use $f(1)=-711$ to identify the slope. The extra datum $f(\\frac{-10}{3})=2370$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-711x$.",
"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 $f(1)=-711$ fixes the function to $f(x)=-711x$. (Here the result is $\boxed{-711x}$.) |
math-016149 | Functional Equations: Additive Maps — Density Argument | 9 | Give reasoning, not just computation: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-471$, and t... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-471x}$.\nBoth methods force linearity and use $f(1)=-471$ to identify the slope. The extra datum $f(\\frac{-21}{8})=\\frac{9891}{8}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-471x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-471$ fixes the function to $f(x)=-471x$. |
math-016150 | 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 $f(1)=689$,... | [
{
"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: $\\boxed{689x}$.\nBoth methods force linearity and use $f(1)=689$ to identify the slope. The extra datum $f(\\frac{20}{11})=\\frac{13780}{11}$ is consistent with $f(x)=kx$ and serves as a built-in check. ... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=689$ fixes the function to $f(x)=689x$. (Here the result is $\boxed{689x}$.) |
math-016151 | 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 $f(1)=-113$, and that $f\!\left(\frac{... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-113x}$.\nBoth methods force linearity and use $f(1)=-113$ to identify the slope. The extra datum $f(\\frac{-8}{7})=\\frac{904}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-113x$.",
"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 $f(1)=-113$ fixes the function to $f(x)=-113x$. (Here the result is $\boxed{-113x}$.) |
math-016152 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Make each step logically reversible (or explain if not): Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-433$,... | [
{
"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: $\\boxed{-433x}$.\nBoth methods force linearity and use $f(1)=-433$ to identify the slope. The extra datum $f(\\frac{7}{5})=\\frac{-3031}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both concl... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-433$ fixes the function to $f(x)=-433x$. (Here the result is $\boxed{-433x}$.) |
math-016153 | 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 $f(1)=-756$, 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-756x}$.\nBoth methods force linearity and use $f(1)=-756$ to identify the slope. The extra datum $f(\\frac{5}{4})=-945$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-7... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-756$ fixes the function to $f(x)=-756x$. (Here the result is $\boxed{-756x}$.) |
math-016154 | Functional Equations: Cauchy — Continuity Implies Linearity | 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 $f(1)=-653$, 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: $\\boxed{-653x}$.\nBoth methods force linearity and use $f(1)=-653$ to identify the slope. The extra datum $f(\\frac{-8}{7})=\\frac{5224}{7}$ 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... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-653$ fixes the function to $f(x)=-653x$. (Here the result is $\boxed{-653x}$.) |
math-016155 | 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: $\\boxed{-643x}$.\nBoth methods force linearity and use $f(1)=-643$ to identify the slope. The extra datum $f(\\frac{-24}{5})=\\frac{15432}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-643x... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-643$ fixes the function to $f(x)=-643x$. |
math-016156 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Solve and sanity-check: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=28$, and that $f\!\left(\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": "The two methods are consistent and must coincide. Final answer: $\\boxed{28x}$.\nBoth methods force linearity and use $f(1)=28$ to identify the slope. The extra datum $f(\\frac{-3}{2})=-42$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=28x$.",
"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 $f(1)=28$ fixes the function to $f(x)=28x$. (Here the result is $\boxed{28x}$.) |
math-016157 | Functional Equations: Regularity Assumptions — Why Needed | 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 $f(1)=-96$, 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: $\\boxed{-96x}$.\nBoth methods force linearity and use $f(1)=-96$ to identify the slope. The extra datum $f(\\frac{13}{12})=-104$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-96$ fixes the function to $f(x)=-96x$. (Here the result is $\boxed{-96x}$.) |
math-016158 | Functional Equations: Additivity — Extension from Q to R | 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": "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: $\\boxed{-775x}$.\nBoth methods force linearity and use $f(1)=-775$ to identify the slope. The extra datum $f(\\frac{19}{11})=\\frac{-14725}{11}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-77... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-775$ fixes the function to $f(x)=-775x$. (Here the result is $\boxed{-775x}$.) |
math-016159 | Functional Equations: Cauchy — Continuity Implies Linearity | 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 $f(1)=173$, 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: $\\boxed{173x}$.\nBoth methods force linearity and use $f(1)=173$ to identify the slope. The extra datum $f(\\frac{-20}{9})=\\frac{-3460}{9}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both concl... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=173$ fixes the function to $f(x)=173x$. (Here the result is $\boxed{173x}$.) |
math-016160 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Compute the requested quantity: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=88$, and that $f\!... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{88x}$.\nBoth methods force linearity and use $f(1)=88$ to identify the slope. The extra datum $f(\\frac{-5}{4})=-110$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=88x$.",
"robustness_... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=88$ fixes the function to $f(x)=88x$. |
math-016161 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Try to avoid pattern-matching; explain why: 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 $0$, th... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{288x}$.\nBoth methods force linearity and use $f(1)=288$ to identify the slope. The extra datum $f(\\frac{17}{8})=612$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=288x$.",
"robustness_anal... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=288$ fixes the function to $f(x)=288x$. (Here the result is $\boxed{288x}$.) |
math-016162 | Functional Equations: Additive Maps — Density Argument | 9 | Start by stating any domain restrictions: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=657$, 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: $\\boxed{657x}$.\nBoth methods force linearity and use $f(1)=657$ to identify the slope. The extra datum $f(\\frac{15}{7})=\\frac{9855}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=657x$.",
"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 $f(1)=657$ fixes the function to $f(x)=657x$. |
math-016163 | Functional Equations: Additive Maps — Density Argument | 9 | Checkpoint: 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 $f(1)=-171$, and that $f\!\left(\frac{3}{4}\... | [
{
"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: $\\boxed{-171x}$.\nBoth methods force linearity and use $f(1)=-171$ to identify the slope. The extra datum $f(\\frac{3}{4})=\\frac{-513}{4}$ 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... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-171$ fixes the function to $f(x)=-171x$. (Here the result is $\boxed{-171x}$.) |
math-016164 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Checkpoint: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-510$, and that $f\!\left(\frac{2}{8}\right)=... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-510x}$.\nBoth methods force linearity and use $f(1)=-510$ to identify the slope. The extra datum $f(\\frac{1}{4})=\\frac{-255}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-510x$.",
"r... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-510$ fixes the function to $f(x)=-510x$. (Here the result is $\boxed{-510x}$.) |
math-016165 | Functional Equations: Additive Maps — Density Argument | 9 | Checkpoint: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=759$, 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: $\\boxed{759x}$.\nBoth methods force linearity and use $f(1)=759$ to identify the slope. The extra datum $f(\\frac{-19}{5})=\\frac{-14421}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. ... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=759$ fixes the function to $f(x)=759x$. |
math-016166 | Real Analysis: Additive Functions — Pathologies Avoided | 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: $\\boxed{-273x}$.\nBoth methods force linearity and use $f(1)=-273$ to identify the slope. The extra datum $f(\\frac{9}{4})=\\frac{-2457}{4}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-273x$.... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-273$ fixes the function to $f(x)=-273x$. (Here the result is $\boxed{-273x}$.) |
math-016167 | Functional Equations: Additive Maps — Density Argument | 9 | Explain what is being counted/optimized: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=542$, and... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{542x}$.\nBoth methods force linearity and use $f(1)=542$ to identify the slope. The extra datum $f(\\frac{11}{9})=\\frac{5962}{9}$ 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 $f(1)=542$ fixes the function to $f(x)=542x$. (Here the result is $\boxed{542x}$.) |
math-016168 | Functional Equations: Additivity — Extension from Q to R | 9 | Provide a rigorous solution: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=288$, and that $f\!\left(\frac{-2}... | [
{
"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: $\\boxed{288x}$.\nBoth methods force linearity and use $f(1)=288$ to identify the slope. The extra datum $f(\\frac{-2}{7})=\\frac{-576}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=288x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=288$ fixes the function to $f(x)=288x$. |
math-016169 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Compute the requested quantity: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-235$, 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: $\\boxed{-235x}$.\nBoth methods force linearity and use $f(1)=-235$ to identify the slope. The extra datum $f(\\frac{1}{10})=\\frac{-47}{2}$ 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 $f(1)=-235$ fixes the function to $f(x)=-235x$. (Here the result is $\boxed{-235x}$.) |
math-016170 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Derive the result step-by-step: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=299$, 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: $\\boxed{299x}$.\nBoth methods force linearity and use $f(1)=299$ to identify the slope. The extra datum $f(\\frac{1}{9})=\\frac{299}{9}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude ... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=299$ fixes the function to $f(x)=299x$. |
math-016171 | Functional Equations: Additive Maps — Density Argument | 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": "Both approaches agree after simplification. Final answer: $\\boxed{410x}$.\nBoth methods force linearity and use $f(1)=410$ to identify the slope. The extra datum $f(\\frac{18}{7})=\\frac{7380}{7}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=410x$.",
"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 $f(1)=410$ fixes the function to $f(x)=410x$. (Here the result is $\boxed{410x}$.) |
math-016172 | Functional Equations: Additivity — Extension from Q to R | 9 | Solve and then verify: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-210$, and that $f\!\left(\frac{20}{11}\... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-210x}$.\nBoth methods force linearity and use $f(1)=-210$ to identify the slope. The extra datum $f(\\frac{20}{11})=\\frac{-4200}{11}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-210x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-210$ fixes the function to $f(x)=-210x$. (Here the result is $\boxed{-210x}$.) |
math-016173 | Functional Equations: Additivity — Extension from Q to R | 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 $f(1)=57$, and that $f\!\l... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{57x}$.\nBoth methods force linearity and use $f(1)=57$ to identify the slope. The extra datum $f(\\frac{1}{4})=\\frac{57}{4}$ 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... | Key idea: 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-016174 | Functional Equations: Regularity Assumptions — Why Needed | 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 $f(1)=759$, and that $f\!\left(... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{759x}$.\nBoth methods force linearity and use $f(1)=759$ to identify the slope. The extra datum $f(\\frac{-7}{5})=\\frac{-5313}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclu... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=759$ fixes the function to $f(x)=759x$. (Here the result is $\boxed{759x}$.) |
math-016175 | Functional Equations: Additivity — Extension from Q to R | 9 | Answer using clear logical steps: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=318$, 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: $\\boxed{318x}$.\nBoth methods force linearity and use $f(1)=318$ to identify the slope. The extra datum $f(\\frac{9}{2})=1431$ 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 $f(1)=318$ fixes the function to $f(x)=318x$. (Here the result is $\boxed{318x}$.) |
math-016176 | Functional Equations: Additivity — Extension from Q to R | 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 $f(1)=-478$, and that $f\!\left(\frac{19}{12}\rig... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-478x}$.\nBoth methods force linearity and use $f(1)=-478$ to identify the slope. The extra datum $f(\\frac{19}{12})=\\frac{-4541}{6}$ is consistent with $f(x)=kx$ and serves as a built-in check.... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-478$ fixes the function to $f(x)=-478x$. |
math-016177 | Functional Equations: Additive Maps — Density Argument | 9 | Solve with verification: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=435$, and that $f\!\left(... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{435x}$.\nBoth methods force linearity and use $f(1)=435$ to identify the slope. The extra datum $f(\\frac{-9}{5})=-783$ 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 $f(1)=435$ fixes the function to $f(x)=435x$. (Here the result is $\boxed{435x}$.) |
math-016178 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Work carefully and justify each inference: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=438$, a... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{438x}$.\nBoth methods force linearity and use $f(1)=438$ to identify the slope. The extra datum $f(\\frac{3}{2})=657$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=438x$.",
"robustness... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=438$ fixes the function to $f(x)=438x$. (Here the result is $\boxed{438x}$.) |
math-016179 | Functional Equations: Regularity Assumptions — Why Needed | 9 | Derive the result step-by-step: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=46$, 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: $\\boxed{46x}$.\nBoth methods force linearity and use $f(1)=46$ to identify the slope. The extra datum $f(-1)=-46$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=46x$.",
"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 $f(1)=46$ fixes the function to $f(x)=46x$. (Here the result is $\boxed{46x}$.) |
math-016180 | Functional Equations: Regularity Assumptions — Why Needed | 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 $f(1)=353$... | [
{
"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: $\\boxed{353x}$.\nBoth methods force linearity and use $f(1)=353$ to identify the slope. The extra datum $f(\\frac{-15}{8})=\\frac{-5295}{8}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=353x$.",
"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 $f(1)=353$ fixes the function to $f(x)=353x$. (Here the result is $\boxed{353x}$.) |
math-016181 | Real Analysis: Additive Functions — Pathologies Avoided | 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 $f(1)=-605$, 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: $\\boxed{-605x}$.\nBoth methods force linearity and use $f(1)=-605$ to identify the slope. The extra datum $f(\\frac{1}{3})=\\frac{-605}{3}$ 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... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-605$ fixes the function to $f(x)=-605x$. (Here the result is $\boxed{-605x}$.) |
math-016182 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Answer using clear logical 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 $f(1)=524$, and that $f\!\left(\fra... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{524x}$.\nBoth methods force linearity and use $f(1)=524$ to identify the slope. The extra datum $f(\\frac{-3}{2})=-786$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=524x$.",
"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... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=524$ fixes the function to $f(x)=524x$. (Here the result is $\boxed{524x}$.) |
math-016183 | Real Analysis: Additive Functions — Pathologies Avoided | 9 | Determine the requested value: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=762$, 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": "The two methods are consistent and must coincide. Final answer: $\\boxed{762x}$.\nBoth methods force linearity and use $f(1)=762$ to identify the slope. The extra datum $f(\\frac{25}{9})=\\frac{6350}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=762x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=762$ fixes the function to $f(x)=762x$. (Here the result is $\boxed{762x}$.) |
math-016184 | 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 $f(1)=-711$, 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: $\\boxed{-711x}$.\nBoth methods force linearity and use $f(1)=-711$ to identify the slope. The extra datum $f(\\frac{8}{11})=\\frac{-5688}{11}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-711x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-711$ fixes the function to $f(x)=-711x$. (Here the result is $\boxed{-711x}$.) |
math-016185 | Functional Equations: Additive Maps — Density Argument | 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 $f(1)=-500$, 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: $\\boxed{-500x}$.\nBoth methods force linearity and use $f(1)=-500$ to identify the slope. The extra datum $f(\\frac{15}{2})=-3750$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclu... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-500$ fixes the function to $f(x)=-500x$. |
math-016186 | Functional Equations: Additive Maps — Density Argument | 9 | Determine the requested value: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-461$, and that $f\!\left(\frac{... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{-461x}$.\nBoth methods force linearity and use $f(1)=-461$ to identify the slope. The extra datum $f(\\frac{-13}{10})=\\frac{5993}{10}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-461x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-461$ fixes the function to $f(x)=-461x$. (Here the result is $\boxed{-461x}$.) |
math-016187 | Functional Equations: Cauchy — Continuity Implies Linearity | 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 $f(1)=-749$, 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": "Both approaches agree after simplification. Final answer: $\\boxed{-749x}$.\nBoth methods force linearity and use $f(1)=-749$ to identify the slope. The extra datum $f(\\frac{-13}{11})=\\frac{9737}{11}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-749x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-749$ fixes the function to $f(x)=-749x$. |
math-016188 | Functional Equations: Cauchy — Continuity Implies Linearity | 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 $f(1)=279$, and that $f\!\left(\frac{-4}{4}\... | [
{
"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: $\\boxed{279x}$.\nBoth methods force linearity and use $f(1)=279$ to identify the slope. The extra datum $f(-1)=-279$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=279x$... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=279$ fixes the function to $f(x)=279x$. |
math-016189 | Functional Equations: Additive Maps — Density Argument | 9 | Give reasoning, not just computation: 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 $f(1)=246$, 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: $\\boxed{246x}$.\nBoth methods force linearity and use $f(1)=246$ to identify the slope. The extra datum $f(\\frac{2}{5})=\\frac{492}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude ... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=246$ fixes the function to $f(x)=246x$. (Here the result is $\boxed{246x}$.) |
math-016190 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Warm-up: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-209$, and that $f\!\left(\frac{16}{6}\right)=\f... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-209x}$.\nBoth methods force linearity and use $f(1)=-209$ to identify the slope. The extra datum $f(\\frac{8}{3})=\\frac{-1672}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. B... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Takeaway: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-209$ fixes the function to $f(x)=-209x$. (Here the result is $\boxed{-209x}$.) |
math-016191 | Functional Equations: Additivity — Extension from Q to R | 9 | Find the exact value: Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-434$, and that $f\!\left(\f... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-434x}$.\nBoth methods force linearity and use $f(1)=-434$ to identify the slope. The extra datum $f(\\frac{-11}{12})=\\frac{2387}{6}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-434x... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-434$ fixes the function to $f(x)=-434x$. |
math-016192 | Functional Equations: Additivity — Extension from Q to R | 9 | Derive the result step-by-step: Given an additive function with a regularity hypothesis, show it must be of the form $f(x)=kx$:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=708$, 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: $\\boxed{708x}$.\nBoth methods force linearity and use $f(1)=708$ to identify the slope. The extra datum $f(\\frac{6}{11})=\\frac{4248}{11}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclu... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=708$ fixes the function to $f(x)=708x$. (Here the result is $\boxed{708x}$.) |
math-016193 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Answer using clear logical 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 $f(1)=-279$, 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: $\\boxed{-279x}$.\nBoth methods force linearity and use $f(1)=-279$ to identify the slope. The extra datum $f(\\frac{5}{2})=\\frac{-1395}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-279x$.",
"... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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 $f(1)=-279$ fixes the function to $f(x)=-279x$. |
math-016194 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Keep the final answer in boxed form: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{-366x}$.\nBoth methods force linearity and use $f(1)=-366$ to identify the slope. The extra datum $f(\\frac{3}{4})=\\frac{-549}{2}$ is consistent with $f(x)=kx$ and serves as a built-in check. Bo... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-366$ fixes the function to $f(x)=-366x$. (Here the result is $\boxed{-366x}$.) |
math-016195 | Functional Equations: Additivity — Extension from Q to R | 9 | Carefully track domains: Determine $f(x)$ for all real $x$. Your proof must show (a) rational values, (b) extension to reals using continuity:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-65$, and ... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{-65x}$.\nBoth methods force linearity and use $f(1)=-65$ to identify the slope. The extra datum $f(\\frac{5}{3})=\\frac{-325}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-65x$.",
... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion 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-016196 | 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 $f(1)=-92$, and that $f\!\left(\frac{1... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-92x}$.\nBoth methods force linearity and use $f(1)=-92$ to identify the slope. The extra datum $f(\\frac{1}{12})=\\frac{-23}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-92$ fixes the function to $f(x)=-92x$. (Here the result is $\boxed{-92x}$.) |
math-016197 | 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 $f(1)=-356$, 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: $\\boxed{-356x}$.\nBoth methods force linearity and use $f(1)=-356$ to identify the slope. The extra datum $f(\\frac{-1}{3})=\\frac{356}{3}$ 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... | Core principle: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-356$ fixes the function to $f(x)=-356x$. (Here the result is $\boxed{-356x}$.) |
math-016198 | Functional Equations: Additivity — Extension from Q to R | 9 | Keep the final answer in boxed form: Cauchy functional equation with regularity. Prove the function is linear and identify it:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that $f(1)=-660$, 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: $\\boxed{-660x}$.\nBoth methods force linearity and use $f(1)=-660$ to identify the slope. The extra datum $f(\\frac{-20}{11})=1200$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conclude $f(x)=-660x$.",
"rob... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-660$ fixes the function to $f(x)=-660x$. |
math-016199 | Functional Equations: Cauchy — Continuity Implies Linearity | 9 | Make each step logically reversible (or explain if not): Solve the functional equation and include a short explanation of why continuity is the key hypothesis:
Let $f:\mathbb{R}\to\mathbb{R}$ satisfy Cauchy's equation $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Assume additionally that $f$ is continuous at $0$, that... | [
{
"method_name": "Boundedness Near 0 ⇒ Linearity",
"approach": "Continuity at 0 implies boundedness on a neighborhood; an additive function bounded on any interval must be linear, then $f(1)=k$ pins down the slope.",
"steps": [
"Step 1: By continuity at 0, there exists $\\delta>0$ such that $|x|<\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{-371x}$.\nBoth methods force linearity and use $f(1)=-371$ to identify the slope. The extra datum $f(\\frac{-21}{5})=\\frac{7791}{5}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both conc... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Remember: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=-371$ fixes the function to $f(x)=-371x$. |
math-016200 | Functional Equations: Additivity — Extension from Q to R | 9 | Solve and then verify: 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 $f(1)=474$, and that $f\!\left(\f... | [
{
"method_name": "Rationals + Density + Continuity",
"approach": "Use additivity to determine values on integers then rationals, then extend to all reals via continuity and density of the rationals.",
"steps": [
"Step 1: Put $x=y=0$ to get $f(0)=f(0)+f(0)$, hence $f(0)=0$.",
"Step 2: For int... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{474x}$.\nBoth methods force linearity and use $f(1)=474$ to identify the slope. The extra datum $f(\\frac{1}{9})=\\frac{158}{3}$ is consistent with $f(x)=kx$ and serves as a built-in check. Both ... | [
{
"error_description": "Concluded $f(x)=kx$ from $f(1)=k$ without explaining why irrationals are determined.",
"why_plausible": "On integers the conclusion is immediate, so it feels like it should extend automatically.",
"why_wrong": "Additivity alone does not determine $f$ on $\\mathbb{R}$; you must us... | Key idea: Cauchy additivity has wild solutions unless you assume regularity. Continuity at 0 forces $f$ to be linear, and $f(1)=474$ fixes the function to $f(x)=474x$. |
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