0
$\begingroup$

Show that $\sqrt{2}$ is irrational using integer root theorem.

Let $P(x)=x^2-2$. Since $\sqrt{2}$ is a root of this polynomial, had it been a rational (suppose $\sqrt{2}=\frac{p}{q}$) no, by integer root theorem $q|1$. Hence $q=\pm1$. Moreover $p|-2$. So $p=\pm1$ or $p=\pm2$. Either way $\sqrt{2}=\pm1$ or $\pm2$ which is absurd.

Is it alright?

$\endgroup$
  • 1
    $\begingroup$ Alright, but slightly overdone I guess. The integer root theorem already says that the rational roots of $x^2-2$ are integers and divisors of $(-)2$. $\endgroup$ – Hagen von Eitzen Aug 6 '14 at 8:03
  • 1
    $\begingroup$ Please avoid creating overly specific tags like "rational root theorem". Nobody is going to subscribe to that, or use it for filtering. I replaced it with the tags (polynomials) and (roots). $\endgroup$ – user147263 Aug 6 '14 at 18:12
  • $\begingroup$ Here's the fifth post (fourth question) on this website: math.stackexchange.com/questions/5. One of its answers solves this problem: math.stackexchange.com/a/16596 $\endgroup$ – Jonas Meyer Aug 6 '14 at 18:20
  • $\begingroup$ math.stackexchange.com/questions/917983/… $\endgroup$ – user117644 Nov 20 '15 at 0:33
1
$\begingroup$

"Absurd" is not quite enough.

You should say that if $q=\pm1$ and $p=\pm1$ or $p=\pm2$ then $\left(\frac{p}{q}\right)^2=1$ or $4$. But $\sqrt{2} ^2=2$, so $\sqrt{2}$ cannot be written as $\frac{p}{q}$.

| cite | improve this answer | |
$\endgroup$

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.