if $a,b \in \mathbb R \backslash \mathbb Q $ then $a \cdot b \in \mathbb R \backslash \mathbb Q $
Okay so the question asks to show, with a counter example, that the above statement is false.
Here is what I have done:
So I'm assuming $\mathbb R \backslash \mathbb Q$ is the set of real but not rational numbers...thus it is the set of real irrational numbers?
With that I assume we just use two irrationals whose product is rational.
Therefore, if $a = \sqrt2 $ and $b = \sqrt8$
$$\sqrt2 , \sqrt8 \in \mathbb R \backslash \mathbb Q \space then \space \sqrt2 \cdot \sqrt8 \in \mathbb R \backslash \mathbb Q$$
We have an error, as $\sqrt2 \cdot \sqrt8$ = 4 which is a rational number.
Thus, the original statement does not hold given the counter example of $a$ = $\sqrt 2$ and $b$ = $\sqrt8$
Is this formal enough? Is this what giving a counter-example actually means? I'd be grateful for any help...and hopefully this thread helps some others :)