In n is an integer, which of the following CANNOT be an integer?

Options

A. `(n - 2)/2`
B. `sqrt(1/(n^2+2))`
C. `2/(n + 1)`
D. `sqrt(n^2 + 3)`

Answer

To solve the problem of determining which of the given expressions cannot be an integer when \( n \) is an integer, we will analyze each option step by step.

### Step-by-Step Solution:

1. **Option 1: \( \frac{n - 2}{2} \)**
- Let’s substitute \( n = 4 \):
\[ \frac{4 - 2}{2} = \frac{2}{2} = 1 \]
- Since \( 1 \) is an integer, this expression can be an integer for some integer values of \( n \).

2. **Option 2: \( \sqrt{\frac{1}{n^2 + 2}} \)**
- Let’s check for \( n = 1 \):
\[ \sqrt{\frac{1}{1^2 + 2}} = \sqrt{\frac{1}{3}} \text{ (not an integer)} \]
- Now check for \( n = 2 \):
\[ \sqrt{\frac{1}{2^2 + 2}} = \sqrt{\frac{1}{6}} \text{ (not an integer)} \]
- It appears that for any integer \( n \), this expression does not yield an integer. Thus, this expression **cannot be an integer**.

3. **Option 3: \( \frac{2}{n + 1} \)**
- Let’s substitute \( n = 1 \):
\[ \frac{2}{1 + 1} = \frac{2}{2} = 1 \]
- Since \( 1 \) is an integer, this expression can be an integer for some integer values of \( n \).

4. **Option 4: \( \sqrt{n^2 + 3} \)**
- Let’s substitute \( n = 1 \):
\[ \sqrt{1^2 + 3} = \sqrt{4} = 2 \]
- Since \( 2 \) is an integer, this expression can also be an integer for some integer values of \( n \).

### Conclusion:
After analyzing all the options, we find that the expression that **cannot be an integer** when \( n \) is an integer is **Option 2: \( \sqrt{\frac{1}{n^2 + 2}} \)**.