关于(R,+)与(Q,+)不同构的证明

hbghlyj

从epsilonify看到一道题:

Prove that the additive groups \(\mathbb{R}\) and \(\mathbb{Q}\) are not isomorphic

Proof. Assume that \(\mathbb{R}\) and \(\mathbb{Q}\) are isomorphic. Then there exists a mapping

\begin{align*}
\phi: \mathbb{Q} \longrightarrow \mathbb{R}
\end{align*}

which is bijective and is a group homomorphism. Let \(x \in \mathbb{R}\) such that \(\phi(2) = x\) and \(q \in \mathbb{Q}\) such that \(\phi(q) = \sqrt{x}\). Then

\begin{align*}
\phi(q)^2 = \phi(q^2) = x = \phi(2).
\end{align*}

Since \(\phi\) is injective, we have that \(q^2 = 2\). So this means that \(q = \sqrt{2}\), which isn’t possible in \(\mathbb{Q}\), a contradiction.

Therefore, the additive groups \(\mathbb{R}\) and \(\mathbb{Q}\) are not isomorphic.

怀疑$\phi(q)^2 = \phi(q^2)$是错的:$\phi$是additive group的同构,和乘法有什么关系

hbghlyj

MSE

Let $\Phi: \mathbb{Q} \rightarrow \mathbb{R}$ homomorphism of additive groups.
Then $\Phi$ is determined by $\Phi(1).\Bigl[$as $\Phi(\frac{a}{b})= \Phi(\frac{1}{b})+ ... + \Phi(\frac{1}{b})$ ($a$ summands) and $\Phi(1)= \Phi(\frac{1}{b}) + ... + \Phi(\frac{1}{b}) $, ($b$ summands)$\Bigr]$

Now, say $\sqrt{2} \cdot \Phi(1)$ doesn't have a preimage.