1
$\begingroup$

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 :)

$\endgroup$
  • 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
3
$\begingroup$

An uncountable number of counterexamples:

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.