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Having trouble with the last part of my proof:

Let f: $\mathbb{Z}\rightarrow \mathbb{Z}$ be a function with $f(x+y)=f(x)+f(y)$ for all $x,y \in \mathbb{Z}$. Prove there exists an element $a\in \mathbb{Z}$ such that $f(n)=na$ for all $n \in \mathbb{Z}$.

What do I do? I was thinking of combining $na$ into $q$ and then $f(n)=q$.

But then that means that $n$ divides $q$. Where do I start?

So if my base case is n=1, then the induction hypothesis must prove true for f(k) = ka. Then in my induction step f(k+1)=(k+1)(a) and what do I do next? I feel like I proved it because there does exist an a.

Shawn S. Rana
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  • So if my base case is n=1, then the induction hypothesis must prove true for f(k) = ka. Then in my induction step f(k+1)=(k+1)(a) and what do I do next? I feel like I proved it because there does exist an a. – Shawn S. Rana Apr 20 '14 at 17:18

1 Answers1

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Hint Let $a = f(1)$. Then $f(n) = na$ holds for $n = 1$. Use that as a base case in an induction proof.

Guest
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  • So if my base case is n=1, then the induction hypothesis must prove true for f(k) = ka. Then in my induction step f(k+1)=(k+1)(a) and what do I do next? I feel like I proved it because there does exist an a. – Shawn S. Rana Apr 20 '14 at 17:19
  • For the induction hypothesis you assume that $f(n) = na$ (for just that particular $n$, not for all $n$). Then you use $f(x + y) = f(x) + f(y)$ to show that $f(n + 1) = (n + 1)a$. After you've done that you're done with the induction and you say that therefore $f(n) = na$ holds for all $n$. – Guest Apr 20 '14 at 17:25