# How do you divide 6/sqrt2?

Mar 30, 2015

Take a one big root as:
$\sqrt{{6}^{2} / 2} = \sqrt{\frac{36}{2}} = \sqrt{18} = \sqrt{9 \cdot 2} = 3 \sqrt{2}$

Mar 30, 2015

It depends on what you mean. If you want the numeric value, you can simply ask a calculator to divide "6" by "$\setminus \sqrt{2}$", and you'll get the answer.

If you're looking for algebraic manipulation, one usually prefers to have the roots at the numerator, and whole numbers at the denominator. In this case, you need to consider these two super-easy facts:

1. Multiply by 1 doesn't change the value of an expression
2. $\setminus \sqrt{n} \setminus \sqrt{n} = \setminus \sqrt{{n}^{2}} = n$, because by definition the square root of a number $n$ is a number which gives $n$ if multiplied by itself.

So, starting from $\frac{6}{\sqrt{2}}$, we multiply the fraction by $\frac{\sqrt{2}}{\sqrt{2}}$, which is 1, and so changes nothing. We have
$\frac{6}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \setminus \frac{6 \setminus \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}}$

For the point 2, $\sqrt{2} \cdot \sqrt{2} = 2$, and so you have
$\setminus \frac{6 \setminus \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \setminus \frac{6 \sqrt{2}}{2} = 3 \setminus \sqrt{2}$