Question

if p is prime number then prove that √p is irrational

Answer 1

If possible,let √p be a rational number.
also a and b is rational.
then,√p = a/b
on squaring both sides,we get,
(√p)²= a²/b²
→p = a²/b²
→b² = a²/p [p divides a² so,p divides a]
Let a= pr for some integer r
→b² = (pr)²/p
→b² = p²r²/p
→b² = pr²
→r² = b²/p [p divides b² so, p divides b]
Thus p is a common factor of a and b.
But this is a contradiction, since a and b have no common factor.
This contradiction arises by assuming √p a rational number.
Hence,√p is irrational.

Answer 2

Answer:

Required to Prove,

√p is an irrational number, for a prime number p

Recall the concept:

A rational number is a number that can be expressed in the form, where p and q are integers and q≠0

Solution:

Let us assume that for a prime number p, √p is a rational number.

Then there exists two coprime integers 'a' and 'b' such that

√p = $\frac{a}{b}$, here a and b have no common terms other than 1 and b ≠0

Squaring on both sides

p = $\frac{a^2}{b^2}$

pb² = a² -----(1)

From the above equation we get, that p divides a².

Since p is a prime number and p divides a²⇒ p divides a

Then we can express 'a' as

a = mp ---------(2), for some integer 'm'

Squaring (1) we get

Substituting a = mp in equation(1) we get

pb²  = m²p²  ⇒ b²  = m²p  ⇒ p divides b²

Since p is prime, p divides b

Hence we have

p divides both a and b, which contradicts the statement that a and b are coprime.

Hence we can conclude that our assumption that √p is a rational number is wrong.

∴ √p is  irrational, for any prime number 'p'

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