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Prove that $2\sqrt 5$ is irrational

My attempt:

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

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

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

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

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

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

How can I procced?

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    $\begingroup$ Hint: $16q^2 + 4q^2=p^2$ has no integer solution by Fermat's last theorem. $\endgroup$ – lEm Mar 8 '16 at 8:56
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    $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ @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
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$$2\sqrt5 = \frac pq, \gcd (p,q=1)$$ $$20q^2=p^2 \Rightarrow 5|p^2 \Rightarrow 5|p \Rightarrow 25|p^2$$ Let $p=5p_1$ $$20q^2=25p_1^2 \Rightarrow 5|q $$ $$\gcd (p,q)\geq 5$$ Сontradiction.

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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.

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  • $\begingroup$ (+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
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    $\begingroup$ You used Euclid's lemma in your question. $\endgroup$ – Ian Miller Mar 8 '16 at 9:22

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