A bijection $\phi:\mathbb R \to \mathbb R$ that must be an identity 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.
 A: 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 $\mathbb{R}$.
A: 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?
