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.

☛ Also Check: Prove that Root 2 is Irrational Number