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 $$J\le x<J+1~.$$
To prove this first suppose that $$x>0$$. Then use well ordering of the natural numbers to find the least positive integer, $$K$$, having the property that $$x<K$$.
Because $$K$$ has that minimal property we see that $$K-1\le x<K$$.
So let $$J=K-1$$. Now you have two more cases: $$x=0\text{ or }x<0~.$$