Prove that for any prime positive integer p, √p is an irrational nu

1.	Prove that for any prime positive integer p, √p is an irrational number.
Answer» Let assume that √p is rational Therefore it can be expressed in the form of \(\frac{a}{b}\), where a and b are integers and b ≠ 0 Therefore we can write √p = \(\frac{a}{b}\) (√p)²= (\(\frac{p}{q}\)) ² P = \(\frac{a^2}{b^2}\) a ² = pb² Since a² is divided by b², therefore a is divisible by b. Let a = kc (kc)² = pb² K ²c² = pb² Here also b is divided by c, therefore b² is divisible by c². This contradicts that a and b are co - primes. Hence √p is an irrational number.

Prove that for any prime positive integer p, √p is an irrational number.

Answer»

Let assume that √p is rational

Therefore it can be expressed in the form of \(\frac{a}{b}\), where a and b are integers and b ≠ 0

Therefore we can write √p = \(\frac{a}{b}\)

(√p)²= (\(\frac{p}{q}\)) ²

P = \(\frac{a^2}{b^2}\)

a ² = pb²

Since a² is divided by b², therefore a is divisible by b.

Let a = kc

(kc)² = pb²

K ²c² = pb²

Here also b is divided by c, therefore b² is divisible by c².

This contradicts that a and b are co - primes.

Hence √p is an irrational number.