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  • $\sqrt{p}$ is rational.
  • $\sqrt{p}=\frac{a}{b}$, where $a,b$ are integers with $\gcd(a,b)=1$.
  • $a^2=b^2p$.

Since $p$ divides $a^2$, $p$ divides $a$.

  • $a=kp$.
  • $a^2=k^2p^2=b^2p$
  • $p=\frac{b^2}{k^2}\Rightarrow\sqrt{p}=\frac{b}{k}\Rightarrow\frac{a}{b}=\frac{b}{k}$
  • $b^2=ak\Rightarrow b=\sqrt{ak}$.
  • $\sqrt{p}=\frac{a}{b}=\frac{a}{\sqrt{ak}}=\frac{\sqrt{a}\sqrt{a}}{\sqrt{a}\sqrt{k}}$ which has $\gcd(a,b)=\sqrt{a}\neq 1$ , a contradiction.

($a\neq 1$ because the only number that divides 1 is 1, but 1 is not a prime from $a=kp$)

Hence, $p$ is irrational. Is this legit?

user3765080
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Nothing
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  • @DietrichBurde, my proof is a little bit different from what you have shared the link if you look through it. – Nothing Mar 08 '18 at 12:50

3 Answers3

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Every square number has an even number of prime divisors. Hence the number of prime divisors of $a^2$ is even where their number in $pb^2$ is odd.

Michael Hoppe
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You can conclude like so, since you have supposed that $gcd(a,b)=1$ at the beginning, then $p$ divides $a$ and $k^2p^2=b^2p$ implies that $pk^2=b^2$ this implies that $p$ divides $b$ contradiction.

Tsemo Aristide
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By way of contradiction, assume $\sqrt{p}$ is rational. Then there exist $a, b \in \mathbb{Z}$ with $b\neq 0$ such that $\sqrt{p} = \frac{a}{b}$. Without loss of generality, we may assume $\text{gcd}(a,b) \neq 1$

We can make this assumption, because we still lose no generality.

Now using $\text{gcd} (a,b) = d \neq 1$. Then we can write $a = d \cdot a'$ and $b = d \cdot b'$, for some relatively prime integers $a'$ and $b'$.

Hence $$ \sqrt{p} = \frac{a}{b} = \frac{d a'}{d b'} = \frac{a'}{b'}, $$ So we have shown that $\sqrt{p}$ is a ratio of two relatively prime integers.