Using contrapositive method prove that, if `n^(2)` is an even integer , then n is also an even integer.

Explanation

To prove that if \( n^2 \) is an even integer, then \( n \) is also an even integer using the contrapositive method, we will follow these steps:

### Step 1: Understand the Statements
Let:
- \( P \): \( n^2 \) is an even integer.
- \( Q \): \( n \) is an even integer.

We want to prove that \( P \implies Q \) using the contrapositive, which states that \( \neg Q \implies \neg P \).

### Step 2: Define the Contrapositive
The contrapositive of the statement \( P \implies Q \) is:
- \( \neg Q \): \( n \) is not an even integer (which means \( n \) is an odd integer).
- \( \neg P \): \( n^2 \) is not an even integer (which means \( n^2 \) is an odd integer).

So, we need to prove that if \( n \) is odd, then \( n^2 \) is also odd.

### Step 3: Assume \( n \) is Odd
Assume \( n \) is an odd integer. By definition, an odd integer can be expressed as:
\[ n = 2k + 1 \]
for some integer \( k \).

### Step 4: Calculate \( n^2 \)
Now, we calculate \( n^2 \):
\[ n^2 = (2k + 1)^2 \]
Expanding this, we get:
\[ n^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1 \]
This shows that \( n^2 \) can be expressed in the form \( 2m + 1 \), where \( m = 2k^2 + 2k \) is an integer.

### Step 5: Conclusion
Since \( n^2 \) can be expressed as \( 2m + 1 \), it follows that \( n^2 \) is odd. Thus, we have shown that if \( n \) is not even (i.e., \( n \) is odd), then \( n^2 \) is also not even (i.e., \( n^2 \) is odd).

### Final Statement
Since we have proven that \( \neg Q \implies \neg P \), by the contrapositive method, we conclude that if \( n^2 \) is an even integer, then \( n \) must also be an even integer.

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