Prove that `sqrt(3)` is an irrational number.

Answer

N/a

Explanation

Let us assume that `sqrt(3)` is rational number.
`So," "sqrt(3)=(p)/(q),p,qinZ,q!=0`
(in simplest form i.e., no factor other 1 is common to both p and q)
`rArr" " p=sqrt(3q)`

Squaring on sides, we get
`p^(2)=3q^(2)"But 3 divides"3q^(2)`
`rArr" ""3 divides "p^(2)` [form (1)]
`rArr" ""3 divides "p" " ((because "3 divides 9" rArr "3 divides 3 also"),("3 divides 36" rArr "3 divides 6 also etc".))`
`"So, let"p=3m,m inZ` . . .(2)
`rArr" "p^(2)=9m^(2)" "("on squaring)"`
`rArr" "3q^(2)=9m^(2)"" "[from (1)]`
`rArr " " q^(2)=3m^(2) " " . . .(3)`
But 3 divides `2m^(2)`
`rArr" ""3 divies"q^(2)` [from (3)]
Let q = 3n, n `in Z` . . . (4)
From (2) and (4), we see that 3 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(3)` is a rational number.
It means our supposition is wrong.
Therefore, `sqrt(3)` is not a rational number.
Hence, `sqrt(3)` is irrational.