MHB Every Number is between Two Consecutuve Integers

[OhMyMarkov]

Hello everyone, I want to prove that every number is between two consecutive integers.

$x\in R$. The archimedean property furnishes a positive integer $m_1$ s.t. $m_1.1>x$.
Apply the property again to get another positive integer $-m_2$ s.t. $-m_2.1>-x$.
Now, we have $-m_2<x<m_1$.

I stopped here, I know there exists an $m\leq m_1$ s.t. $m-1<x<m$, but I don't know how to continue.

Any help is appreciated!

 

[Sudharaka]

Hello everyone, I want to prove that every number is between two consecutive integers.

$x\in R$. The archimedean property furnishes a positive integer $m_1$ s.t. $m_1.1>x$.
Apply the property again to get another positive integer $-m_2$ s.t. $-m_2.1>-x$.
Now, we have $-m_2<x<m_1$.

I stopped here, I know there exists an $m\leq m_1$ s.t. $m-1<x<m$, but I don't know how to continue.

Any help is appreciated!

Hi OhMyMarkov, :)

Every number does not lie between two consecutive integers. You can easily verify this by taking any integer. :)

Kind Regards,
Sudharaka.

 

[Plato2]

Hello everyone, I want to prove that every number is between two consecutive integers.

As pointed out in reply #2, the way you worded this is problematic.
This is the correct problem: Given x\in\mathbb{R} there is an integer J such that J\le x&lt;J+1~.
To prove this first suppose that x&gt;0. Then use well ordering of the natural numbers to find the least positive integer, K, having the property that x&lt;K.
Because K has that minimal property we see that K-1\le x&lt;K.
So let J=K-1. Now you have two more cases: x=0\text{ or }x&lt;0~.