Question

prove that √5 is irrational number

Answer 1

Our assumption was wrong.

So, √5 is an irrational number.

Answer is in the pic

Answer 2

If possible, let √5 be rational and let its simplest form be a\b.

Then, a and b are integers having no common factor other than 1, and b is not equal to 0.

Now,

$\sqrt{5} = \frac{a}{b} \: = > \: \frac{ {a}^{2} }{ {b}^{2} } \: \: \: (on \: squaring \: both \: sides) \\ \\ = > \: 5 {b}^{2} = {a}^{2} \: \: \: ...(1) \\ \\ = > \: 5 \: divides \: {a}^{2} \: \: \: ( \: hererfore \: 5 \: divides \: {5b}^{2} ) \\ \\ = > \: 5 \: divides \: a \\ (as \: 5 \: is \: prime \: and \: 5 \: divides \: {a}^{2} \: = > \: 5 \: divides \: a)$

Let a = 5c for some integer c.

Putting a = 5c in (1), we get

${5b}^{2} = 25 {c}^{2} \: = > \: {b}^{2} = {5c}^{2} \\ \\ = > \: 5 \: divides \: {b}^{2} \: \: \: \: (herefore \: 5 \: divides \: 5 {c}^{2} ) \\ \\ = > \: 5 \: divides \: b \\ (as \: 5 \: is \: prime \: and \: 5 \: divides \: {b}^{2} \: = > \: 5 \: divides \: b)$

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

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

The contradiction arises by assuming that √5 is rational.

Hence, √5 is Irrational.