Question

Show work and explain why the irrational number (the square root of 99) is between 9 and 11

Answer 1

To determine why the irrational number \(\sqrt{99}\) is between 9 and 11, we can start by squaring the endpoints (9 and 11) to see where \(\sqrt{99}\) lies in relation to these two integers.

1. **Square 9**:
\[
9^2 = 81
\]

2. **Square 11**:
\[
11^2 = 121
\]

Now we know:
- \(9^2 = 81\)
- \(11^2 = 121\)

Next, we'll determine where 99 fits in relation to these squares:

\[
81 < 99 < 121
\]

This inequality shows that \(99\) is greater than \(81\) and less than \(121\). We can now take the square root of the numbers in the inequalities:

\[
\sqrt{81} < \sqrt{99} < \sqrt{121}
\]

Substituting back the values of the square roots we just calculated:

\[
9 < \sqrt{99} < 11
\]

This proves that \(\sqrt{99}\) is indeed between 9 and 11.

### Conclusion
Thus, we conclude that \(\sqrt{99}\) is an irrational number that lies strictly between 9 and 11, as shown by the inequalities.