Question

prove that √5 is an irrational number.​

Answer 1

Answer:

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

Answer 2

Given:-

• A Irrational Number is given to us.
• The number is √5.

To Prove:-

• √5 is a Irrational Number.

Proof:-

Given irrational number to us is √5 . On on the contrary let us suppose that √5 is a Rational number.

So ,it can be expressed in the form of p /q where p and q are integers and q is not equal to zero. Also p and q are coprimes which means HCF of p and q is one.

As per our assumption:-

⟹ √5 = p/q .

⟹ (√5)² = (p/q)².

⟹ 5 = p²/q² .

⟹ p² = 5q². ............. (i)

➤ This implies that 5 is a factor of p² so by the Fundamental theorem of arithmetic we can say that 5 is also a factor of p.

⟹ p = 5k .

Put this value in equation 1:-

⟹ (5k)² = 5q².

⟹ 25k² = 5q².

⟹ q² = 25k²/5.

⟹ q² = 5k².

➤ This implies that 5 is a factor of q² , so by the fundamental theorem of arithmetic we can say that 5 is also a factor of q .

But this violates our assumption that p and q are co primes since we find that 5 is also a factor of p and q . So our assumption was wrong √5 is a irrational number.

Hence Proved.