Question

which one of the following is the greatest?
(A). $\sqrt{15}-\sqrt{13}$
(B). $\sqrt{13}-\sqrt{11}$
(C). $\sqrt{11}-\sqrt{9}$
(D). $\sqrt{9}-\sqrt{7}$

Answer 1

Hint: Take each of the options and rationalize it. If any no. x is greater than y then their reciprocal is $\dfrac{1}{x}$ will be less than $\dfrac{1}{y}$ . If the denominator is greater than the no. will be smaller. Thus find the option which has the least denominator that is the greatest no. among the options.

Complete step-by-step solution -
Now, from the question, we have been given four options out of which we need to find the greatest one. If any number x is greater than any number y the, i.e. $x>y$ then $\dfrac{1}{x}<\dfrac{1}{y}$ .
It is quite logical that if the number is bigger than it will take less quotient to divide a fixed quantity then the number which is less.
For example: $3>2$ but $\dfrac{1}{2}>\dfrac{1}{3}$ .
Now let us do the same for $\sqrt{15}-\sqrt{13},\sqrt{13}-\sqrt{11},\sqrt{11}-\sqrt{9},\sqrt{9}-\sqrt{7}$ .
We need to rationalize their values.
In $\sqrt{15}-\sqrt{13}$ , multiply by $\left( \sqrt{15}+\sqrt{13} \right)$ in the numerator and denominator
$\dfrac{\left( \sqrt{15}-\sqrt{13} \right)\left( \sqrt{15}+\sqrt{3} \right)}{\left( \sqrt{15}+\sqrt{13} \right)}=\dfrac{{{\left( \sqrt{15} \right)}^{2}}-{{\left( \sqrt{13} \right)}^{2}}}{\left( \sqrt{15}+\sqrt{13} \right)}$ We know that, $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$
$=\dfrac{15-13}{\sqrt{15}+\sqrt{13}}=\dfrac{2}{\sqrt{15}+\sqrt{13}}$
$\therefore \sqrt{15}-\sqrt{13}=\dfrac{2}{\sqrt{15}+\sqrt{13}}$ …………………………… (1)
Now let us rationalize $\left( \sqrt{13}-\sqrt{11} \right)$ by multiplying the numerator and denominator by $\left( \sqrt{13}+\sqrt{11} \right)$
$\therefore \dfrac{\left( \sqrt{13}-\sqrt{11} \right)\left( \sqrt{13}+\sqrt{11} \right)}{\left( \sqrt{13}+\sqrt{11} \right)}=\dfrac{{{\left( \sqrt{13} \right)}^{2}}-{{\left( \sqrt{11} \right)}^{2}}}{\sqrt{13}+\sqrt{11}}=\dfrac{13-11}{\sqrt{13}+\sqrt{11}}=\dfrac{2}{\sqrt{13}+\sqrt{11}}$ .
$\sqrt{13}-\sqrt{11}=\dfrac{2}{\sqrt{13}+\sqrt{11}}$ ……….(2)
Similarly let us rationalize $\left( \sqrt{11}-\sqrt{9} \right)$ by multiplying the narrator and denominator by $\left( \sqrt{11}+\sqrt{9} \right)$ .
$\therefore \dfrac{\left( \sqrt{11}-\sqrt{9} \right)\left( \sqrt{11}+\sqrt{9} \right)}{\left( \sqrt{11}+\sqrt{9} \right)}=\dfrac{{{\left( \sqrt{11} \right)}^{2}}-{{\left( \sqrt{9} \right)}^{2}}}{\sqrt{11}+\sqrt{9}}=\dfrac{11-9}{\sqrt{11}+\sqrt{9}}=\dfrac{2}{\sqrt{11}+\sqrt{9}}$
$\therefore \sqrt{11}-\sqrt{9}=\dfrac{2}{\sqrt{11}+\sqrt{9}}$ ………………………….. (3)
Now let us rationalize $\left( \sqrt{9}-\sqrt{7} \right)$ by multiplying the numerator and denominator by $\left( \sqrt{9}+\sqrt{7} \right)$ .
$\therefore \dfrac{\left( \sqrt{9}-\sqrt{7} \right)\left( \sqrt{9}+\sqrt{7} \right)}{\left( \sqrt{9}+\sqrt{7} \right)}=\dfrac{{{\left( \sqrt{9} \right)}^{2}}-{{\left( \sqrt{7} \right)}^{2}}}{\sqrt{9}+\sqrt{7}}=\dfrac{9-7}{\sqrt{9}+\sqrt{7}}=\dfrac{2}{\sqrt{9}+\sqrt{7}}$ .
$\sqrt{9}-\sqrt{7}=\dfrac{2}{\sqrt{9}+\sqrt{7}}$ …………………..(4)
Now let us compare all the 4 expression we got
$\sqrt{15}-\sqrt{13}=\dfrac{2}{\sqrt{15}+\sqrt{13}}$, $\sqrt{13}-\sqrt{11}=\dfrac{2}{\sqrt{13}+\sqrt{11}}$,
$\sqrt{11}-\sqrt{9}=\dfrac{2}{\sqrt{11}+\sqrt{9}}$, $\sqrt{9}-\sqrt{7}=\dfrac{2}{\sqrt{9}+\sqrt{7}}$.
The denominator of $\left( \sqrt{9}+\sqrt{7} \right)$ is smaller than the denominator of $\left( \sqrt{11}+\sqrt{9} \right)$ , which is smaller than $\left( \sqrt{13}+\sqrt{11} \right)$ . the largest denominator is $\left( \sqrt{15}+\sqrt{13} \right)$ all of the four expression.
In the beginning we told that if x is greater than y in $x>y$ then the reciprocal of x will be less than y.
i.e. $\dfrac{1}{x}<\dfrac{1}{y}$.
Hence $\left( \sqrt{15}+\sqrt{13} \right)$ has the largest denominator, hence its value will be small. Whereas $\left( \sqrt{9}+\sqrt{7} \right)$ has the smallest denominator. Thus it will have the largest value, thus we can write,
\[\dfrac{2}{\sqrt{15}+\sqrt{13}}<\dfrac{2}{\sqrt{13}+\sqrt{11}}<\dfrac{2}{\sqrt{11}+\sqrt{9}}<\dfrac{2}{\sqrt{9}+\sqrt{7}}\] .
Hence we can also write it as,
$\sqrt{15-}\sqrt{13}<\sqrt{13}-\sqrt{11}<\sqrt{11}-\sqrt{9}<\sqrt{9-}\sqrt{7}$ .
Thus the greatest value out of three $=\sqrt{9}-\sqrt{7}$ .
Hence we found the greatest value from among the four options as $\sqrt{9}-\sqrt{7}$
Therefore, option (D) is the correct answer.

Note: we can also find the largest value if you know the square roots of $\sqrt{15},\sqrt{13},\sqrt{11},\sqrt{9},$ and $\sqrt{7}$ .
$\sqrt{15}=3.873,\sqrt{13}==3.606,\sqrt{11}=3.317,\sqrt{9}=3,$ and $\sqrt{7}=2.646$ .
$\begin{align}
  & \sqrt{15}-\sqrt{13}=3.873-3.606=0.267 \\
 & \sqrt{13}-\sqrt{11}=3.606-3.317=0.289 \\
 & \sqrt{11}-\sqrt{9}=3.317-3=0.317 \\
 & \sqrt{9}-\sqrt{7}=3-2.646=.354 \\
\end{align}$
From this you will get $\sqrt{9}-\sqrt{7}$has the largest value.