Tag: rational and irrational numbers

SInce, we know that prime numbers are those which are never perfect square and not divisible by any other number except by itself.
which are $2,3,5,7,...$
Clearly, if $p$ is prime then $\sqrt p $ is irrational number.
Option $D$ is correct. 

$3\sqrt{18}=9\sqrt{2},$ which is the product of a rational and an irrational number and so an irrational number.

Let's assume that $6+\sqrt2$ is rational..... 

then 

$6+\sqrt2 = p/q $

$\sqrt2 =( p-6q)/(q) $ 

now take $p-6q$ to be P and $q$ to be Q........where P and Q are integers 

which means, $\sqrt2= P/Q$...... 

But this contradicts the fact that $\sqrt2$ is rational 

So our assumption is wrong and $6+\sqrt2$ is irrational.

   $\sqrt[8]{80}, \sqrt[4]{40}$
$={80}^{\frac{1}{8}}, {40}^{\frac{1}{4}}$
$={80}^{\frac{1}{8}}, {40}^{\frac{2}{8}}$
$={80}^{\frac{1}{8}}, {1600}^{\frac{1}{8}}$
Now,
   $80<1600$
$=>{80}^{\frac{1}{8}}<{1600}^{\frac{1}{8}}$
$=>\sqrt[8]{80}< \sqrt[4]{40}$

Rationalizing factor of  $\displaystyle \frac{2\sqrt{3}}{\sqrt{5}}$ is $\sqrt{5}$

$\dfrac{3\sqrt{3}}{2\sqrt{2}}=\dfrac{3\times 1.73214}{2\times 1.41429} = \dfrac{5.19642}{2.85828}
=1.82151$
$\dfrac{2\sqrt{2}}{3\sqrt{3}}=\dfrac{2\times 1.41429}{3\times 1.73214}=\dfrac{2.85828}{5.19642}=0.55004$
$\therefore \dfrac{3\sqrt{3}}{2\sqrt{2}} >\dfrac{2\sqrt{2}}{3\sqrt{3}}$