Let, if possible, `sqrt(2)` be rational and its simplest form be`(a)/(b).`
Then a and b are integers and having no common factor other than 1 and `b ne 0`.
Now, `sqrt(2) = (a)/(b)`
`rArr` `2 = (a^(2))/(b^(2))`
`rArr` `a^(2) = 2b^(2)" "`...(1)
As `2b^(2)` is divisible by 2.
`therefore` `a^(2)` is divisible by 2.
`rArr` a is divisible by 2.`" "` (Theorem)
Let a = 2c, for some integer c.
`therefore` From equation (1)
`(2c)^(2) = 2b^(2)`
`rArr` `b^(2) = 2c^(2)`
But `2c^(2)` is divisible by 2.
`therefore b^(2)` is divisible by 2.
`rArr` b is divisible by 2.
Let b = 2d, for some integer d.
Thus, 2 is a common factor of a and b both.
But it contradicts the fact that a and b have no common factor other than 1.
So, our supposition is wrong.
Hence, `sqrt(2)` is irrational. `" "`Hence Proved.
