prove that √5 is an irrational number
Answers
Answer:
Given: √5
We need to prove that √5 is irrational
✍︎✍︎Proof:
Let us assume that √5 is a rational number.
Sp it t can be expressed in the form p/q where p,q are co-prime integers and q≠0
⇒√5=p/q
On squaring both the sides we get,
⇒5=p²/q²
⇒5q²=p² —————–(i)
p²/5= q²
So 5 divides p
p is a multiple of 5
⇒p=5m
⇒p²=25m² ————-(ii)
From equations (i) and (ii), we get,
5q²=25m²
⇒q²=5m²
⇒q² is a multiple of 5
⇒q is a multiple of 5
Hence, p,q have a common factor 5. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number
√5 is an irrational number
Hence proved
✍︎✍︎✍︎Some more info ✍︎✍︎✍︎
- A rational number is a number that is expressed as the ratio of two integers, where the denominator should not be equal to zero,
- An irrational number cannot be expressed in the form of fractions. Rational numbers are terminating decimals but irrational numbers are non-terminating.
is irrational.
as a rational number.
We can find integers a and b such = .
Suppose a and b have common factors other than 1, then we can divide by the common factor, and assume that a and b are co-primes.
So,
Squaring on both the sides, we get :-
By substituting for a, we get :-
Since,
a and b have at least 3 as a common factor.
But,
this contradicts the fact that a and b have no common factors other than 1.
Reason,
this contradiction arised because of our wrong assumption that is rational.