# How do you order the following from least to greatest 4 4/5, sqrt19, -sqrt5, -21, sqrt5?

Jan 19, 2017

$- 21 < - \sqrt{5} < \sqrt{5} < 4 \frac{4}{5} < \sqrt{19}$

#### Explanation:

Notice that $- 21$ and $- \sqrt{5}$ are negative

or

$- 21 < 0$ and $- \sqrt{5} < 0$

One way to think of this when it comes to the order of these two negatives is that

$- 21 < - 20$ therefore $- \sqrt{21} < - \sqrt{20} = - \left(\sqrt{4 \left(5\right)}\right) = - \left(\sqrt{4} \left(\sqrt{5}\right)\right) = - 2 \sqrt{5}$

So

$- 21 < - 2 \sqrt{5} < - \sqrt{5}$

Then

$- 21 < - \sqrt{5}$

That takes care of the negative numbers, but we still have the three positive numbers.

First notice

$4 \frac{4}{5} = \frac{16}{5} = 3.5$

and also

$4 < 5 < 9$

$\iff$

$\sqrt{4} < \sqrt{5} < \sqrt{9}$ Take the square root of all sides

$\iff$

$2 < \sqrt{5} < 3$

Therefore $\sqrt{5} < 4 \frac{4}{5}$

$4 \frac{4}{5} = 3.5 \implies {\left(4 \frac{4}{5}\right)}^{2} = {\left(3.5\right)}^{2}$

$= {\left(3 + 0.5\right)}^{2} = 9 + 2 \left(3\right) \left(0.5\right) + 0.25 = 9 + 3 + 0.25 = 12.25$

So we know that

$4 \frac{4}{5} = \sqrt{12.25}$

and since

$12.25 < 19$

then

$\sqrt{12.25} < \sqrt{19}$

So we get

$\underline{- 21 < - \sqrt{5} < \sqrt{5} < 4 \frac{4}{5} < \sqrt{19}}$