An odd integer minus an even integer is odd. Prove or Disprove: An odd integer minus an even integer is odd.
I am assuming you would define an odd integer and an even integer. than you would use quantifiers which shows your solution to be odd or even. I am unsure on how to show this...
 A: An even number is an integer which is divisible by $2$. In other words, $n$ is if and only if $n=2m$ for some integer $m$.
An odd number is a number which is $1$ more (or less) than an even number. In other words, $n$ is odd if and only if $n=2m+1$ for some integer $m$.
So suppose $n$ is odd and $n$ is odd. Write $n=2m+1$ and $n'=2m'$.
What can you say about $n-n'$?
A: Instead of a pure algebraic argument, which I don't dislike, it's also possible to see visually. Any even number can be represented by an array consisting of 2 x n objects, where n represents some number of objects. An odd number will be represented by a "2 by n" array with an item left over. Odd numbers don't have "evenly matched" rows or columns (depending on how you depict the array). Removing or subtracting an even number of items, then, will remove pairs of objects or items in the array. The leftover item (the "+1") will still remain and cannot have a "partner." Thus, an odd minus an even (or plus an even, for that matter) must be an odd result.
A: Let $m$ be odd, and let $n$ be even.


*

*An odd number $m$ is not divisible by $2$, and can be expressed in
the form $m = 2j + 1$, where $j$ is some integer.

*An even number $n$ is divisible by $2$ and can be expressed in the
form $n = 2k$, where $k$ is some integer.
Now: Subtract $n$ from $m$: express $m - n = (2j + 1) - (2k),\,$ and what is the resulting form of this difference?
A: Proof by contradiction: Let $j, k$  and $m$ are integers, suppose the difference between the even number $2j$ and odd number $2k+1 is even$, viz., $2j-(2k+1) = 2m$.  Then $2(j-k-m) = 1$.  But the right side of that equation is odd while the left side is even, which is a contradiction. QED.
A: Proof using the Peano axioms and related definitions:
$S(n)$ is the successor of $n$.
Definitions of even and odd:
$even(1) = false$, $odd(1) = true$,
$even(S(n)) = odd(n)$,
$odd(S(n)) = even(n)$.
Theorems proved by induction:
$even(n) = not(odd(n))$,
$odd(n) = not(even(n))$,
$even(S(S(n))) = even(n)$,
$odd(S(S(n))) = odd(n)$.
We want to show that
$odd(n)$ and $even(m)$ and $greater(n, m)$
implies $odd(minus(n, m))$.
Base case:
$odd(S(m), m)$.
Proof:
$even(m) \implies odd(S(m))$.
$minus(S(m),m) = 1
\implies odd(minus(S(m), m))$.
Induction step:
Suppose $odd(n)$ and $even(m)$ and $greater(n, m)$
and $odd(minus(n, m))$.
We want to show
$odd(minus(S(S(n)), m)$.
Since $greater(n, m)$,
$greater(S(n), m)$ and
$greater(S(S(n)), m)$.
Since $minus(S(n), m) = S(minus(n, m))$
and
$minus(S(S(n)), m) = S(S(minus(n, m)))$,
$odd(minus(n, m)) \implies even(minus(S(n), m))
\implies odd(minus(S(S(n)), m))
 $.
I think that does it,
except for having to prove that
all odd number greater than $m$
are taken care of.
It is surprising to me
how difficult it is
to reason about even and odd
Peanoishly.
