Prove that $2\sqrt 5$ is irrational

My attempt:

Suppose $$2\sqrt 5=\frac p q\quad\bigg/()^2$$

$$\Longrightarrow 4\cdot 5=\frac{p^2}{q^2}$$

$$\Longrightarrow 20\cdot q^2=p^2$$

$$\Longrightarrow q\mid p^2$$


$$\Longrightarrow \text{gcd}(p^2,q)=1$$

How can I procced?

  • 3
    $\begingroup$ Hint: $16q^2 + 4q^2=p^2$ has no integer solution by Fermat's last theorem. $\endgroup$ – lEm Mar 8 '16 at 8:56
  • 2
    $\begingroup$ I did not learn yet this theorem $\endgroup$ – 3SAT Mar 8 '16 at 8:57
  • $\begingroup$ @user319071, If $20|p^2$ we can't immediately say that $20|p$ and in general it is not true for composite numbers. $20|p^2$, thus $2|p^2$ and $5|p^2$, hence $2|p\wedge 5|p$, hence $10|p$. For example $20|100$, but $20$ doesn't divide $10$. $\endgroup$ – Galc127 Mar 8 '16 at 9:02
  • 1
    $\begingroup$ @Nehorai Teachers don't teach Fermat's last theorem. Its a proof that took 358 years to be proven and hence made the proof quite famous especially as it was only published in 1994-95. en.wikipedia.org/wiki/Fermat%27s_Last_Theorem $\endgroup$ – Ian Miller Mar 8 '16 at 9:19
  • 1
    $\begingroup$ @user319071 How does Fermat's Last Theorem prove that $16q^2 + 4q^2 = p^2$ has no solutions? $\endgroup$ – Erick Wong Mar 8 '16 at 9:47

$$2\sqrt5 = \frac pq, \gcd (p,q=1)$$ $$20q^2=p^2 \Rightarrow 5|p^2 \Rightarrow 5|p \Rightarrow 25|p^2$$ Let $p=5p_1$ $$20q^2=25p_1^2 \Rightarrow 5|q $$ $$\gcd (p,q)\geq 5$$ Сontradiction.

| cite | improve this answer | |

You have $20|p^2$, thus $2|p^2\wedge 5|p^2$, hence by Euclid's lemma $2|p\wedge 5|p$, hence $10|p$, so we can write $p=10k$ for $k\in\mathbb{Z}$ and then $20q^2=100k^2\Rightarrow q^2=5k^2$, hence $5|q^2$, and by the same lemma $5|q$, thus $\text{gcd}(p,q)=5$, so we got a contradiction.

| cite | improve this answer | |
  • $\begingroup$ (+1) for the effort, I did not learn this lemma yet $\endgroup$ – 3SAT Mar 8 '16 at 9:10
  • $\begingroup$ Do you know how to show that $\sqrt{5}$ is irrational without this lemma? Please notice that Roman83 answer also uses it. $\endgroup$ – Galc127 Mar 8 '16 at 9:14
  • $\begingroup$ I have a proof for $\sqrt 2$ in my textbook $\endgroup$ – 3SAT Mar 8 '16 at 9:15
  • $\begingroup$ And this sketch of proof will hold for any number which is not a perfect square, but it uses the mentioned lemma. $\endgroup$ – Galc127 Mar 8 '16 at 9:16
  • 1
    $\begingroup$ You used Euclid's lemma in your question. $\endgroup$ – Ian Miller Mar 8 '16 at 9:22

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.