Question

prove that root 3 is irrational number​

Answer 1

Step-by-step explanation:

The number √3 is irrational ,it cannot be expressed as a ratio of integers a and b. To prove that this statement is true, let us Assume that it is rational and then prove it isn't (Contradiction).

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Answer 2

Let us assume √3 is rational. So,

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

Where a and b are Co - primes....

$b \sqrt{3 } = a$

Squaring on both sides , we get

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

$3 {b}^{2} = {a}^{2} ........eq(1)$

here 3 divdies a^2 . So ,3 divides a

let a = 3c , we get by substituting...

${3b}^{2} = {(</strong><strong>3</strong><strong>c)}^{2}$

${3b}^{2} = 9 {c}^{2}$

${b}^{2} = {3c}^{2} ......eq(2)$

here 3 divides b^2 . So ,3 divides b

By eq (1) and (2) , we Get

3 is the common factor for a and b other 1....

This contradict the fact that a and b are Co- primes...

That means our assumption is wrong..

Therefore √3 is irrational

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