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# Prove that √5 is irrational and hence prove that (2 - √5) is also irrational.

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 √5 and (2 - √5) are irrational numbers

Let's find if (2 - √5) is irrational

**Explanation:**

Let us assume that √5 is a rational number with p and q as co-prime integers and q ≠ 0

⇒ √5 = p / q

On squaring both sides we get,

⇒ 5q^{2} = p^{2}

⇒ p^{2} is a prime number that divides q. Therefore, p is a prime number that divides q

Let p = 5x where x is a whole number

By substituting the value of p in 5q^{2} = p^{2}, we get

⇒ 5q^{2} = (5x)^{2}

⇒ 5q^{2} = 25 x^{2}

⇒ q^{2} = 5 x^{2}

⇒ q^{2} is a prime number that divides x. Therefore, q is a prime number that divides x

Since p and q both are prime numbers with 5 as a common multiple which means that p and q are not co-prime numbers as their HCF is 5

This leads to a contradiction that root 5 is a rational number in the form of p / q with p and q both co-prime numbers and q ≠ 0

Now, to prove that (2 - √5) is an irrational number, we will again use the contradiction method

Let us assume that (2 - √5) is a rational number with p and q as co-prime integers and q ≠ 0

⇒ (2 - √5) = p / q

⇒ √5 = 2 - p/ q

⇒ √5 = (2q - p) / q

⇒ (2q - p) / q is a rational number

However,√5 is an irrational number

This leads to a contradiction that (2 - √5) is a rational number

### Thus, (2 - √5) is an irrational number by contradiction method

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