Question

Question: Prove that the additive group $\mathrm{\mathbb{Q}}$is not isomorphic to the multiplicative group ${\mathrm{\mathbb{Q}}}^{**}$of positive rational numbers, even though $\mathrm{ℝ}$and${\mathrm{ℝ}}^{**}$are isomorphic.

Short answer

Answer

It is proved that the additive group$\mathrm{\mathbb{Q}}$ is not isomorphic to the multiplicative group ${\mathrm{\mathbb{Q}}}^{**}$of positive rational numbers even though $\mathrm{ℝ}$and${\mathrm{ℝ}}^{**}$ are isomorphic.

Step-by-step solution

Isomorphic Groups

Two groups G and H are said to be isomorphic if there exist an isomorphism F from G onto H .

Step 2: Prove that the additive group  ℚ is not isomorphic to the multiplicative group ℚ**  of positive rational numbers even though ℝ  and ℝ**  are isomorphic

Claim that the additive group $\mathrm{\mathbb{Q}}$is not isomorphic to the multiplicative group ${\mathrm{\mathbb{Q}}}^{**}$of positive rational numbers.

The required result can be proved by proving the contradictory statement.

Since $f:\mathrm{\mathbb{Q}}\to {\mathrm{\mathbb{Q}}}^{**}$is an isomorphism and let $\frac{a}{b}=f^{-1}\left(2\right)$. Therefore,

$$\begin{gathered} 2=f\left(\frac{a}{b}\right) \\ =f\left(2\frac{a}{2b}\right) \\ =f\left(\frac{a}{2b}+\frac{a}{2b}\right) \\ =f{\left(\frac{a}{2b}\right)}^2 \\ And \\ \sqrt{2}=f\left(\frac{a}{2b}\right) \end{gathered}$$

As$f\left(\frac{a}{2b}\right)$is a rational number, $f\left(\frac{a}{2b}\right)=\sqrt{2}$implies that $\sqrt{2}$is rational number. This gives a contradiction as $\sqrt{2}$is irrational number.

Thus, the assumption is wrong and hence, the claim follows.

Therefore, the required statement is proved.