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# Prove that $\sqrt5$ is irrational.

__To find:__

We have to prove that $\sqrt5$ is irrational.

__Solution:__

We know that,

If $p$ is a prime number and if $p$ divides $a^2$, then $p$ divides $a$, where $a$ is a positive integer.

Now,

Let us assume, to the contrary, that $\sqrt5$ is rational.

So, we can find integers $a$ and $b (≠ 0)$ such that $\sqrt5= \frac{a}{b}$.

Where $a$ and $b$ are co-prime.

⇒ $(\sqrt5)^2= (\frac{a}{b})^2$

⇒ $5 = \frac{a^{2}}{b^{2}}$

⇒ $5b^2 =a^2$

Therefore, $5$ divides $a^2$

This implies,

$5$ divides $a$.

So, we can write $a = 5c$ for some integer $c$.

⇒ $a^2 = 25c^2$

⇒ $5b^2 = 25c^2$ (Using, $5b^2= a^2$)

⇒ $b^2 = 5c^2$

Therefore, 5 divides $b^2$.

This implies,

$5$ divides $b$.

Therefore, $a$ and $b$ have at least $5$ as a common factor.

But this contradicts the fact that $a$ and $b$ have no common factors other than 1.

This contradiction has arisen because of our incorrect assumption that $\sqrt5$ is rational.

So, we conclude that $\sqrt5$ is irrational.

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