prove: if n is even, then n+1 is not even This proof seems so simple that it's hard (if that makes any sense.)
based on the definition, n is even iff there exists k such that n = 2k.
What I really want to say is (big picture)
By definition,  let $n = 2k.\;$ Then $n+1 = 2k + 1$.
$2k + 1$ is not divisible by $2$, therefore $n + 1$ is not even.
I can't seem to figure out how to show the work.  Any help would be appreciated.
 A: You'd need to add a step to relate to your definition of even number.
Suppose $2k+1$ is even. then there exist some $k_1$ such that $2k+1=2k_1$. What does this mean?
Well if that is the case, then simple algebric manipulations get us to 
$$
1=2(k_1-k)
$$
which means $(k_1-k)=1/2$. But $k,k_1$ are integers, and $1/2$ is not, $1$ being smaller than $2$. We get a contradiction.
A: What's missing so far is a rigorous proof that $1$ is not even. Here is one that starts from the Peano arithmetic.
Suppose $1=S(0)$ is even. That means that for some $a$ we have that $S(0)=a+a$. Now either $a=0$ or $a=S(b)$ for some $b$. (This can be proved by induction on $a$ if you don't already know it).
In the case where $a=0$, we have $S(0)=0+0=0$, but $S(x)\ne 0$ for all $x$ by axiom, so this is a contradiction.
On the other hand, if $a=S(b)$, then $$S(0)=S(b)+S(b)=S(S(b)+b)=S(b+S(b))=S(S(b+b))$$
where in the middle step I'm using commutativity of addition which I assume we have already proved. But the successor function is required to be injective, so the extremes of this equation gives $0=S(b+b)$ which again contradicts the $S(x)\ne 0$ axiom.
Therefore there is no $a$ such that $1=a+a$ -- in other words, $1$ is not even.
A: I love contradiction. Here is how I would do it:
Let n and n+1 both be even,
Therefore, $n=2k$ for some k and $n+1=2j$ for some  $j,k \in \mathbb{I}$
Subtracting, $n+1-n=2j-2k$.
$$1 = 2(j-k)$$
$$\frac{1}{2} = j-k$$
But, 1<2 so, the fraction is not an integer and the difference of 2 integers is necessarily an integer. Thus, Contradiction!
A: If $n$ is even, then 
$$n\equiv 0 \quad(\mod2) \quad\text  {and} \quad n+1 \equiv 1\quad (\mod2)$$
which implies that $n+1$ is not even.
A: For a given pair of even numbers $2a>2b$ it is the case that $2a-2b=2(a-b)$. Thus the difference between two even numbers is even. However, the difference between $n$ and $n+1$ is $1$, which is not an even number. Thus it cannot be the case that both $n$ and $n+1$ are even.
This is a slightly different take that proves that, in general, no two consecutive numbers can be even.
A: Here is another short proof that that $1$ is not even. It does not go down to the foundations like Henning's, but may be appropriate depending on the tools you are working with.

Consider the expression $1 \div 2$. By the Division Algorithm, there are unique $q$ and $r < 2$ such that
$$
1 = 2q + r,
$$
Since $1 < 2$, it must be that $q = 0$ and $r = 1$, that is
$$
1 = 2\cdot 0 + 1.
$$
Since we have a nonzero remainder, it follows that $2$ does not divide $1$. That is, $1$ is not even.
A: Proving an axiom is comparitively hard. For that, we have to use another axiom which is much more complicated than the hypothesis. 
So, I have two proofs.


*

*Contradiction: Assume that $n$ and $n + 1$ are both even. This implies that for some $p \in \mathbb Z$, $ \ \ n = 2p$ and so $n + 1 = 2p + 1$. But  all numbers in the form $2 p + 1$ are of the odd parity. Q.E.D.

*Modular arithmetic: We know that if $n \pmod{2} \equiv0$, it follows that $n+1\pmod2 \equiv1$ which implies that $n + 1$ is odd.

