Question

hi guys

how to prove that root 3 is irrational and how to to moderate ​

Answer 1

proof
let us assume

$\sqrt{3} \: be \: rational$

Then, it will be of the form

$\frac{a}{b}$

where a,b are integers and b is not equal to 0.

Again, let a and b have no common factor other than 1.

Now,

$\sqrt{3 \: = \frac{a}{b} }$

on squaring both sides ,we get

$3 = \frac{ {a}^{2} }{ {b}^{2} }$

$3 {b}^{2} = {a}^{2} \: \: \: \: \: \: \: \: \: \: ...eq1$

3divides

${a}^{3}$

3 divides a ......from theorem 1

Then, a can be written as 3m, where m is an integer.
On putting a=3m in eq (i),we get
3b sq =(3m)sq
3b sq =9m sq
b sq =3m sq
3 divides b sq.
3 divides b
Thus, 3 is abcommon factor of a and b .
But this contradicts the fact that a and b have no common factor other than 1. The contradiction arises by assuming that root 3 is rational .
Hence, root 3 is irrational .
hence proved .

Answer 2

$\huge {Solution}$

Let us assume to the contrary that $\sqrt{3}$is a rational no. So,

$\sqrt{3} = \frac{a}{b}$

where a and b are coprime numbers .

By squaring both sides , we get ...

${( \sqrt{3}) }^{2} = { (\frac{a}{b}) }^{2}$

$= 3 = {a}^{2} \div {b}^{2}$

$= {a}^{2} = 3 {b}^{2}$

It means that 3 is divisible by (a)^2 .

=> So it will also be divisible by a .

Now ,

Let

$a = 3n$

(for some integer n .)

=>( 3n)^2 = 3(b)^2

=> 9(n)^2 = 3(b)^2

=> 3(n)^2 = (b)^2

It means( b)^2 is divisible by 3 .

=> So b is also divisible by 3 .

So our assumption is incorrect because here a and b are having a common factor .

$= > \sqrt{3} \: is \: an \: irrational \: number$