Tag: properties of irrational numbers

$Here,\quad we\quad will\quad carry\quad out\quad rationalization.\quad \ \frac { 3-\sqrt { 3 }  }{ 3+\sqrt { 3 }  } =\frac { 3-\sqrt { 3 }  }{ 3+\sqrt { 3 }  } x\frac { 3-\sqrt { 3 }  }{ 3-\sqrt { 3 }  } =\frac { { (3-\sqrt { 3 } ) }^{ 2 } }{ (3+\sqrt { 3) } (3-\sqrt { 3 } ) } =\frac { 9+3-6\sqrt { 3 }  }{ 9-3 } =\frac { 12-6\sqrt { 3 }  }{ 6 } =\frac { 2-\sqrt { 3 }  }{ 1 } \ Since\quad \sqrt { 3 } is\quad irrational\quad number\quad and\quad subtraction\quad of\quad rational\quad and\quad irrational\quad is\quad irrational.\ The\quad given\quad expression\quad is\quad irrational.\ \quad $

Let be the Number are $\sqrt{5}  and  -\sqrt{5}$
Sum of Number  $\left(\sqrt{5}\right) + \left(-\sqrt{5}\right)$
$\sqrt{5}-\sqrt{5} = 0$
Which is a rational number

Let be the Number are $4\sqrt{3}  and  2\sqrt{3}$
Difference of Number  $4\sqrt{3} - 2\sqrt{3} = 2\sqrt{3}$
Which is a irrational number

Let be the Number are $\sqrt{15}  and  \sqrt{5}$
Quotient of Numbers  $\frac{\sqrt{15}}{\sqrt{5}} = \sqrt{\frac{15}{5}} = \sqrt{3} $
Which is a irrational number

Let be the Number are $2\sqrt{5}  and  3\sqrt{5}$
Sum of Number  $2\sqrt{5} + 3\sqrt{5} = 5\sqrt{5}$
Which is a irrational number

Let be the Number are $\sqrt{8}  and  \sqrt{2}$
Quotient of Numbers  $\frac{\sqrt{8}}{\sqrt{2}} = \sqrt{\frac{8}{2}} = \sqrt{4} = 2 $
Which is a rational number

Let be the Number are $\sqrt{8}  and  \sqrt{2}$
Product of Numbers  $\sqrt{8}\times \sqrt{2} = \sqrt{16} = 4$
Which is a rational number

Let be the Number are $\sqrt{2}  and  \sqrt{3}$
Product of Numbers  $\sqrt{2}\times \sqrt{3} = \sqrt{6} $
Which is a irrational number

$  Log AB = log A + log B $
Also, $ log a^b = b log a $

So, $ log _{4} 18 = \frac { log 2 \times 3^2}{log 4}  = \frac { log 2 \times 3^2}{log 2^2}  = \frac { log 2}{2log 2} + \frac {2 log 3}{2 log 2}  = \frac {1}{2} +  \frac {log 3}{log 2}   $

As both $ log 2 $ and $ log 3 $ are irrational numbers, $ log _{x} 18 $ is an irrational number too. 

For a number to be divisible by $\sqrt { 2 } $, it must be an irrational number. An integer is not an irrational,