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December 27th, 2017, 06:08 AM   #1
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prime number

This is an axiom or a theorem:
"There is a number that divide x (that is integer) exist, between 2 to sqrt-x, 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.
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December 27th, 2017, 11:02 AM   #2
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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.
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December 27th, 2017, 11:04 AM   #3
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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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