# Every Number is between Two Consecutuve Integers

#### OhMyMarkov

##### Member
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

##### Well-known member
MHB Math Helper
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.

#### Plato

##### Well-known member
MHB Math Helper
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: $$x=0\text{ or }x<0~.$$