Question

Prove that √p is an irrational number , if p is not a perfect square.​

Answer 1

$\huge \frak \green{question}$

prove that √p is an irrational number , if p is not a perfect square.

$\huge \frak \red{answer}$

Let us assume, to the contrary, that √p is rational. So, we can find coprime integers a and b(b ≠ 0) such that :-

=> √p = a/b

=> √p b = a

=> pb² = a² ….(i) [Squaring both the sides]

=> a² is divisible by p

=> a is divisible by p So, we can write a = pc for some integer c.

Therefore,

a² = p²c² ….[Squaring both the sides]

=> pb² = p²c² ….[From (i)]

=> b² = pc²

=> b² is divisible by p

=> b is divisible by p

=> p divides both a and b.

=> a and b have at least p as a common factor.

But this contradicts the fact that a and b are coprime. This contradiction arises because we have assumed that √p is rational. Therefore, √p is irrational.

_____________________________________

$\bold \green{i \: hope \: it \: is \: helpful}$

________________________________