For an integer n, a student states the following :
I. If n is odd, `(n+1)^(2)` is even .
II. If n is even `(n-1)^(2)` is odd.
III. If n is even `sqrt((n-1))` is irrational .
Which of the above statements would be true ?

Options

A. Both I and III
B. Both I and II
C. All I, II and III
D. Both II and III

Answer

To determine which statements about the integer \( n \) are true, we will analyze each statement one by one.

### Statement I: If \( n \) is odd, \( (n+1)^2 \) is even.

1. **Assume \( n \) is odd.**
- An odd integer can be expressed as \( n = 2k + 1 \) for some integer \( k \).
2. **Calculate \( n + 1 \):**
- \( n + 1 = (2k + 1) + 1 = 2k + 2 = 2(k + 1) \).
- This shows that \( n + 1 \) is even.
3. **Calculate \( (n + 1)^2 \):**
- \( (n + 1)^2 = (2(k + 1))^2 = 4(k + 1)^2 \).
- Since \( 4(k + 1)^2 \) is a multiple of 4, it is even.

**Conclusion:** Statement I is true.

### Statement II: If \( n \) is even, \( (n-1)^2 \) is odd.

1. **Assume \( n \) is even.**
- An even integer can be expressed as \( n = 2m \) for some integer \( m \).
2. **Calculate \( n - 1 \):**
- \( n - 1 = 2m - 1 \).
- This shows that \( n - 1 \) is odd.
3. **Calculate \( (n - 1)^2 \):**
- \( (n - 1)^2 = (2m - 1)^2 = 4m^2 - 4m + 1 \).
- The expression \( 4m^2 - 4m \) is even, and adding 1 makes it odd.

**Conclusion:** Statement II is true.

### Statement III: If \( n \) is even, \( \sqrt{n-1} \) is irrational.

1. **Assume \( n \) is even.**
- Again, let \( n = 2m \) for some integer \( m \).
2. **Calculate \( n - 1 \):**
- \( n - 1 = 2m - 1 \), which is odd.
3. **Check if \( \sqrt{n - 1} \) is irrational:**
- The square root of an odd integer can be rational or irrational depending on the integer. For instance, if \( n = 2 \) (so \( n - 1 = 1 \)), then \( \sqrt{1} = 1 \), which is rational.
- Therefore, \( \sqrt{n - 1} \) can be rational.

**Conclusion:** Statement III is false.

### Final Conclusion:
- Statements I and II are true.
- Statement III is false.

**Correct Answer:** Statements I and II are true.

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