Show that the following statement is true by the method of contrapositive. P: If x is an integer and `x^(2)` is even, then x is also even.

Answer

To show that the statement "If x is an integer and x² is even, then x is also even" is true using the method of contrapositive, we will first restate the original statement in terms of its contrapositive.

### Step-by-step Solution:

1. **Identify the Original Statement**:
The original statement is:
- If \( P \): "x is an integer and \( x^2 \) is even"
- Then \( Q \): "x is even"

In logical terms, this can be expressed as:
\[ P \implies Q \]

2. **Formulate the Contrapositive**:
The contrapositive of the statement \( P \implies Q \) is:
\[ \neg Q \implies \neg P \]
Where:
- \( \neg Q \): "x is not even" (which means x is odd)
- \( \neg P \): "x is not an integer or \( x^2 \) is not even"

Thus, the contrapositive becomes:
- If \( x \) is odd, then \( x^2 \) is odd.

3. **Assume \( x \) is Odd**:
Let’s assume \( x \) is an odd integer. By definition, an odd integer can be expressed as:
\[ x = 2k + 1 \]
for some integer \( k \).

4. **Calculate \( x^2 \)**:
Now, we will calculate \( x^2 \):
\[ x^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 4k(k + 1) + 1 \]
Notice that \( 4k(k + 1) \) is even (since it is a multiple of 4), and adding 1 to an even number results in an odd number.

5. **Conclusion**:
Therefore, we conclude that if \( x \) is odd, then \( x^2 \) is odd. This means:
\[ \neg Q \implies \neg P \]
Hence, we have shown that the contrapositive is true, which implies that the original statement is also true.

### Final Statement:
Thus, we have proven that if \( x \) is an integer and \( x^2 \) is even, then \( x \) is also even.

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