Prove that root 5 is irrational.
Answers
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 √5 be a rational number.
√5 = p/q
Squaring Both sides
(√5)² = (p/q)²
5 = p²/q²
5q² = p²
Here,
5 is a factor of
AND 5 is also the factor of p²....(¡)
let p be 5
5q² = p²
5q² = (5)²
5q² = 25
q² = 5
here,
we see that
5 is a factor of q
AND 5 is also a factor of q².....(¡¡)
FROM (¡) AND (¡¡) WE see that
5 is a common factor of p and q.
So, Our assumption is wrong.
√5 is an irrational number.
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