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.

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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