If possible,let √p be a rational number.

also a and b is rational.

then,√p = a/b

on squaring both sides,we get,

(√p)²= a²/b²

→p = a²/b²

→b² = a²/p [p divides a² so,p divides a]

Let a= pr for some integer r

→b² = (pr)²/p

→b² = p²r²/p

→b² = pr²

→r² = b²/p [p divides b² so, p divides b]

Thus p is a common factor of a and b.

But this is a contradiction, since a and b have no common factor.

This contradiction arises by assuming √p a rational number.

Hence,√p is irrational.


...it's from Google

**Proof by Contradiction:**

Let's assume that √p is a rational number for a prime positive integer p.

**Definition of Rational Number:**
A rational number can be expressed as the quotient of two integers, where the denominator is not equal to zero.

So, if √p is rational, it can be written as √p = a/b, where a and b are integers with b ≠ 0, and a/b is in its simplified form.

**Square both sides:**
√p = a/b
p = (a/b)^2
p = a^2/b^2

Since p is a prime number, it cannot be written as the square of any integer other than 1. Therefore, a^2 and b^2 must be equal to p.

**Prime Factorization of p:**
p can be written as a product of prime factors. Since p is a prime number, its prime factorization will have only one prime factor, which is p itself. Therefore, p = p * p.

**Substitute into the equation:**
p = a^2/b^2
p = p * p

**Cancel out the common factors:**
Since p is a prime number, it cannot be canceled out from the equation.

So, a^2 and b^2 cannot be equal to p, which contradicts our assumption.

**Conclusion:**
Since the assumption that √p is rational leads to a contradiction, we can conclude that √p is an irrational number for any prime positive integer p.