Question

prove.
$\sqrt{2}$
is irrational number​

Answer 1

Answer :

Given :

✓2 an irrational number

Required to prove :

1. ✓2 is an irrational number .

Solution :

Let assume on the contradictory that ✓2 is an rational number .

So,

Let ✓2 is equal to p and q

That is ,

✓2 = p/q ( where p and q are integers , q ≠ 0 , p a and are co - primes ).

So, let's solve this problem

Here ,

✓2 = p/q

( do cross multiplication on both sides )

✓2q = p

Now let's do squaring on both sides

we get,

$( \sqrt{2} q {)}^{2} = (p {)}^{2} \\ \\ 2 {q}^{2} = {p}^{2}$

Here ,

2 divides p^2

So,

2 divides p also

This is because according to Fundamental Theorem of Arithmetic if an integer p divides a^2 then p divides a too.

Now ,

Let consider p to be 2k

( where k is an integer )

Now substitute this in the above results

That is ,

2q = p

2q = 2k ( from the above consideration )

so,

Let's do squaring on both sides.

$( \sqrt{2} q {)}^{2} = (2k {)}^{2} \\ \\ 2 {q}^{2} = 4 {k}^{2} \\ \\ {q}^{2} = \frac{4 {k}^{2} }{2} \\ \\ {q}^{2} = 2 {k}^{2} \\ \\ 2 {k}^{2} = {q}^{2}$

So,

here we can say that

2 divides q^2

similarly, 2 divides q also .

Hence,

Therefore we can say that

p and q have more than 1 factor .

But according to Rational number property they should have only one common factor that is one.

This states the fact that our contradict is wrong .

Hence , our assumption is wrong .

Therefore,

✓2 is an irrational number .