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Every Number is between Two Consecutuve Integers

OhMyMarkov

Member
Mar 5, 2012
83
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
Feb 5, 2012
1,621
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
Jan 27, 2012
196
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 [tex]x\in\mathbb{R}[/tex] there is an integer [tex]J[/tex] such that [tex]J\le x<J+1~.[/tex]
To prove this first suppose that [tex]x>0[/tex]. Then use well ordering of the natural numbers to find the least positive integer, [tex]K[/tex], having the property that [tex]x<K[/tex].
Because [tex]K[/tex] has that minimal property we see that [tex]K-1\le x<K[/tex].
So let [tex]J=K-1[/tex]. Now you have two more cases: [tex]x=0\text{ or }x<0~.[/tex]