Let $f$ be a real function such that, for $x,y\in \mathbb{R}$ $f(x+y)=f(x)+f(y)$ Prove that (a) $f(0)=0$ (b) $f(n)=nf(1)$

Question

Let $f$ be a real function such that, for $x,y\in \mathbb{R}$ $$f(x+y)=f(x)+f(y)$$ Prove that

(a) $f(0)=0$

(b) $f(n)=nf(1)$

I have no clue how to prove this, would anyone give me some primer to start with?

Answer 1

Here some primers:

For (a), plug in $x=y=0$ into your equation.

For (b), plug in $x=y=1$, then $y=2,y=1$ and use the first case, e.t.c and try to find a pattern inductively.

Answer 2

Since $f(1)=f(0)+f(1)$, $f(0)=0$.

Then assuming $f(n-1)=(n-1)f(1)$, we have $f(n)=f(1)+f(n-1)=nf(1)$.

Answer 3

For part a) $f(0)=f(0)+f(0)=2f(0)$. If $f(0)=2f(0)$, then $f(0)=0$.

For part b)

We have that $f(2)=f(1+1)=f(1)+f(1)=2f(1)$

We have that $f(3)=f(2)+f(1)=2f(1)+f(1)=3f(1)$

Using induction, if $f(n)=nf(1)$, then $f(n+1)=f(n)+f(1)=nf(1)+f(1)=(n+1)f(1)$.

We also have $f(0)=f(1)+f(-1)$, so $f(-1)=-f(1)$.

We have $f(-1)=f(1)+f(-2)$, so $f(-2)=-f(1)-f(1)=-2f(1)$.

Then, if $f(-n)=-nf(1)$, then $f(-n-1)=f(-n)+f(-1)=-nf(1)-f(1)=(-n-1)f(1)$