To find an irrational number between 4 and 8, we can start by calculating the square roots given in the options and see which one lies in that range.
-
\(\sqrt{17}\):
- We know that \(4^2 = 16\) and \(5^2 = 25\).
- Since \(16 < 17 < 25\), we have \(4 < \sqrt{17} < 5\).
- Therefore, \(\sqrt{17}\) is between 4 and 5.
-
\(\sqrt{12}\):
- We know that \(3^2 = 9\) and \(4^2 = 16\).
- Since \(9 < 12 < 16\), we have \(3 < \sqrt{12} < 4\).
- Therefore, \(\sqrt{12}\) is between 3 and 4.
-
\(\sqrt{5}\):
- We know that \(2^2 = 4\) and \(3^2 = 9\).
- Since \(4 < 5 < 9\), we have \(2 < \sqrt{5} < 3\).
- Therefore, \(\sqrt{5}\) is between 2 and 3.
-
\(\sqrt{8}\):
- We know that \(2^2 = 4\) and \(3^2 = 9\).
- Since \(4 < 8 < 9\), we have \(2 < \sqrt{8} < 3\).
- Therefore, \(\sqrt{8}\) is between 2 and 3.
None of option 2 or 3 contained irrational numbers in the desired range. The only one that is an irrational number and between 4 and 8 is \(\sqrt{17}\).
So, the answer is:
- \(\sqrt{17}\), which is an irrational number between 4 and 8.
1 answer