Prove that 5 is an irrational number
Answers
We were told that the rational numbers will be in the “form” of form Where “p, q” are integers. So, ... So, our consideration is false and is an “irrational number”.
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To prove that √5 is irrational number
Let us assume that √5 is rational
Then √5 =
(a and b are co primes, with only 1 common factor and b≠0)
⇒ √5 =
(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