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I saw this question where I had to prove/disprove that:

Ques. Product of two irrational number is irrational.


I tried 'Proof by Contraposition'.

Product of two irrational number is irrational.

p : Product of two irrational number

q : Irrational number.

Thus, given statement is : p -> q

Contraposition of p : ¬q -> ¬p

Rational number -> Can be broken down into product of two rational number.

Proof :

Let m be a rational number such that m = p/q.

Then I can always write m as (p/1)*(1/q)

where (p/1) and (1/q) are both rational numbers. Hence proved.

But it turns out that books disproves the statement saying $\sqrt2\cdot\sqrt2=2$ which is a rational number and hence Product of two irrational number need not always be irrational. Which I find convincing.

Can someone please point out where am I going wrong in my proof?

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    $\begingroup$ The contrapositive is $\neg q \implies \neg p$. $\endgroup$ – user61527 Sep 25 '13 at 4:50
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    $\begingroup$ Where you're going wrong is that you're trying to prove a false statement... $\endgroup$ – Zev Chonoles Sep 25 '13 at 4:51
  • $\begingroup$ Your method won't work because the product of two irrational numbers could be rational or irrational, it depends on the two numbers. $\endgroup$ – Michael Albanese Sep 25 '13 at 4:51
  • $\begingroup$ @T.Bongers Ahh, that was a typo, just corrected that. $\endgroup$ – Amit Tomar Sep 25 '13 at 5:00
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    $\begingroup$ @Zev No, trying to prove false statements is kosher and actually the OP was perfectly fine until the step the first comment pointed to. Now that the post is modified, the OP is OK until the line just after that point: $\lnot q\to\lnot p$ does not mean [Rational number -> Can be broken down into product of two rational number]. In fact, $\lnot q\to\lnot p$ means [Rational number -> Cannot be broken down into product of two irrational numbers] (which is a false statement, naturally, and not easier to disprove than the original statement, but this is another matter). $\endgroup$ – Did Sep 25 '13 at 5:06
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The negation of the assertion [Is the product of two irrational numbers] is the assertion [Is not the product of two irrational numbers]. There is no a priori reason to expect that the assertion [Is not the product of two irrational numbers] is equivalent to the assertion [Is the product of two rational numbers] (and in fact these last two are not equivalent).

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Disprove:

Let $\sqrt{2}$ be the irrational number. Then $\sqrt{2}\times \sqrt{2}=|2|$, which is rational. So, the product of two irrational numbers is not always irrational

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  • $\begingroup$ +1 for the simplest constructive proof, without diving into deep logical arguments of proof by contrapositive. Your way is how I'd do it myself: a constructive proof by counter-example. Of course it requires the prior knowledge that $sqrt{2}$ is irrational. But this is whole another story... $\endgroup$ – SasQ Jan 23 '15 at 14:34
  • $\begingroup$ @SasQ You know, math is simple! $\endgroup$ – Hassan Muhammad Jan 26 '15 at 22:42
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Some more examples are

$$\sqrt{8}×\sqrt2=\sqrt{16} =4$$

$$\sqrt2×\sqrt{32}=\sqrt{64} =8$$

$$\sqrt5×\sqrt5=\sqrt{25} =5$$

In this way product of two irrational number is rational.

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    $\begingroup$ $\sqrt4=2$ and $\sqrt9=3$. Both perfectly rational. $\endgroup$ – Asaf Karagila Jul 25 '15 at 18:02

protected by Asaf Karagila Jul 25 '15 at 18:02

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