Prove that there is no bijective homomorphism from $\left(\mathbb{Q},\ +\right)$ to $\left(\mathbb{Q_+^*},\ \times \right)$

Question

I need to prove that there does not exist any bijective homomorphism from $\left(\mathbb{Q},\ +\right)$ to $\left(\mathbb{Q_+^*},\ \times \right)$

Here is a way to prove it:

Let $f$ be a homomorphism from $\left(\mathbb{Q},\ +\right)$ to $\left(\mathbb{Q_+^*},\ \times \right)$

$\Leftrightarrow \left(\forall\ \left(x,\ y\right) \in \mathbb{Q}^2\right)\ f\left(x +y\right) = f\left(x\right)\times f\left(y\right)$

$0$ is the identity element of $\left(\mathbb{Q},\ +\right)$ and $\left(\forall\ x \in \mathbb{Q}\right)\ x^{-1}=-x$ is the $\times-$inverse of $x$

We also know that $1$ is the identity element of $\left(\mathbb{Q_+^*},\ \times \right)$

So $$\left(\forall\ x \in \mathbb{Q}\right)\ f\left(x + x^{-1}\right) = f\left(x\right)\times f\left(x^{-1}\right)$$

$$\Leftrightarrow f\left(x + (-x)\right) = f\left(x\right)\times f\left(-x\right)$$ $$\Leftrightarrow f\left(0\right) = f\left(x\right)\times f\left(-x\right)$$

Since $f$ is a homomorphism from $\left(\mathbb{Q},\ +\right)$ to $\left(\mathbb{Q_+^*},\ \times \right)$ then $f\left(0\right)=1$

So $$f\left(x\right)\times f\left(-x\right) = 1$$

Case 1: $f$ is even

Then $$f\left(-x\right) = f\left(x\right)$$

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

Case 2: $f$ is odd

Then $$f\left(-x\right) = -f\left(x\right)$$

So $$f^2\left(x\right) = -1$$

Since $f\left(x\right) \in \mathbb{Q_+^*},$ then $f\left(x\right) > 0 \Rightarrow f^2\left(x\right) > 0$

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

$$f\left(x\right)\times f\left(-x\right) = 1$$

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

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

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

$$\Rightarrow f_1(x) = \sqrt{f_2^2(x) + 1}$$

$$\Rightarrow f_1(x) \notin \mathbb{Q} \Leftrightarrow f_1(x) + f_2(x) \notin \mathbb{Q} \Leftrightarrow f(x) \notin \mathbb{Q} \Rightarrow f(x) \notin \mathbb{Q_*^+}$$

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?

Answer 1

There is a shorter way. By assumption there is an $a \in \mathbb{Q}$ with $f(a) = 2$. Then $$ 2=f(a) = f\left(\frac{a}{2} +\frac{a}{2}\right) = f\left(\frac{a}{2}\right)^2. $$ But $f(a/2)$ must be rational and $\sqrt{2}$ is not.