Question

Prove that root 3+root7 is irrational

Answer 1

QUESTION:-

Prove that √3+√7 is an irrational number.

SOLUTION:-

Lets, assume that √3+√7 is an irrational number,

Then lets equate it by 'x'

Therefore,

=> x= √7+√3

Squarring both the sides .

=> x²=(√7+√3)²

=> x²= 10+2√21

Solving them by keeping R.H.S and L.H.S

=> x²-10/2= √21

Now,

since,

x²= Rational

-10/2= rational...

but, we can say that √21 is an irrational number as it cant be represented it at p/q form,

So,

by this method of contradiction,

We contradict that √7+√3 is an irrational number.

Ans.

Answer 2

Let us suppose that √3 + √7 be a rational no.

A rational number can be written in the form of p/q where 'p' and 'q' are co-prime rational integers and q≠0.

Then, √3 + √7 = p/q.

√3=p/q-√7

squaring both sides:

3 = p²/q² + 7 - 2*p/q*√7

⇒p²/q² + 7 - 2*p/q*√7-3 = 0

⇒p²/q²+4 = 2√7-p/q

⇒$\frac{(p^2+4q^2)q}{pq^2} =2\sqrt{7}$

⇒$\frac{qp^2+4q^3}{2pq^2} =\sqrt{7}$

In LHS, p and q are already taken as rational integers, 2 & 4 are also rational, and denominator ≠0. So, the LHS is rational. But we know that √7 is irrational. So, this contradicts us.

Thus proved that √3+√7 is irrational.