Proving that if $x$ is irrational, then $\frac1x$ is irrational

Question

I was given the question:

Prove that if $x$ is irrational, then $\frac1x$ is irrational

I did this through contraposition, namely:

if $\frac1x$ is rational, it can be written in the form $\frac ab$ where b is nonzero.

Thus, $\frac1x = \frac ab$ which gives $ x = \frac ba$, so $x$ is also rational, hereby indirectly proving that the original statement is true.

I couldn't find this proven in the way I did when looking for it on the internet, so where did I go wrong?

Things can be proven in more than one way. In a couple course I took I found a proof the professor hadn't seen before. Being new to the subject doesn't mean you can't find a new way forward. — CyclotomicField, Apr 27 '23 at 12:23
I thought I was wrong because the next question in the book was asking for basically the same answer if it was right. — Willem, Apr 27 '23 at 12:25
It may be worth taking a brief moment to explain why $a\neq 0$. — JMoravitz, Apr 27 '23 at 12:43
$ a$ ≠ 0 because $\frac1x$ = $\frac ab$ is equal to $x$= $\frac 1\frac ab$ where $\frac ab$ is nonzero, right? — Willem, Apr 27 '23 at 12:58

score 1 · Answer 1 · answered Apr 30 '23 at 18:58

Your reasoning is correct,
and merits a bit more precision: You should not ignore zero as mentioned in a comment, e.g. by writing
If $\frac1x$ is a nonzero rational, it can be written in the form $\frac ab$ where both $a,b$ are nonzero.

Multiplicative inversion $\iota:\mathbb R\backslash\{0\}\to\mathbb R\backslash\{0\}, x\mapsto\tfrac1x$ is invertible, being its own inverse: $\iota^{-1}=\iota\,$.
It maps $\,\mathbb Q\backslash\{0\}\,$ to itself, whence it maps the irrationals $\,\mathbb R\backslash\mathbb Q\,$ to itself as well.

I am unfortunately not nearly far enough in my studies to transcribe that second paragraph — Willem, May 01 '23 at 20:11

Proving that if $x$ is irrational, then $\frac1x$ is irrational

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