Integer numbers

Printable View

March 19th 2012, 08:48 AM
Fernando

Integer numbers

Hi everyone.

I can't show this exercise:

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

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

Thanks
March 19th 2012, 09:07 AM
Fernando

Re: Integer numbers

Well, I've done following:

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

Therefore $a+1\leq{b}$

Is it correct?
March 19th 2012, 09:21 AM
princeps

Re: Integer numbers

Quote:

Originally Posted by Fernando

Hi everyone.

I can't show this exercise:

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

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

Thanks

Suppose :

$a+1 > b$

hence :

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

so we have contradiction because :

$a < b$
March 19th 2012, 09:27 AM
Fernando

Re: Integer numbers

Quote:

Originally Posted by princeps

Suppose :

$a+1 > b$

hence :

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

so we have contradiction because :

$a < b$

Thanks! But is my answer correcto ?
March 19th 2012, 09:45 AM
princeps

Re: Integer numbers

Quote:

Originally Posted by Fernando

Thanks! But is my answer correcto ?

Since :

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

cannot be true at the same time your reasoning is correct .
March 19th 2012, 11:47 AM
Fernando

Re: Integer numbers

OK! thanks again !
March 19th 2012, 12:51 PM
Plato

Re: Integer numbers

Quote:

Originally Posted by Fernando

Well, I've done following:
Suppose that $a+1>b$ , but $a<b\Leftrightarrow{a+1<b+1}$ hence it's a contraction because $a+1>b$ and $a+1<b+1$ .
Therefore $a+1\leq{b}$
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 $b-a>1$ then $(\exists j\in\mathbb{Z})[a<j<b].$
That depends upon Archimedes' principle and well ordering.