0
$\begingroup$

Prove that $\sqrt{27} \not\in\Bbb Q$

I have to prove that $\not\exists m, n\in \Bbb Z$ that satisfies the following equation: $m,n \in \Bbb Z : \sqrt{27} = {m\over n} \implies 27 = {{m^2}\over{n^2}} \implies 27n^2 = m ^2$

So (...)

$n=p_1^{2\alpha_1}\times p_2^{2\alpha_2} \times ... \times p_s^{2\alpha_s}$ and $m=q_1^{2\varphi_1}\times q_2^{2\varphi_2} \times ... \times q_r^{2\varphi_r}$

Then $27 p_1^{2\alpha_1}\times p_2^{2\alpha_2} \times ... \times p_s^{2\alpha_s} = q_1^{2\varphi_1}\times q_2^{2\varphi_2} \times ... \times q_r^{2\varphi_r}$

But I don't know how to follow with this. I really appreciate that someone can guide me with some details.

$\endgroup$
  • 3
    $\begingroup$ Perhaps you can use $\;\sqrt{27}=3\sqrt3\;$ and work directly with $\;\sqrt3\;$ which is easier to handle? $\endgroup$ – DonAntonio Nov 7 '13 at 20:37
  • $\begingroup$ m² is a multiple of 3 and so m is a multiple of 3. So write m=3k only to discover that n will also be al multiple of 3. Fron here you can draw your conlcusion $\endgroup$ – imranfat Nov 7 '13 at 20:39
  • $\begingroup$ One of the $q_i$, say $q_1$, is equal to $3$. This appears to an even power of the right. Show that it appears to an odd power on the left, contradicting Unique Factorization. $\endgroup$ – André Nicolas Nov 7 '13 at 20:39

Your Answer

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

Browse other questions tagged or ask your own question.