showing $\nexists\;\beta\in\mathbb N:\alpha<\beta<\alpha+1$ I want to prove that $\nexists\; \beta\in\mathbb N$ such that $\alpha<\beta<\alpha+1$ for all $\alpha\in\mathbb N$. I just want to use the Peano axioms and $+$ and $\cdot$
If $\alpha<\beta$ then there is a $\gamma\in\mathbb N$ such that $\beta=\alpha+\gamma$.
If $\beta<\alpha+1$ then there is a $\delta\in\mathbb N$ such that $\alpha+1=\beta+\delta$.
Now I tried to equalize the two equations and I got $\gamma\le0$ which is  contradictory to $\gamma\in\mathbb N$. But I used $\alpha+1-\delta=\beta$ in which the $-$ is problematic because I am not allowed to use it.
Anbody knows a better solution? Thanks a lot!

I assume that $0\notin \mathbb N$ and $x<y:\Leftrightarrow \exists z\colon x+z=y$.
You may already have proved 
$$\tag1\forall x,y\colon \exists z\colon x+y=S(z)$$
and
$$\tag2 \forall x,y,z\colon (x+y=x+z\rightarrow y=z).$$
Now the assumption $\alpha<\beta \land \beta<\alpha+1$ implies that there are $x,y$ such that $\alpha+x=\beta$ and $\beta+y=S(\alpha)$, hence $\alpha+x+y=S(\alpha)$.
Using $(1)$, there exists $z$ such that 
$\alpha+1=\alpha+S(z)=$ and by $(2)$, we conclude $1=S(z)$ contrary to $\neg\exists x\colon S(x)=1$.

In case you don't have $(1)$ or $(2)$ yet, they are readily proved by induction:


*

*If $x=1$, then $x+y=S(y)$. For the induction step, $S(x)+y=S(x+y)$ gives us $z=x+y$ as candidate immediately.

*If $x=1$, then $S(y)=x+y=x+z=S(z)$, i.e. $S(y)=S(z)$, implies $y=z$. Otherwise $S(x+y)=S(x)+y=S(x)+z=S(x+z)$ implies $x+y=x+z$ and thus $x=z$ by induction hypothesis.

