Additive Group of Real Numbers is Not Isomorphic to Multiplicative Group of Real Numbers/Proof 3

From ProofWiki
Jump to navigation Jump to search


Let $\struct {\R, +}$ denote the additive group of real numbers.

Let $\struct {\R_{\ne 0}, \times}$ denote the multiplicative group of real numbers.

Then $\struct {\R, +}$ is not isomorphic to $\struct {\R_{\ne 0}, \times}$.


There are two element of $\struct {\R_{\ne 0}, \times}$ which are self-inverse, that is, $-1$ and $1$.

However, there is only one element of $\struct {\R, +}$ which is self-inverse, that is, $0$.

Aiming for a contradiction, suppose there exists an isomorphism $f: \struct {\R_{\ne 0}, \times} \to \struct {\R, +}$.


\(\ds 0\) \(=\) \(\ds \map f 1\)
\(\ds \) \(=\) \(\ds \map f {-1 \times -1}\)
\(\ds \) \(=\) \(\ds \map f {-1} + \map f {-1}\)
\(\ds \) \(=\) \(\ds 2 \map f {-1}\)
\(\ds \leadsto \ \ \) \(\ds \map f {-1}\) \(=\) \(\ds 0\)
\(\ds \) \(=\) \(\ds \map f 1\)

So we have demonstrated that $f$ is not an injection.

Hence $f$ is not a bijection and so not an isomorphism.

It follows from Proof by Contradiction that there can be no such isomorphism.

Hence the result.