Question

i will mark as the brainliest

sorry frnds there was a mistake in the previous que

if p is a prime number then prove that root p is an irrarional number

Answer 1

hey mate

here's the answer

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 assumi

hope it helps

please mark it as brainliest answer

Answer 2

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.

HOPE IT HELPS YOU!!