Prove that √5 is irrational
Answers
Step-by-step explanation:
To prove that √5 is irrational number
Let us assume that √5 is rational
Then √5 = \frac{a}{b}
b
a
(a and b are co primes, with only 1 common factor and b≠0)
⇒ √5 = \frac{a}{b}
b
a
(cross multiply)
⇒ a = √5b
⇒ a² = 5b² -------> α
⇒ 5/a²
(by theorem if p divides q then p can also divide q²)
⇒ 5/a ----> 1
⇒ a = 5c
(squaring on both sides)
⇒ a² = 25c² ----> β
From equations α and β
⇒ 5b² = 25c²
⇒ b² = 5c²
⇒ 5/b²
(again by theorem)
⇒ 5/b-------> 2
we know that a and b are co-primes having only 1 common factor but from 1 and 2 we can that it is wrong.
This contradiction arises because we assumed that √5 is a rational number
∴ our assumption is wrong
∴ √5 is irrational number
Let us assume that √5 is rational.
√5 = a/b where a and b are co primes and integers, b is not equal to 0.
√5 = a/b
Squaring on both sides,
(√5)^2 = (a/b)^2
25 = a^2/b^2
25b^2 = a^2
a^2 is a factor of 5.
Therefore, a is also a factor of 5.
It contradicts the fact that√5 is irrational.
Therefore, our assumption is wrong.
Therefore, √5 is irrational.