# 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 with 2 as a common multiple.

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,

⇒ 2q^{2} = p^{2}

⇒ p^{2} is an even number that divides q^{2}. Therefore, p is an even number that divides q.

Let p = 2x where x is a whole number.

By substituting this value of p in 2q^{2} = p^{2}, we get

⇒ 2q^{2} = (2x)^{2}

⇒ 2q^{2} = 4x^{2}

⇒ q^{2} = 2x^{2}

⇒ q^{2} is an even number that divides x^{2}. Therefore, q is an even number that divides x.

Since p and q both are even numbers with 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.

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