# Integer numbers

• Mar 19th 2012, 08:48 AM
Fernando
Integer 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 AM
Fernando
Re: 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 AM
princeps
Re: Integer numbers
Quote:

Originally Posted by Fernando
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

Suppose :

$\displaystyle a+1 > b$

hence :

$\displaystyle a>b-1 \Rightarrow a \geq b$

so we have contradiction because :

$\displaystyle a < b$
• Mar 19th 2012, 09:27 AM
Fernando
Re: Integer numbers
Quote:

Originally Posted by princeps
Suppose :

$\displaystyle a+1 > b$

hence :

$\displaystyle a>b-1 \Rightarrow a \geq b$

so we have contradiction because :

$\displaystyle a < b$

Thanks! But is my answer correcto ?
• Mar 19th 2012, 09:45 AM
princeps
Re: Integer numbers
Quote:

Originally Posted by Fernando
Thanks! But is my answer correcto ?

Since :

$\displaystyle a+1 > b ~\text{and}~ a+1 < b+1$

cannot be true at the same time your reasoning is correct .
• Mar 19th 2012, 11:47 AM
Fernando
Re: Integer numbers
OK! thanks again !
• Mar 19th 2012, 12:51 PM
Plato
Re: Integer numbers
Quote:

Originally Posted by Fernando
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?

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.