Question

$\text{Rationalize the denominator of}\frac{1}{\sqrt{7}}$$\text{and express the equivalent number}.$

• A.
  $\frac{7}{\sqrt{7}}$
  
• B.
  $\frac{\sqrt{7}}{7}$
  
• C.
  $\left(a\right)$ and $\left(b\right)$ above.
• D.
  None of the above

Answer 1

The correct option is B

$\frac{\sqrt{7}}{7}$

$\text{To rationalize the denominator of}\frac{1}{\sqrt{7}}$, multiply the denominator with its conjugate pair.

The conjugate pair of $\sqrt{7}\text{is}(-\sqrt{7})\text{or}\sqrt{7}$.

Let's multiply both the numerator and denominator with $\sqrt{7}$.
$\frac{1}{\sqrt{7}}=\frac{1}{\sqrt{7}}\times 1$
$=\frac{1}{\sqrt{7}}\times \frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{7}}{7}$

($\because \sqrt{7}\times \sqrt{7}=(\sqrt{7}{)}^2=7$)

Let's multiply both the numerator and denominator with $(-\sqrt{7})$.
$\frac{1}{\sqrt{7}}=\frac{1}{\sqrt{7}}\times 1$
$=\frac{1}{\sqrt{7}}\times \frac{-\sqrt{7}}{-\sqrt{7}}$
$=\frac{-\sqrt{7}}{-7}=\frac{\sqrt{7}}{7}$

$(\because \sqrt{7}\times -\sqrt{7}=-(\sqrt{7}{)}^2=-7)$

$\therefore$ $\frac{1}{\sqrt{7}}=\frac{\sqrt{7}}{7}$