Prove that √5 is an irrational number.
Answers
Given: √5
We need to prove that √5 is irrational
Proof:
Let us assume that √5 is a rational number.
So it 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.
Step-by-step explanation:
Let us assume on the contradiction that √5 is a rational number.
So it 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. ( Fundamental theorem of Arithmetic )
Thus,
p is a multiple of 5
Now,
⇒ p = 5m for some integer m
Squaring both sides,
⇒ p² = 25m² ————-(ii)
From equations (i) and (ii), we get,
5q² = 25m²
⇒ q² = 5m²
⇒ q² is a multiple of 5
Thus,
⇒ 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
Hope it helps you