# I want to show that $f(x)=x.f(1)$ where $f:R\to R$ is additive. [duplicate]

I know that if $f$ is continuous at one point then it is continuous at every point. From this i want to show that $f(x)=xf(1).$ Can anybody help me to proving this?

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## marked as duplicate by Arturo Magidin, Martin Sleziak, t.b., Leonid Kovalev, Jennifer Dylan Aug 17 '12 at 19:40

Start with integer $x$. Then try rational $x$. –  Hurkyl Apr 14 '12 at 4:13
The magic words are "Cauchy functional equation". See here and here. –  Arturo Magidin Apr 14 '12 at 4:16

HINTS:

1. Look at $0$ first: $f(0)=f(0+0)=f(0)+f(0)$, so $f(0)=0=0\cdot f(1)$.

2. Use induction to prove that $f(n)=nf(1)$ for every positive integer $n$, and use $f(0)=0$ to show that $f(n)=nf(1)$ for every negative integer as well.

3. $f(1)=f\left(\frac12+\frac12\right)=f\left(\frac13+\frac13+\frac13\right)=\dots\;$.

4. Once you’ve got it for $f\left(\frac1n\right)$, use the idea of (2) to get it for all rationals.

5. Then use continuity at a point.

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