Prove that the additive group $ℚ$ is not isomorphic with the multiplicative group $ℚ^*$. [duplicate]

For example, the equation $x+x=a$ has a solution in $ℚ$, unlike $x \cdot x=a$ in $ℚ^*_{>0}$.

In addition to other good answers, it might be worthwhile to observe that not only are these groups not exactly isomorphic, but, in fact, they are wildly different: the positive multiplicative group is free on the primes, while the additive group is (uniquely-) divisible, meaning that for every integer $n$ and $a$ in the additive group, $n\cdot x=a$ has a (unique) solution $x$. That is, the positive multiplicative group is projective (implied by free-ness), while the additive group is injective (implied by divisibility... this is Baer's criterion). So one really should feel that they are completely different.

If $t$ means the torsion subgroup of a group so $$t(\mathbb Q,+)=\{0\}\neq \{\pm1\}=t(\mathbb Q^*,\cdot)$$

If $\phi$ were such mapping (i.e. $\phi :\mathbb{Q} \to \mathbb{Q^*}$ is an isomorphism it means that it must satisfy all the properties of isomorphism), there would be a rational number $a$ such that $\phi(a)=-1$.

$$-1=\phi(a)=\phi \bigg(\frac{1}{2}a+\frac{1}{2}a\bigg)=\phi \bigg(\frac{1}{2}a\bigg)\phi \bigg(\frac{1}{2}a\bigg)=\bigg[\phi\bigg(\frac{1}{2}a\bigg)\bigg]^{2}$$

Is there any rational number whose square is $-1$?

If you think about it, it is the simplest solution by contradiction.