Question

The greatest number among √11-√5,√19-√13 is

Answer 1

Step-by-step explanation:

hope you understand

please refer to the attachment

Answer 2

Answer:

$\sqrt{11}$ - $\sqrt{5}$ is greater than  $\sqrt{19}$ - $\sqrt{13}$

Step-by-step explanation:

The given numbers are: $\sqrt{11}$ - $\sqrt{5}$  and  $\sqrt{19}$ - $\sqrt{13}$

To make the comparison easy, we rationalize the numbers

=>  $\sqrt{11}$ - $\sqrt{5}$ * $\frac{\sqrt{11}+\sqrt{5} }{\sqrt{11}+\sqrt{5} }$   and  $\sqrt{19}$ - $\sqrt{13}$ * $\frac{\sqrt{19}+\sqrt{13} }{\sqrt{19}+\sqrt{13} }$

=>   $\frac{(\sqrt{11})^2-(\sqrt{5})^2 }{\sqrt{11}+\sqrt{5} }$  and   $\frac{(\sqrt{19})^2-(\sqrt{13})^2 }{\sqrt{19}+\sqrt{13} }$

=>   $\frac{11-5 }{\sqrt{11}+\sqrt{5} }$  and   $\frac{19-13 }{\sqrt{19}+\sqrt{13} }$

=>   $\frac{6 }{\sqrt{11}+\sqrt{5} }$  and   $\frac{6 }{\sqrt{19}+\sqrt{13} }$

Now as the numerators are equal, we can compare the numbers by looking at the denominator.

We know that the number with higher denominator will be less and vice versa.

We know that  $\sqrt{11}$  <  $\sqrt{19}$

and  $\sqrt{5}$ < $\sqrt{13}$

On adding these, we get

$\sqrt{11}$ + $\sqrt{5}$  <  $\sqrt{19}$ + $\sqrt{13}$

Therefore as the denominator of  $\frac{6 }{\sqrt{19}+\sqrt{13} }$ is greater, the number will be smaller.

Therefore,   $\frac{6 }{\sqrt{11}+\sqrt{5} }$ is the greater number.

=>  $\sqrt{11}$ - $\sqrt{5}$ is greater than  $\sqrt{19}$ - $\sqrt{13}$