Question

prove that √3 is irrational​

Answer 1

Answer:

$\huge\tt\orange{Answer✿}$

Step-by-step explanation:

We have to prove that √3 is irrational .

therefore it can be written in the form of

$\frac{a}{b}$

where a and b (b is not equal to 0)

and co prime ( no common factor other than 1)

Hence,

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

Squaring both side

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

Hence 3 is divided by a²

So, let's divide 3 also --------(1)

Hence we can say that

$\frac{a}{3} = c$

Where c is the common factor

So,

$a = 3c$

now we know that

3b² = a²

Putting a= 3c

${3b}^{2} = ( {3c}^{2} ) \\ 3 {b}^{2} = 9 {c}^{2} \\ {b}^{2} = \frac{1}{3} \times {9c}^{2} \\ {b}^{2} = (3 {c}^{2} ) \\ \frac{ {b}^{2} }{3} = {c}^{2}$

Hence 3 divides b²

So, 3 divides b also ------(2)

So a and b both are the factors of a and b

a and b are not co prime

Hence our assumptions are wrong

By contradiction

$\sqrt{3} \: \: is \: \:irrational$