Q. 16 B5.0( 3 Votes )

# Prove that is irrational.

Answer :

Let us consider to be rational

where a & b are integers (b≠0)

Rearranging we get

The R.H.S of the above expression is a rational number since it can be expressed as a numerator by a denominator

Let L.H.S = where p and q are integers (q≠0)

⇒

⇒

Squaring both sides we get

2q^{2} = p^{2}…Equation 1

Since 2 divides p^{2} so it must also divide p

so p is a multiple of 2

let p = 2k where k is an integer

Putting in Equation 1 the value of p we get

2q^{2} = 4k^{2}

⇒ q^{2} = 2k^{2}

Since 2 divides q^{2} so it must also divide q

so q is a multiple of 2

But this contradicts our previously assumed data since we had considered p & q has been resolved in their simplest form and they shouldn't have any common factors.

So √2 is irrational and hence

is also irrational

__Hence Proved__

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