Proof that an integer is odd if its square is odd
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Hello everyone. Today we will prove that an integer is odd if its square is odd. We will use proof by contraposition as proof method.
Theorem: Let \(n \) be an integer. If \(n ^ 2 \) is an odd number, then \(n \) is also an odd number.
Proof:
The contrapositive of this statement is: Let \(n \) be an integer. If \(n \) is an even number, then \(n ^ 2 \) is also an even number.
Let us now prove the contrapositive. Since \(n \) is an even number, we can write it in the form \(n = 2k \) where \(k \) is an integer. By squaring this equation we get: \[n ^ 2 = (2k) ^ 2 \] \[n ^ 2 = 4k ^ 2 \] \[n ^ 2 = 2 \left(2k ^ 2 \right) \] Since \(k \) is an integer then \(2k ^ 2 \) must be an integer due to the closedness of multiplication and exponentiation operations on the set of integers. Denote \(2k ^ 2 \) by \(r \), then we have \(n ^ 2 = 2r \), from which we conclude that \(n ^ 2 \) is an even number. Since we have proved that the contrapositive of the original statement is correct, it means that the original statement must be correct as well.
\(\blacksquare \)
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