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December 27th, 2017, 06:08 AM  #1 
Member Joined: Dec 2017 From: Tel Aviv Posts: 45 Thanks: 3  prime number
This is an axiom or a theorem: "There is a number that divide x (that is integer) exist, between 2 to sqrtx, if x is not a prime number". If this theorem, how it can be proved? Last edited by skipjack; December 27th, 2017 at 11:45 AM. 
December 27th, 2017, 11:02 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,967 Thanks: 807 
First, if x is not a prime number, then there exist integers, a and b, such that $\displaystyle ab= x= (\sqrt{x})(\sqrt{x})$. Second, If one of those numbers is greater than $\displaystyle \sqrt{x}$ the other must be less than $\displaystyle \sqrt{x}$. Last edited by Country Boy; December 27th, 2017 at 11:04 AM. 
December 27th, 2017, 11:04 AM  #3 
Senior Member Joined: May 2016 From: USA Posts: 922 Thanks: 368 
It is a theorem based primarily on definitions and the lemma that $1 < \sqrt{m} \implies \sqrt{m} < m.$ $u \text { is a divisor of } w \iff u,\ w \in \mathbb N^+ \text { and } \exists \ v \in \mathbb N^+ \text { such that } u * v = w.$ Do you buy that as a definition? $n \in \mathbb N^+ \text { and } n > 1 \implies n \text { has at least two distinct divisors.}$ That is a theorem easily proved from $n * 1 = n \text { and } n > 1 \implies n \ne 1.$ Now let's define a prime. $p \text { is a prime } \iff p \in \mathbb N^+, \ p > 1,\ \text { and } p \text { has exactly two distinct divisors, namely } 1 \text { and } p.$ OK so far? $\text {Given: } x, y \in \mathbb N+ \text { and } y \text { is a divisor of } x \text { such that } 2 \le y \le \sqrt{x}.$ $\therefore 1 < y < x \implies y \ne 1 \text { and } y \ne x \implies$ $x \text { has at least three distinct divisors } \implies x \text { is not a prime.}$ Last edited by JeffM1; December 27th, 2017 at 11:07 AM. 

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