Prove that `sqrt(2)` is an irrational number.

Answer

N/a

Explanation

We will use the division method to find the square root of 2.

`sqrt(2)=1.4142135 . . .`
Alternative Method : We are going to prove it by the technique called "method of contradiction".
Let us assume that `sqrt(2)` is rational.
Then `sqrt(2)` can be expressed as
`sqrt(2)=(p)/(q),p,qinZ,q!=0`
(in simplest form i.e., no factor other 1 is common to both p and q)
Squaring on both sides, we get
`2=(p^(2))/(q^(2))rArrp^(2)=2q^(2)` . . .(1)
`"But 2 divides"2q^(2)`
`rArr"""2 divides "p^(2)` [form (1)]
`rArr" ""2 divides "p" "((because"2 divides 64" rArr "2 divides 8 also"),("2divides 36" rArr "2 divides 6 also etc".))`
`"So, let"p=2m,m inZ` . . .(2)
`rArr" "p^(2)=4m^(2)" "("on squaring)"`
`rArr" "2q^(2)=4m^(2)"" "[from (1)]`
`rArr " " q^(2)=2m^(2) " " . . .(3)`
But 2 divides `2m^(2)`
`rArr" ""2 divies"q^(2)` [from (3)]
`rArr " ""2 divides q"`
Let q = 2n, n `in Z` . . . (4)
From (2) and (4), we see that 2 divides p and q both. But we have already assumed that p and q has no common factor other than 1.
So, this contradicts our assumption that `sqrt(2)` is a rational number.
It means our supposition is wrong.
Therefore, `sqrt(2)` is not a rational number.
Hence, `sqrt(2)` is irrational.