Question

prove that root 3 is an irrational number​

Answer 1

Answer:

Let us assume to the contrary that √3 is a rational number.

It can be expressed in the form of p/q

where p and q are co-primes and q≠ 0.

⇒ √3 = p/q

⇒ 3 = p2/q2 (Squaring on both the sides)

⇒ 3q2 = p2………………………………..(1)

It means that 3 divides p2 and also 3 divides p because each factor should appear two times for the square to exist.

So we have p = 3r

where r is some integer.

⇒ p2 = 9r2………………………………..(2)

from equation (1) and (2)

⇒ 3q2 = 9r2

⇒ q2 = 3r2

Where q2 is multiply of 3 and also q is multiple of 3.

Then p, q have a common factor of 3. This runs contrary to their being co-primes. Consequently, p / q is not a rational number. This demonstrates that √3 is an irrational number.

Answer 2

Let us assume that √3 is an irrational number

So,

√3 = p/q   [In which 'p' and 'q' are integer, 'q ≠ 0', and HCF(p,q) = 1]

Now squaring both the side, we get

⇒ 3 = p²/q²

⇒ 3q² = p²    -----(i)

Here,

3 is a factor of p²

3 is also a factor of p

Now,

Let p = 3k

So, p² = (3k)²

⇒ p² = 9k² ------(ii)

Now,

Comparing equations (i) and (ii) we get,

3q² = 9k²

⇒ q² = (9k²)/3

⇒ q² = 3k²

Here,

3 is a factor of q²

3 is also a factor of q

Thus, we get,

HCF(p,q) = 3

But it should be 1

It is due to our wrong assumption.

Hence,

The assumption is wrong.

So,

• √3 is not a rational number.

∴ √3 is an irrational number