A mathematics post on APOP is long overdue. In recognition of a current favorite topic of mine in Discrete Mathematics (proving or disproving propositions), I invite you to join me on a fun trip through evens and odds. My professor calls these baby proofs; they’re minor league mental preparation.

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Definitions: For a \in \mathbb{Z}, a is \textbf{even} if a has the form a = 2m , where m \in \mathbb{Z} .
For b \in \mathbb{Z}, b is \textbf{odd} if b has the form b = 2m + 1 , again where m \in \mathbb{Z} .


Proposition: The sum of two even integers is even.
Pre-Proof Thoughts: Testing with a few pairs of even integers (2 + 4, 10 + 20, 104 + 320), we find that the sums are even. However, examples are not sufficient to prove a proposition. We want to confirm that it holds for all even integers.
Proof: Let a, b \in \mathbb{Z} be even integers. Then a = 2m and b = 2n , where m, n \in \mathbb{Z} . Then we have:

a + b = 2m + 2n , which is equivalent to
a + b = 2 \cdot (m + n) , which is equivalent to
a + b = 2p , where p = m + n \in \mathbb{Z} by Closure of Addition

The result, 2p , fulfills the definition for an even number. QED.


Proposition: The sum of two odd integers is even.
Pre-Proof Thoughts: We follow the same line of thought as before.
Proof: Let a, b \in \mathbb{Z} be odd numbers. Then a = 2m + 1 and b = 2n + 1 , where m, n \in \mathbb{Z} . Then we have:

a + b = (2m + 1) + (2n + 1) , which is equivalent to
a + b = 2 \cdot (m + n + 1) , which is equivalent to
a + b = 2p , where p = m + n + 1 \in \mathbb{Z} by Closure of Addition

The result, 2p , fulfills the definition for an even number. QED.


Proposition: The sum of an even integer and an odd integer is odd.
Proof: Let a \in \mathbb{Z} be even and b \in \mathbb{Z} be odd. Then a = 2m and b = 2n + 1 , where m, n \in \mathbb{Z} . Then we have:

a + b = (2m) + (2n + 1)
a + b = 2 \cdot (m + n) + 1
a + b = 2p + 1 , where p = m + n \in \mathbb{Z} by Closure of Addition

The result, 2p + 1 , fulfills the definition for an odd number. QED.


Proposition: The product of an even integer and an odd integer is even.
Proof: Let a \in \mathbb{Z} be even and b \in \mathbb{Z} be odd. Then a = 2m and b = 2n + 1 , where m, n \in \mathbb{Z} . Then we have:

a \cdot b = (2m) \cdot (2n + 1) , which is equivalent to
a \cdot b = 4mn + 2m , which is equivalent to
a \cdot b = 2 \cdot (2mn + m) , which is equivalent to
a \cdot b = 2 \cdot (p + m) , where p = 2mn \in \mathbb{Z} by Closure of Multiplication,
which is equivalent to
a \cdot b = 2q , where q = p + m \in \mathbb{Z} by Closure of Addition

The result, 2p + 1 , fulfills the definition for an odd number. QED.


Proposition: An integer cannot be both even and odd.
Proof (by Contradiction): Suppose a is both even and odd. Then a = 2m and a = 2n + 1. where m, n \in \mathbb{Z} . Setting these equations equal to one another gives:

a = a , which is equivalent to
2m = 2n + 1

There is no m \in \mathbb{Z} and n \in \mathbb{Z} such that 2m = 2n + 1 . Thus, a cannot be both even and odd, contradicting the proposition. QED.

 

Perhaps you disagree, but I for one find this proving business in post-Calculus undergraduate courses quite fun. Many aren’t as tidy as the even and odd proofs above, but once you go from proposition to end-of-proof symbol the satisfaction is unmatched.

For proofs that draw illogical conclusions because of logical fallacies, Dr. Math FAQ has some examples for the infamous 1 = 2.

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