To find the largest integer whose square root is an irrational number between 3 and 4, we first identify the range of numbers we are interested in.
The square root of a number \( n \) is between 3 and 4 if: \[ 3 < \sqrt{n} < 4 \]
Squaring all parts of the inequality gives us: \[ 3^2 < n < 4^2 \] or \[ 9 < n < 16 \]
This means \( n \) must be an integer in the range from 10 to 15, inclusive. The integers in this range are:
- 10
- 11
- 12
- 13
- 14
- 15
Next, we need \( \sqrt{n} \) to be an irrational number. The square root is irrational for any integer that is not a perfect square. The perfect squares within our range are 9 (which is less than 10) and 16 (which is greater than 15), meaning there are no perfect squares between 10 and 15. Therefore, all integers from 10 to 15 have irrational square roots.
Among the integers from 10 to 15, the largest integer is 15. We need to confirm that \( \sqrt{15} \) is indeed irrational and falls within the earlier specified range:
Calculating the square root, \[ 3^2 = 9 \quad \text{and} \quad 4^2 = 16 \] Thus, \[ 3 < \sqrt{15} < 4 \]
Since we have confirmed that \( \sqrt{15} \) is irrational and lies between 3 and 4, we conclude that the largest integer whose square root is an irrational number between 3 and 4 is:
\[ \boxed{15} \]