1
$\begingroup$

A homework question recently asked for me to simplify:

$\frac{1}{\sqrt{7}} \div {7}$

It's easy to see that this becomes

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

But according to wolfram alpha this is also equal to $\frac{\sqrt{7}}{49}$.

What sequence of steps can I use to get the second representation of this quantity from the first?

$\endgroup$
3
  • 6
    $\begingroup$ $$\frac{1}{7\sqrt{7}}\cdot \frac{\sqrt{7}}{\sqrt{7}}.$$ $\endgroup$ Dec 8, 2013 at 14:52
  • $\begingroup$ Multiply both numerator and denominator bt Sqrt[7] and simplify the denominator [7 Sqrt(7) Sqrt(7)] = 7 7 = 49 $\endgroup$ Dec 8, 2013 at 14:52
  • $\begingroup$ If you replace 7 by x, you have the same solution $\endgroup$ Dec 8, 2013 at 15:04

3 Answers 3

7
$\begingroup$

$\dfrac{1}{\sqrt{7}}\div7=\dfrac{1}{7\sqrt{7}}=\dfrac{1}{7\sqrt{7}}\cdot\dfrac{\sqrt{7}}{\sqrt{7}}=\dots$

$\endgroup$
3
$\begingroup$

$$\frac{1}{7\sqrt{7}} =\frac{\sqrt{7}}{7\sqrt{7}\cdot\sqrt{7}}=\frac{\sqrt{7}}{7\cdot7}=\frac{\sqrt{7}}{49}$$

$\endgroup$
2
$\begingroup$

This is called denominator rationalization. You multiply numerator and denominator with the same root, so you effectively move from denominator into the numerator.

$\endgroup$
1
  • $\begingroup$ FWIW, a similar technique is used for dividing complex number - you multiply both numerator and denominator by the conjugate of the denominator, to make the denominator real. $\endgroup$
    – Trang Oul
    Feb 21, 2023 at 14:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .