Question

is it rational or irrational please proove it
$\frac{6}{3 \sqrt{2} }$

Answer 1

$\frac{6}{3 \sqrt{2} } = \frac{2 * 3}{3 \sqrt{2} } = \frac{ \sqrt{2} * 3\sqrt{2} }{3\sqrt{2}} = \sqrt{2}$
So we are getting $\sqrt{2}$
note:
$x = \sqrt{x} \sqrt{x} , x \ \textgreater \ =0$

So now we need to prove $\sqrt{2}$ is irrational

Note:
Any rational number can be expressed as 
$\frac{x}{y}$

where
x and y are in lowest form, (i mean they are co prime)

( by coprime we mean x and y do not have any common factor other than 1)

For example 2 can written as
$\frac{2}{1}$

Now Assume 
$\sqrt{2}$ is rational,
 i.e. it can be expressed as a rational fraction of the form $\frac{x}{y}$, where x and y are two relatively prime integers.

Now, since $\sqrt{2} = \frac{x}{y}$ ,
 we have $2 = \frac{x^2}{y^2}$,
or $2y^2 = x^2$.

Since $2y^2$  is even, 
$x^{2}$ must be even,
 and since $x^{2}$ is even,
 so is $x$ .

 Let $x = 2c$.
We have $4c^2 = 2y^2$ 
and thus $y^2 = 2 c^2$ .

Since  $2c^2$ is even, 
$y^2$  is even,
and since y   is even, so is x.

 However, two even numbers cannot be co prime,
 so $\sqrt{2}$  cannot be expressed as a rational fraction;

hence $\sqrt{2}$  is irrational