# How do you order the following from least to greatest without a calculator 0, -sqrt2, sqrt5, 13/4?

color(blue)(ul(bar(abs(color(black)(-sqrt2, 0, sqrt5, 13/4))))

#### Explanation:

No calculators are being used in this answer I'm writing!

We have four numbers to order from least to greatest:

$0 , - \sqrt{2} , \sqrt{5} , \frac{13}{4}$

Let's first see that there is only one negative number - that has to be the smallest. And then there's 0, which is certainly smaller than the remaining two positive numbers, so we have so far:

$- \sqrt{2} , 0$

So the question is now which of the two remaining numbers, $\sqrt{5} , \frac{13}{4}$, is smaller.

Let's estimate the value of each of these.

$\sqrt{5}$ is going to be more than $\sqrt{4}$ and less than $\sqrt{9}$, so we can write that as:

$\sqrt{4} < \sqrt{5} < \sqrt{9}$

We know that $\sqrt{4} = 2 \mathmr{and} \sqrt{9} = 3$, and so:

$2 < \sqrt{5} < 3$

Is $\frac{13}{4}$ bigger or smaller than that? Let's do something similar with the fraction - we know that $\frac{13}{4}$ is bigger than $\frac{12}{4}$ and smaller than $\frac{16}{4}$, so we can write:

$\frac{12}{4} < \frac{13}{4} < \frac{16}{4}$

We know that $\frac{12}{4} = 3 \mathmr{and} \frac{16}{4} = 4$, and so:

$3 < \frac{13}{4} < 4$

With $\sqrt{5} < 3 \mathmr{and} \frac{13}{4} > 3$, we can now finish ordering our numbers:

color(blue)(ul(bar(abs(color(black)(-sqrt2, 0, sqrt5, 13/4))))