just consider if √q is rational no ...then √q=a/b .....so squaring b.s...we get..a^2=b^2q........every square no.has an even no. of prime factors..since a^2=(factors of a)(factors of a).........and b^2also have even no. of factors....but p is extra..it means 1 extra factor ..odd no.of factors ....it means a^2 is not equals to b^2p..... this contradicts d fact that...√q is a rational no..

Proof that √q is irrational when q is prime:

Introduction:
To prove that √q is irrational when q is a prime number, we need to show that √q cannot be expressed as a fraction of two integers. In other words, we need to demonstrate that √q is not a rational number.

Assumption:
Let's assume that √q is rational, which means it can be written as a fraction p/q, where p and q are integers and q ≠ 0. We will then proceed to show that this assumption leads to a contradiction, proving the irrationality of √q.

Proof by Contradiction:

Step 1: Assume √q is rational and can be expressed as a fraction p/q, where p and q are integers and q ≠ 0.

Step 2: Square both sides of the equation to eliminate the square root:
(√q)^2 = (p/q)^2
q = (p^2)/(q^2)

Step 3: Multiply both sides of the equation by q^2 to get rid of the denominator:
q * q^2 = p^2
q^3 = p^2

Step 4: The equation q^3 = p^2 implies that p^2 is divisible by q^3. This means that p is also divisible by q.

Step 5: Rewrite p as p = kq, where k is an integer. Substituting it into the equation q^3 = p^2:
q^3 = (kq)^2
q^3 = k^2 * q^2

Step 6: Divide both sides of the equation by q^2 (since q ≠ 0) to get:
q = k^2

Step 7: The equation q = k^2 indicates that q is a perfect square. However, since q is a prime number, it cannot be expressed as the square of any integer other than 1 or itself.

Conclusion:
From Step 7, we can see that the assumption that √q is rational leads to a contradiction. Therefore, our initial assumption is false, and √q is irrational when q is a prime number.