Vtyjkyuk

I need to prove that there does not exist any bijective homomorphism from $left(mathbbQ, +right)$ to $left(mathbbQ_+^*, times right)$

Here is a way to prove it:

Let $f$ be a homomorphism from $left(mathbbQ, +right)$ to $left(mathbbQ_+^*, times right)$

$Leftrightarrow left(forall left(x, yright) in mathbbQ^2right) fleft(x +yright) = fleft(xright)times fleft(yright)$

$0$ is the identity element of $left(mathbbQ, +right)$ and $left(forall x in mathbbQright) x^-1=-x$ is the $times-$inverse of $x$

We also know that $1$ is the identity element of $left(mathbbQ_+^*, times right)$

So $$left(forall x in mathbbQright) fleft(x + x^-1right) = fleft(xright)times fleft(x^-1right)$$

$$Leftrightarrow fleft(x + (-x)right) = fleft(xright)times fleft(-xright)$$
$$Leftrightarrow fleft(0right) = fleft(xright)times fleft(-xright)$$

Since $f$ is a homomorphism from $left(mathbbQ, +right)$ to $left(mathbbQ_+^*, times right)$ then $fleft(0right)=1$

So $$fleft(xright)times fleft(-xright) = 1$$

Case 1: $f$ is even

Then $$fleft(-xright) = fleft(xright)$$

$Rightarrow f$ is not injective $Rightarrow f$ is not bijective

Case 2: $f$ is odd

Then $$fleft(-xright) = -fleft(xright)$$

So $$f^2left(xright) = -1$$

Since $fleft(xright) in mathbbQ_+^*,$ then $fleft(xright) > 0 Rightarrow f^2left(xright) > 0$

Therefore $f$ is not odd.

Case 3: $f$ is the sum of two functions such that one is even and the other is odd.

so let $f(x) := f_1(x) + f_2(x)$ such that $f_1$ is even and $f_2$ is odd

$$fleft(xright)times fleft(-xright) = 1$$

$$Leftrightarrow left(f_1(x) + f_2(x)right) times left(f_1(x) - f_2(x)right) = 1$$

$$Leftrightarrow f_1^2(x) - f_2^2(x) = 1$$

$$Leftrightarrow f_1^2(x) = f_2^2(x) + 1$$

$$Rightarrow f_1(x) = sqrtf_2^2(x) + 1$$

$$Rightarrow f_1(x) notin mathbbQ Leftrightarrow f_1(x) + f_2(x) notin mathbbQ Leftrightarrow f(x) notin mathbbQ Rightarrow f(x) notin mathbbQ_*^+$$

Therefore the 3th assumption is not correct.

Thus just the first case is correct which implies that $f$ is not bijective.

Is there a shorter proof?

There is a shorter way. By assumption there is an $a in mathbbQ$ with $f(a) = 2$. Then
$$
2=f(a) = fleft(fraca2 +fraca2right) = fleft(fraca2right)^2.
$$
But $f(a/2)$ must be rational and $sqrt2$ is not.