Geometry

Question Analysis:
This is a contradiction-based logic problem involving parity (even/odd).
The main focus is assuming the negation of the conclusion, while keeping the hypothes is true.

Key Concept Explanation:
To prove "If p → q" by contradiction, we assume p is true and q is false — that is, both happen together, which is logically impossible.
Original:
p: x is odd
q: x² is odd
Contradiction: assume x is odd AND x² is not odd → that means x² is even

Step-by-Step Solution:
Assume x is an odd integer
Also assume x² is even
But the square of an odd number is always odd → contradiction
So original statement is true

Option Analysis:
A. ✅ Correct: the negation of "odd" is "even"
B. ❌ Irrelevant: primality not part of ...