Prove that square root5 is irrational.
Answers
Answer:
Solution: Let us consider that √5 is a “rational number”. We were told that the rational numbers will be in the “form” of form Where “p, q” are integers. So, our consideration is false and √5 is an “irrational number”."
Step-by-step explanation:
Answer:
8
Step-by-step explanation:
:
√5
To prove:
√5 is a rational number
Solution:
Let us consider that √5 is a “rational number”.
We were told that the rational numbers will be in the “form” of \frac {p}{q}
q
p
form Where “p, q” are integers.
So, \sqrt { 5 } = \frac {p}{q}
5
=
q
p
p = \sqrt { 5 } \times qp=
5
×q
we know that 'p' is a “rational number”. So 5 \times q should be normal as it is equal to p
But it did not happens with √5 because it is “not an integer”
Therefore, p ≠ √5q
This denies that √5 is an “irrational number”
So, our consideration is false and √5 is an “irrational number”."