Question

prove that prove that root 5 is irrational ​

Answer 1

Proving √5 as an irrational number.

Proof:

Let us assume that √5 is a rational number

So,

⇒ √5 = p/q

[∴ In which p and q are integers and q ≠ 0, HCF(p,q) = 1.]

⇒ (√5)² = (p/q)²

[Squaring both the sides]

⇒ 5 = p²/q²

⇒ 5q² = p²   ----(i)

So,

Here,

5 is a factor of p²

5 is also a factor of p

Now,

Let p = 5k

And (p)² = (5k)²

⇒ p² = 25k²   ----(ii)

Now,

By comparing both the equations(i) and (ii), we will get,

⇒ 5q² = 25k²

⇒ q² = (25k²)/5

⇒ q² = 5k²

So, Here,

5 is a factor of q²

5 is a factor of q

So,

From the above discussion, we found that 5 is a factor (common factor) of p and q i.e HCF(p,q) = 5

But it should be HCF(p,q) = 1

Thus,

This contradiction arises due to our wrong assumption.

Hence,

√5 is an irrational number.

PROVED.

Answer 2

Proving √5 as an irrational number.

Proof:

Let us assume that √5 is a rational number

So,

⇒ √5 = p/q

[∴ In which p and q are integers and q ≠ 0, HCF(p,q) = 1.]

⇒ (√5)² = (p/q)²

[Squaring both the sides]

⇒ 5 = p²/q²

⇒ 5q² = p²   ----(i)

So,

Here,

5 is a factor of p²

5 is also a factor of p

Now,

Let p = 5k

And (p)² = (5k)²

⇒ p² = 25k²   ----(ii)

Now,

By comparing both the equations(i) and (ii), we will get,

⇒ 5q² = 25k²

⇒ q² = (25k²)/5

⇒ q² = 5k²

So, Here,

5 is a factor of q²

5 is a factor of q

So,

From the above discussion, we found that 5 is a factor (common factor) of p and q i.e HCF(p,q) = 5

But it should be HCF(p,q) = 1

Thus,

This contradiction arises due to our wrong assumption.

Hence,

√5 is an irrational number.

PROVED.