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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?

3SAT
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    Hint: $16q^2 + 4q^2=p^2$ has no integer solution by Fermat's last theorem. – lEm Mar 08 '16 at 08:56
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    I did not learn yet this theorem – 3SAT Mar 08 '16 at 08:57
  • @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 $10$. – Galc127 Mar 08 '16 at 09:02
  • Remark that $2\sqrt{5}$ is rational if and only if $\sqrt{5}$ is rational and then go see this question. – user37238 Mar 08 '16 at 09:15
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    @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. https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem – Ian Miller Mar 08 '16 at 09:19
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    @user319071 How does Fermat's Last Theorem prove that $16q^2 + 4q^2 = p^2$ has no solutions? – Erick Wong Mar 08 '16 at 09:47
  • Using Fermat here would be like using a nuclear warhead to open a soda can. – zz20s Mar 23 '16 at 14:58

2 Answers2

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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.

user26977
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Roman83
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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.

Galc127
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