Negation of a statement for proof by contradiction.

Question

**Question:**Let 𝑎 be a rational number and 𝑏 an irrational number. Prove the following:
𝒂 ≠ 𝟎 → 𝒂𝒃 is irrational.

Proof by Contradiction:

Edit

I am not sure if my proof is correct. I am having trouble with the proper negation of the statement, do I only need to negate the ($\rightarrow$ ab is irrational) part in the question for the proof?

If yes, then is line 1 in the proof correct, or would it suffice to say 'Suppose $ab$ $ \in$ $\mathbb{Q}$' in line 1, omitting the conditions for a and b, since they are the stated in the question?

Also, in line 4, do I need to write that $a$ $\in \mathbb{Q}\ \land a \ne 0$ in order to divide by $a$ or can I just write divide $ab$ by $a$ to get $b = \frac{md}{nc}$

Thank you.

Answer 1

Your argument is correct. The theorem can be restated as follows: if $a$ is a non-zero rational, and $b$ is irrational, then $ab$ is irrational, which is essentially your first line. Note that it’s not necessary to argue by contradiction: the theorem is logically equivalent to the assertion that if $a$ is a non-zero rational, and $ab$ is rational, then $b$ is rational, which is what you’ve proved directly.

What you need to write depends on the requirements under which you’re working. As an instructor I’d be happy to see something like this:

Suppose that $a$ is a non-zero rational and that $ab\in\Bbb Q$. Then $a^{-1}$ exists and is rational, so $b=a^{-1}(ab)$ is rational. Taking the contrapositive, we see that if $b$ is irrational, $ab$ must also be irrational.

A proof by contradiction would also be fine, but I’d rather see it in something like this form:

Let $a$ be a non-zero rational and $b$ an irrational, and suppose that $ab\in\Bbb Q$. Then $a^{-1}$ exists and is rational, so $b=a^{-1}(ab)$ is rational. This contradiction shows that $ab$ cannot rational and must therefore be irrational.

I would prefer not to see an argument in the format that you used, with numbered statements and reasons — what I think of as high school geometry style — because it is nothing like normal mathematical writing. (I realize that you may not have a choice in the matter.)