Hi everyone.

I can't show this exercise:

Let $\displaystyle a,b\in{\mathbb{Z}$ be and $\displaystyle a<b$ then $\displaystyle a+1\leq{b}$

I've tried by contraction, but I haven't gotten anything.

Thanks

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- Mar 19th 2012, 08:48 AMFernandoInteger numbers
Hi everyone.

I can't show this exercise:

Let $\displaystyle a,b\in{\mathbb{Z}$ be and $\displaystyle a<b$ then $\displaystyle a+1\leq{b}$

I've tried by contraction, but I haven't gotten anything.

Thanks - Mar 19th 2012, 09:07 AMFernandoRe: Integer numbers
Well, I've done following:

Suppose that $\displaystyle a+1>b$, but $\displaystyle a<b\Leftrightarrow{a+1<b+1}$ hence it's a contraction because $\displaystyle a+1>b$ and $\displaystyle a+1<b+1$.

Therefore $\displaystyle a+1\leq{b}$

Is it correct? - Mar 19th 2012, 09:21 AMprincepsRe: Integer numbers
- Mar 19th 2012, 09:27 AMFernandoRe: Integer numbers
- Mar 19th 2012, 09:45 AMprincepsRe: Integer numbers
- Mar 19th 2012, 11:47 AMFernandoRe: Integer numbers
OK! thanks again !

- Mar 19th 2012, 12:51 PMPlatoRe: Integer numbers
It is difficult to answer that. We don't even know what set of axioms are in use here.

Here is a standard theorem in the development of number systems.

Given any two numbers, if $\displaystyle b-a>1$ then $\displaystyle (\exists j\in\mathbb{Z})[a<j<b].$

That depends upon Archimedes' principle and well ordering.