Answer:
√5 and 3√5 are irrational numbers.
Explanation:
Given number = √5
Let us assume that √5 is a rational number.
It must be in the form of p/q
Where, p and q are integers and q≠0
Let √5 = a/b (a,b are co primes)
On squaring both sides
⇛ (√5)² = (a/b)²
⇛ 5 = a²/b²
⇛ 5b² = a² – – – Eqⁿ(1)
⇛ b² = a²/5
⇛ 5 divides a²
⇛5 divides a also
⇛ 5 is a multiple of a – – – Eqⁿ(2)
Put a = 5c in (1) then
5b² = (5c)²
⇛ 5b² = 25c²
⇛ b² = 5c²
⇛ c² = b²/5
⇛ 5 divides b²
⇛ 5 divides also b
⇛ 5 is a multiple of b – – – – Eqⁿ(3)
From Eqⁿ(2)& Eqⁿ(3)
We have,
5 is a multiple of both a and b
⇛ 5 is a common multiple of a and b
but a and b are co primes
They have only one common factor that is 1
This contradicts to our assumption that is√5 is a rational number.
So, √5 is not a rational number.
√5 is an irrational number.
Hence, Proved.
and
Given number = 3√5
Let us assume that 3√5 is a rational number.
It must be in the form of p/q
Where, p and q are integers and q≠0
Let 3√5 = a/b (a,b are co primes)
⇛ √5 = (a/b)/3
⇛ √5 = a/3b
⇛ √5 is in the form of p/q
⇛√5 is a rational number
But √5 is not a rational number.
It is an irrational number
This contradicts to our assumption that is 3√5 is a rational number.
So, 3√5 is not a rational number.
3√5 is an irrational number.
Hence, Proved.
Note-
The product of a rational and an irrational is an irrational number.