Q. 16 B5.0( 3 Votes )

# Prove that is irrational.

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

2q2 = p2…Equation 1

Since 2 divides p2 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

2q2 = 4k2

q2 = 2k2

Since 2 divides q2 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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