Question

Prove that root of N is not a rational number if N is not a perfect square

Answer 1

We have to prove that √n is not a rational number, if n is not a perfect square. Ex: Let we take the number is √5 Here 5 is not a perfect square. Proof: Let us suppose that √5 is a rational number.

So √5 = p/q

=> 5 = p2 /q2

=>5q2 = p2 ..............1

So p2 is divisible by 5

=> p is divisible by 5

Let p =5x (x is a positive integer)

Now p2 = 25c2

from equation 1

5q2 = 25c2

=> q2 = 5c2

So q is divisible by 5

Thus p and q has a common factor 5. It is contradiction of our assumption.

So, √5 is not a rational number

Answer 2

Step-by-step explanation:

The only way to solve is the cotradictory method.

rational no can pe written in p/q form where q=0; and p and q are coprime( i.e. they have no common factor unless 1)

let root n be the rational no

$\sqrt{n} = \frac{p}{q} \\ n = \frac{ {p}^{2} }{{q}^{2} } \\ n{q}^{2} = {p}^{2} \\if \: n \: is \: a \: factor \: of \: {p}^{2} \\ \: then \: n \: will \: be \: the \: factor \: of \\ \: p \\ now \: let \: p \: = nm \\ n{q}^{2} = {p}^{2} \\ n{q}^{2} = {n}^{2} {m}^{2} \\ {q}^{2} = n {m}^{2} \\if \: n \: is \: a \: factor \: of \: {q}^{2} \\ \: then \: n \: will \: be \: the \: factor \: of \\ \: q \\ this \: contradicts \: that \: p \: and \: q \: are</p><p> \\ \: coprime. \: hence \: \sqrt{n} is \: a \: irrational \: no$