Proving there is no natural number which is both even and odd

Question

I've run into a small problem while working through Enderton's Elements of Set Theory. I'm doing the following problem:

Call a natural number even if it has the form $2\cdot m$ for some $m$. Call it odd if it has the form $(2\cdot p)+1$ for some $p$. Show that each natural number number is either even or odd, but never both.

I've shown most of this, and along the way I've derived many of the results found in Arturo Magidin's great post on addition, so any of the theorems there may be used. It is the 'never both' part with which I'm having trouble. This is some of what I have:

Let $$ B=\{n\in\omega\ |\neg(\exists m(n=2\cdot m)\wedge\exists p(n=2\cdot p+1))\}, $$ the set of all natural numbers that are not both even and odd. Since $m\cdot 0=0$, $0$ is even. Also $0$ is not odd, for if $0=2\cdot p+1$, then $0=(2\cdot p)^+=\sigma(2\cdot p)$, but then $0\in\text{ran}\ \sigma$, contrary to the first Peano postulate. Hence $0\in B$. Suppose $k\in B$. Suppose $k$ is odd but not even, so $k=2\cdot p+1$ for some $p$. Earlier work of mine shows that $k^+$ is even. However, $k^+$ is not odd, for if $k^+=2\cdot m+1$ for some $m$, then since the successor function $\sigma$ is injective, we have $$ k^+=2\cdot m+1=(2\cdot m)^+\implies k=2\cdot m $$ contrary to the fact that $k$ is not even.

Now suppose $k$ is even, but not odd. I have been able to show that $k^+$ is odd, but I can't figure a way to show that $k^+$ is not even. I suppose it must be simple, but I'm just not seeing it. Could someone explain this little part? Thank you.

Answer 1

HINT $\ $ Here's the inductive step: $\rm\ 2m \ne 2n+1\ \Rightarrow\ 2m+1 \ne 2(n+1)$

Answer 2

Here is a complete proof. We assume that $n^+=m^+$ implies $n=m$ (*) and say that $n$ is even if it is $m+m$ for some natural number $m$ and odd if it is $(m+m)^+$ for some natural number $m$.

We know that $\phi=\phi+\phi$ so $\phi$ is even and $\phi\neq p^+$ for any $p$ and so is therefore not odd. Now assume that $k\neq\phi$ and $k$ is never both odd and even. Now consider $k^+.$
Suppose $k^+$ is odd. Then $k^+=(n+n)^+$ for some $n$. So by (*), $k=n+n$ so $k$ is even.
Suppose $k^+$ is even. Then $k^+=(m+m)$ for some $m$. We know $m\neq\phi$ as otherwise $k^+=\phi$, so $m=p^+$ for some $p$. Hence $k^+=(p^++p^+)=(p^++{p)}^+$. So by (*), $k=p+p^+=(p+p)^+$, which is odd. Hence if $k^+$ is even and odd then $k$ is odd and even, a contradiction to our induction hypothesis. Hence $k^+$ is never both even and odd, and we have the result by induction.

Answer 3

Suppose there exists some $n \in \mathbb{N}$ which is both even and odd. Then $n= 2m = 2p+1$. So $2m = 2p+1$ or $2(m-p) = 1$. Contradiction.

Answer 4

Referring to OP's question on proving "if $k$ is not odd, then $k^+$ is not even", here is an attempt.

We prove the contrapositive of the statement, i.e. we prove "if $k^+$ is even, then $k$ is odd". Let $k^+=2n$ for some $n\in\mathbb{N}$. Note that $n\neq 0$; for otherwise $k^+=2\cdot 0=0$, contradicting the fact that $0$ is not the successor of any natural number. Since $n\neq 0$, we can write $n=m^+$ for some $m\in\mathbb{N}$, as every natural number other than $0$ is the successor of some other natural number. Then $$k^+=2m^+=2m+2=2m+1^+=(2m+1)^+.$$ Since the successor operator $\sigma:\mathbb{N}\to\mathbb{N}$ is injective, $k=2m+1$.