5

A bijection $\phi:\mathbb R \to \mathbb R$ that satisfies:

$$\phi(a+b)=\phi(a) +\phi(b)\text{, }$$ $$\phi(ab)=\phi(a)\phi(b)$$ $$\forall a,b\in \mathbb R$$

must be the identiy function. What I've done is to show the obvious $\phi(0)=0$ and that the function is the identity $\forall a,b\in \mathbb Z$, which follows from $\phi(\underbrace{a+\dots +a}_n)=n\phi(a)=\phi(a)\phi(n) \implies \phi(n)=n \text{ }\forall n\in \mathbb N$. Now it is easy to show that $\phi(-1)=-1$ and to extend the identity to the integers. I can't seem to find a way to the reals.

blundered_bishop
  • 946
  • 2
    For what it's worth, though I'm not sure it's quite at your level given your phrasing, this kind of function $\phi$ is a field homomorphism. Or in particular, a field isomorphism, since it's bijective, or even more specifically an automorphism, since it's from a field to itself. So what you're looking to prove is that only field automorphism on $\Bbb R$ is that of $\phi(x) = x$. You can see an argument for this here: link – PrincessEev Oct 19 '19 at 22:27
  • See https://math.stackexchange.com/questions/449404/is-an-automorphism-of-the-field-of-real-numbers-the-identity-map – James Oct 19 '19 at 22:28
  • 1
    Show $\phi(q)=q$ for rational $q$, then show $\phi$ is nonnegative at nonnegative reals, and deduce that $\phi$ is nondecreasing. – Aphelli Oct 19 '19 at 22:28
  • 2
    Possible duplicate of Is an automorphism of the field of real numbers the identity map? –  Oct 19 '19 at 22:32

2 Answers2

5

From what you've done, it's easy to see that $\phi(q) = q$ for any rational number $q$.

Now the key point is to show that $\phi$ is monotone.

In fact, if $x\in\mathbb{R}$ is non-negative, then there exists $y\in\mathbb{R}$ such that $x = y^2$. It follows that $\phi(x) = \phi(y)^2 \geq 0$. Thus we have shown that $\phi$ maps non-negative real numbers to non-negative real numbers.

Hence if $u \geq v$ are real numbers, then we have $u - v \geq 0$ and therefore $\phi(u) - \phi(v) = \phi(u - v) \geq 0$, or $\phi(u)\geq\phi(v)$.

Once this is done, it only remains to note that $\mathbb{Q}$ is dense in $\mathbb{R}$.

WhatsUp
  • 22,201
  • 19
  • 48
0

You can now state that $\phi(0)=0$.

If $a>0$, then $a=b^2$, for some $b>0$; therefore $$ \phi(a)=\phi(b^2)=(\phi(b))^2>0 $$ by the assumed multiplicative property. Conversely, if $\phi(a)>0$, then $\sqrt{\phi(a)}=\phi(b)$ for some $b\ne0$, so $$ \phi(b^2)=\phi(b)^2=\phi(a) $$ implying $a=b^2>0$; therefore $a>0$ if and only if $\phi(a)>0$. In particular, $\phi(1)=\phi(1^2)=(\phi(1))^2$, so $\phi(1)=1$.

Also $\phi(a-b)=\phi(a)-\phi(b)$ (easy to prove) and consequently $$ \phi(a)<\phi(b)\text{ if and only if }a<b $$

Consider now a rational $q=m/n$: then $$ m\phi(a)=\phi(ma)=\phi(n(qa))=n\phi(qa) $$ so that $\phi(qa)=q\phi(a)$.

Can you finish?

egreg
  • 238,574