Question

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

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.

$hope \: it \: help \: you$

Answer 2

Answer:

Here ur ans ate:-

Step-by-step explanation:

We have to prove √3 is a irrational

Let assume the opposite,

i.e., √3 is a rational

Hence root 3 can be written in the form a/b

where a and b (b not equal to 0) are co-prime

(no common factors other than one)

Hence,√3 = a/b

√3b = a

Sqauring both sides

(√3b) 2 = (a) 2

3b2 = a

a2/3 = b2

Hence 3 divides a2,

by theorem - if p is a prime number,and p divides a2 , then p divides a, where a is a positive integer.

So,3 shall divides a also

Hence, we can say

a/3 = c where c is some integer

So, a = 3c

Hope it helps u mate...

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