To solve for the largest integer whose square root is an irrational number between 5 and 7, we start by identifying the range of integers whose square roots fall within the specified limits.
First, we can express the condition mathematically:
\[ 5 < \sqrt{n} < 7 \]
Next, we square all parts of the inequality to eliminate the square root:
\[ 5^2 < n < 7^2 \]
This simplifies to:
\[ 25 < n < 49 \]
This means \( n \) must be an integer within the range of 26 to 48.
Now, we need \( \sqrt{n} \) to be an irrational number. This will be the case if \( n \) is not a perfect square. We identify the perfect squares in this range:
- The perfect squares between 25 and 49 are \( 36 \) (which is \( 6^2 \)) and \( 49 \) (which is \( 7^2 \)). However, since we are looking for \( n < 49 \), we'll treat \( 36 \) as the only relevant perfect square here.
Thus, the integers \( n \) that are not perfect squares are from 26 to 48, with the exception of \( 36 \):
\[ 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48 \]
Out of these integers, the largest integer is \( 48 \). We need to confirm that \( \sqrt{48} \) is indeed irrational:
- Since \( 48 \) is not a perfect square, \( \sqrt{48} \) cannot be expressed as a simple fraction or integer.
- We can also express \( \sqrt{48} \) as follows: \[ \sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3} \] Since \( \sqrt{3} \) is irrational, \( 4\sqrt{3} \) is also irrational.
Thus, the largest integer \( n \) whose square root is an irrational number between 5 and 7 is:
\[ \boxed{48} \]