First, we’ll assume that p and q is rational , where p and q are distinct primes

p+q=x, where x is rational

Rational numbers are closed under multiplication, so if we square both sides, we still get rational numbers on both sides.

(p+q)2=x2

p+2pq+q=x2

2pq=x2−p−q

pq=2(x2−p−q)

Now, x, x2, p, q, & 2 are all rational, and rational numbers are closed under subtraction and division.

So, 2(x2−p−q) is rational.

But since p and q are both primes, then pq is not a perfect square and therefore pqis not rational. But this is contradiction. Original assumption must be wrong.

So, p and q is irrational, where p and q are distinct primes.

## Answers ( )

Step-by-step explanation:Prime number has factors one and itself.

So prime number cannot be a square of a number.

Therefore,the square root of a prime number is irrational.

If p is prime then √p is irrational.

If q is prime then √q is irrational.

Sum of two irrational numbers is always irrational.

So,√p+√q is always irrational

Hence proved

Answer:First, we’ll assume that p and q is rational , where p and q are distinct primes

p+q=x, where x is rational

Rational numbers are closed under multiplication, so if we square both sides, we still get rational numbers on both sides.

(p+q)2=x2

p+2pq+q=x2

2pq=x2−p−q

pq=2(x2−p−q)

Now, x, x2, p, q, & 2 are all rational, and rational numbers are closed under subtraction and division.

So, 2(x2−p−q) is rational.

But since p and q are both primes, then pq is not a perfect square and therefore pqis not rational. But this is contradiction. Original assumption must be wrong.

So, p and q is irrational, where p and q are distinct primes.