2
$\begingroup$

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!

  • 0
    So you have from the definition of addition that $S(\alpha)=\alpha+1$ and so if $\beta > \alpha$ then $\exists \gamma$ s.t. $\alpha+\gamma=\beta$ so then if $\gamma =1$ then$\beta=\alpha+1$ and if $\gamma > 1$ then $\beta > \alpha+1$ which are both contradictions2012-10-20

3 Answers 3