Question

prove root3 is irrational​

Answer 1

Answer:

Prove that sqaure root of 3 is an irrational number.

Step-by-step explanation:

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 = p^2/q^2 (Squaring on both the sides)

⇒ 3q^2 = p^2………………………………..(1)

This means that 3 divides p2. This means that 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.

⇒ p^2 = 9r^2………………………………..(2)

from equation (1) and (2)

⇒ 3q^2 = 9r^2

⇒ q^2 = 3r^2

Where q^2 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

Answer:

To Prove root (3) is irrational.

Step-by-step explanation:

Let root(3) be rational

root(3) = p/q where p and q are co-prime and q is not equal t zero

squaring both sides - 3 = p*p/q*q

p*p=3q*q------1

as 3 divides p*p

    3 divides p as well

p = 3c (where c is an no zero integer)------2

substitute 2 in 1

9c^2 = 3q^2

3c^2 = q^2

as 3 divides q^2

3 divides q as well

therefore p and q have a common factor 3

but this is not possible as p and q are co-prime

this problem occured due to our wrong assumption that root(3) is rationla

therefore root(3) is irrational

HOPE IT HELPS:););)