Question

Question 1
Prove that $\sqrt{5}$ is irrational.

Answer 1

Let's take $\sqrt{5}$ as rational number.
We can write $\sqrt{5}=\frac{a}{b}$ such that,
a and b are integers, $b\ne 0$ and there are no factors common to a and b.
Multiply by "b" on both sides we get;
$b\sqrt{5}=a$
To remove root, squaring on both sides, we get,
$5b^2=a^2$ … (i)
That means, 5 is a factor of $a^2$.
For any prime number p which is a factor $a^2$ then it will be the factor of 'a' also.
So, 5 is a factor of 'a'. ------(ii)
Hence, we can write a = 5c for some integer 'c'.
Putting value of 'a' in equation (i) we get,
$5b^2=(5c{)}^2$
$5b^2=25c^2$
Divide by 5; we get,
$b^2=5c^2$
It means 5 is a factor of $b^2$
Thus, 5 is a factor of b. ---(iii)
From (ii) and (iii), we can say that 5 is the factor of both 'a' and 'b'.

This contradicts our theory, as we stated that a and b have no factors in common

Thus, our assumption that $\sqrt{5}$ is rational is wrong.

Hence $\sqrt{5}$ is not a rational number, it is irrational.