1

In this answer by Hagen von Eitzen they said that, we can conclude that, $f(x)=xf(1)$ also holds for $x\in\mathbb{Q}$. How to show this?

I just tried like this: $f(q\frac{1}{q})=f(1)$ that implies $f(q)f(\frac{1}{q})=f(1)$ that implise $qf(1)f(\frac{1}{q})=f(1)$ But from this we can't conclude that, $f(1/q)=\frac{1}{q}f(1)$ unless we must already have $f(1)=1$.

user26857
  • 52,094
Akash Patalwanshi
  • 3,269
  • 5
    $f(1)=f(1^2)=f(1)f(1)$, so $f(1)$ equals 0 or 1. – Andrew Hubery Aug 21 '21 at 05:43
  • @AndrewHubery so when $f(1)=0$ then $f(x)=f(x1)=f(x)f(1)=f(x)0=0=xf(1)$ and hence $f(x)=xf(1)$ for all $x\in\mathbb{Q}$? – Akash Patalwanshi Aug 21 '21 at 06:46
  • 1
    See also https://math.stackexchange.com/questions/449404/is-an-automorphism-of-the-field-of-real-numbers-the-identity-map – lhf Aug 21 '21 at 10:08
  • If $f : G \to H$ is a group homomorphism (use additive notation on both $G$ and $H$), then $f(nx) = nf(x)$ for all $n \in \mathbb Z$ and $x \in G$. In your case, if $m, n \in \mathbb Z$ and $n \neq 0$, then $nf(m/n) = f(n(m/n)) = f(m) = mf(1)$, so dividing both sides by $n$ we get… – azif00 Aug 21 '21 at 19:41

0 Answers0