# Explain why the square root of 89 must be between 9 and 10?

Jun 1, 2016

Evaluating the square of $9$ and $10$.

#### Explanation:

Recall that the square root is the inverse operation of the square.
Then we can see that the square of $9$ is

${9}^{2} = 81$

and the square of $10$ is

${10}^{2} = 100$

comparing with $89$ we have that $89 > 81$ and $89 < 100$.
Doing the square root we have

$\sqrt{89} > \sqrt{81}$ that is $\sqrt{89} > 9$

$\sqrt{89} < \sqrt{100}$ that is $\sqrt{89} < 10$.
We proved that the square root of $89$ is greater than $9$ and smaller than $10$.

Jun 1, 2016

$9 < \sqrt{89} < 10$

#### Explanation:

We have $81 = {9}^{2} < 89 < {10}^{2} = 100$
and $\sqrt{x}$ is monotonic for $0 < x < \infty$
which means that if $0 < {x}_{1} < {x}_{2} \to \sqrt{{x}_{1}} < \sqrt{{x}_{2}}$
then $9 < \sqrt{89} < 10$