Question

proof that √ 3 is an irrational number​

Answer 1

Answer:

Its an irrational Number.

Step-by-step explanation:

U know tha √3 can be written in the form of a/b

Hence √3 = a/b

            √3b=a  ( I multiplied b towards the other side)

             squarring both sides we get,

             3b^2 =a^2

              a^2/3 = b^2 ( i divided 3 on the right side)

              hence 3 divides a^2

              So 3 shall divide a also

Hence, we can say

a/3=c where c is some integer

so, a=3c (i multiplied 3 on the right side)

Hope it helps :)

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Answer 2

Answer:

${\huge{\bf{\purple{\mathbb{\underline {Question:-}}}}}}$

Proof that √3 is irrational

${\huge{\bf{\purple{\mathbb{\underline {Proof:--}}}}}}$

Let us assume on the contrary that

√3 is a rational number.

Then, there exist positive integers a and b such that

√3=a/b

where, a and b, are co-prime i.e. their HCF is 1

Now,

on squaring both side,

3=a²/b²

3b²=a²

3 divides a². [∵3 divides 3b²]

3 divides a. ..(eq i)

a=3c for some integer c

a²=9c²

3b²=9c². [a²=3b²]

b²=3c²

3 divides b².

3 divides b. ..(eq ii)

From (i) and (ii), we observe that a and b have at least 3 as a common factor. But, this contradicts the fact that a and b are co-prime. This means that our assumption is not correct.

Hence,

√3 is an irrational number.