Question

Prove that sqaure root of 3 is an irrational number... and​

Answer 1

Answer:

Let us assume to the contrary that v3 is a rational number.

It can be expressed in the form of p/q

where p and q are co-primes and q 0.

V3 = p/q

3 = p?/q? (Squaring on both the sides)

3q? = p?.

..(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? = 9r2.

from equation (1) and (2) 3q? = 9r2

q3 r2

0.6KB/s *

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 prime.

Consequently, p/q is not a rational number. This demonstrates that v3 is an irrational number.

$hope \: this \: help \: you$

Answer 2

Answer:

$\huge{\underline{\mathtt{\red{❥A}\pink{N}\green{S}\blue{W}\purple{E}\orange{R}}}}$

Let us assume to the contrary that v3 is a rational number.

It can be expressed in the form of p/q

where p and q are co-primes and q 0.

V3 = p/q

3 = p?/q? (Squaring on both the sides)

3q? = p?.

..(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? = 9r2.

from equation (1) and (2) 3q? = 9r2

q3 r2

0.6KB/s *

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 prime.

Consequently, p/q is not a rational number. This demonstrates that v3 is an irrational number.