Prove that √2 is irrational

Prove that root 2 is an irrational number. Rational numbers are integers that are expressed in the form of p / q where p and q are both co-prime numbers and q is non-zero. Answer: Hence proved that √2 is an irrational number. Let's find if √2 is irrational. Explanation: To prove that √2 is an irrational number, we will use the contradiction method. Let us assume that √2 is a rational number with p and q as co-prime integers and q ≠ 0 ⇒ √2 = p/q On squaring both sides we get, ⇒ 2q2 = p2 ⇒ Here, 2q2 is a multiple of 2 and hence it is even. Thus, p2 is an even number. Therefore, p is also even. So we can assume that p = 2x where x is an integer. By substituting this value of p in 2q2 = p2, we get ⇒ 2q2 = (2x)2 ⇒ 2q2 = 4x2 ⇒ q2 = 2x2 ⇒ q2 is an even number. Therefore, q is also even. Since p and q both are even numbers, they have 2 as a common multiple which means that p and q are not co-prime numbers as their HCF is 2. This leads to the contradiction that root 2 is a rational number in the form of p/q with "p and q both co-prime numbers" and q ≠ 0. Thus, √2 is an irrational number by the contradiction method.