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 :)

  • 2
    $\begingroup$ Yes, this is enough. :) Another possible counterexemple is to choose b = 1/a (for any a irrationnal) $\endgroup$ – Tryss Apr 4 '16 at 4:41
  • 1
    $\begingroup$ This is enough, and will fetch you all the marks you like in an exam. $\endgroup$ – астон вілла олоф мэллбэрг Apr 4 '16 at 4:42
  • $\begingroup$ Terrific. Thanks guys! $\endgroup$ – Rubicon Apr 4 '16 at 4:45

An uncountable number of counterexamples:

If $x \in \mathbb R \backslash \mathbb Q$, then $\dfrac1{x} \in \mathbb R \backslash \mathbb Q$, but $x\cdot \dfrac1{x} =1 \not \in \mathbb R \backslash \mathbb Q$


Yes, it is formal enough.

The statement implies a $ for all $. The denial of that $for all$ element, a proposition $p"$is true, is that $exists$ a case that does not fulfill.

You have shown that there is a case for $a$ and $b$ that does not fulfill.


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