Question

Proof by Contradiction : Show that √ 2 is irrational.

Answer 1

Explanation:

Let's assume that
`√(2)` is rational,

So, it should be expressed in the form of
`(a)/(b)`

where a and b are two prime integers.

So, we can write,

`√(2)=(a)/(b)`

`=>\ 2=(a^2)/(b^2)`

`=>\ a^2=2b^2.`

As
`2b^2` represents a even number,
`a^2` must also be even.

Since
`a^2` is even number, so a is also an even number.

Let assume again that a=2c

`=>\ a^2\ =\ 4c^2`

`=>\ 2b^2=\ 4c^2`

`=>\ b^2=2c^2.`

As
`2c^2` is even, so
`b^2` is also an even number, and so is b.

And since and b both are even number here, two even numbers can not be relatively prime, so
`√(2)` can not be expressed in the form of
`(a)/(b)`. So,
`√(2)` is considered as an irrational number.