Question

explain the irrationality of root p where p is a prime number​

Answer 1

Answer:-

Irrational square root of a prime number

$\sqrt{p} \: is \: a \: irrationl \: \: prime \: number$We can think that the square root of the pA (p) is rational.

Therefore,

$\sqrt{p} =ab$

Note

The prime number must be an irrational number.

Answer 2

Let us consider, to the contrary, that √p is rational.

So, √p = a/b, where a and b are co-primes rational numbers and b is not equal to 0.

$\sqrt{p} = \frac{a}{b} \\ \\ b \sqrt{p} = a$

Squaring on both sides:

pb² = a².

So, p divides a². And according to theorem 1.3 [ If x is a factor of a², then x is a factor of a] p divides a.

Now, let a = pc, for some integer c.

Substituting for a:

pb² = p²c².

b² = pc²

Similarly, according to the theorem mentioned above, since p divides b², p also divides b.

But, this is contradicting to the fact that a and b were co-primes.

So, our assumption that √p is rational is wrong. [Proof by contradiction]

Therefore, √p is irrational where p is a prime number.