Question

By using the method of contradiction verify that $\sqrt{5}$ is irrational.

Answer 1

If possible, $let\sqrt{5}$ be rational and let its simplest form be$\frac{a}{b}$.

Then a and b are integers having no common factor other than 1 and $b\ne 0.$

Now, $\sqrt{5}=\frac{a}{b}\Rightarrow 5=\frac{a^2}{b^2}$ [on squaring both sides]

$\Rightarrow 5b^2=a^2$ ...(i)

$\Rightarrow 5$ divides $a^2[\because 5divides5b^2]$

$\Rightarrow 5$ divides $a[\because 5$ is prime and divides $a^2\Rightarrow 5dividesa].$

Let a=5c for some integer C.

Putting a =5 c in (i), we get

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

$\Rightarrow 5$divides $b^2[\because 5divides5c^2]$

$\Rightarrow 5$divides $b[\because$5 is prime and divides $b^2\Rightarrow$ 5 divides b.]

Thus, 5 is a common factor of a and b.

But, this contradicts the fact a and b have no common factor other than 1.

Hence $\sqrt{5}$ is irrational.