Question

Prove that root of p - root of q are irrational numbers where p and q are prime numbers

Answer 1

Let us assume that √p - √q is rational.

√p - √q = x where x is a rational

(√p-√q)² = x²
p + q - 2√pq =x²

2√pq = p + q - x²

√pq = (p+q-x²)/2

Here, x,x²,p,q,2 all are rational. So, (p+q-x²)/2 is rational

But since p,q are primes, pq is not a perfect square. So, √pq is not rational.
This is a contradiction. Our original assumption is wrong.

So, √p-√q must be irrational where p,q are primes.

Answer 2

√p - √q is irrational as √p and √q are irrational and we know the difference of two irrational no. is always an irrational no.
Proof ---

Let √p - √q be a rational no.

Then. ,it must ne in the form of x/y where x and y are integers and y is not equal to 0

√p-√q = x/y

Squaring both the sides

(√p - √q )^2 =( x/y)^2

p + q - 2√p √q = x^2 /y^2

√p √q = - 1/2 {x^2 /y^2} - p - q

We know that , 1/2 , x^2/y^2 , -p and -q ar e integers , therefore this whole term will be rational but this contradicts with this fact that √p and √q are irrational.

Hence , our supposition was wrong .

√p - √q is an irrational no.
Hence proved

✌Hope it helps .^_^