Question

Prove $\surd 5$ is irrational.

Answer 1

We have to prove $\sqrt{5}$ is irrational.

Let us assume the opposite,

Hence, $\sqrt{5}$ can be written in the form $\frac{a}{b}$ where $a$ and $b$ are co-prime.

Hence $\sqrt{5}=\frac{a}{b}$

$\Rightarrow 5b^2=a^2$

$\Rightarrow \frac{a^2}{5}=b^2$

Hence,$5$ divides $a^2$

By theorem:If $p$ is a prime number, and $p$ divides $a^2$, then $p$ divides $a$, where $a$ is a positive number.

So, $5$ shall divide $a$ also. .........$\left(1\right)$

Hence, we can say $\frac{a}{5}=c$ where $c$ is some integer.

So,$a=5c$

Now we know that $5b^2=a^2$

Put $a=5c$

$\Rightarrow 5b^2={\left(5c\right)}^2$

$\Rightarrow 5b^2=25c^2$

$\Rightarrow b^2=5c^2$

$\Rightarrow \frac{b^2}{5}=c^2$

Hence $5$ divides $b^2$

So, $5$ divides $b$ also ........$\left(2\right)$

By $\left(1\right)$ and $\left(2\right)$

$5$ divides both $a$ and $b$

Hence $5$ is a factor of $a$ and $b$

So,$a$ and $b$ have a factor $5$

$\therefore$, $a$ and $b$ are not co-prime.

Hence, our assumption is wrong.

$\therefore$ by contradiction, $\sqrt{5}$ is irrational.

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