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Ok so my theorem goes like this

For any Z = X/Y

Where X,Y are positive integers & Y > 0

A1,A2.... are factors of X & B1,B2.... are factors of Y

There exists a Bn of Y such that Ai = Bn * Z

If Bn < Y then X must be composite

If no such Bn exists for any Y where 1 < Y < X then X must be prime

It's something so basic that I feel like it must exist since it simply describes division but at the same time it shows things like this

   27|6 = 4.5,    4.5 * 2(a factor of 6) = 9,    9 must be a factor of 27

So does anybody know if this is already a proven theory that I'm just rediscovering?

Note: I've edited this to make it more clear, sorry if that makes any of the comments seem weird

Samantha Clark
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  • I feel bad since it does seem like something really basic and I'm just wasting everyone's time but I haven't taken any deep math classes or anything like that and I'm not sure how to check for myself if this exists – Samantha Clark May 05 '17 at 02:28
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    Can you explain what $X|Y=Z$ means? Is the vertical line a division bar, so $X=Y*Z$? – Eric Stucky May 05 '17 at 02:34
  • "There exists a $B_n$ of $Y$ such that $B_n\cdot Z = A_n$", what is $A_n$ here? Do you mean there exists a $A_i$ such that $B_n\cdot Z = A_i$? – Lazy Lee May 05 '17 at 02:36
  • sorry, my teacher told me to use a vertical bar to represent division so X|Y = Z is division like 10|5 = 2 Yes, Bn and An represent some factor of X and some factor of Y – Samantha Clark May 05 '17 at 02:39
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    Good on you for playing with math! To help us understand your theorem, make sure you are as clear as possible. X|Y is not standard notation for division. Do you just mean division or something else? what is $n$? Is it the largest of the factors or any factor? Also when you say factors of number X, do you include 1 and X itself? (they are usually included, but what do you mean?) – Thanassis May 05 '17 at 02:40
  • Yup, a factor of X can include 1 and X itself. Should I change my original post to make it more clear? – Samantha Clark May 05 '17 at 02:41
  • I would encourage you to edit the question and write it as clearly as you can. Make the description so detailed that you believe no one can have a question about it (not doable in reality, but that's your goal). Also note that this part $B_n \cdot Z = A_n$ is not true. You have to use a different index for $A_n$. Maybe write them as $B_i$ and $A_j$ – Thanassis May 05 '17 at 02:49
  • I feel that you might be rediscovering some formulation of Euclid's lemma – Alex May 05 '17 at 02:53

1 Answers1

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Claim 1: Let $X,Y$ be positive integers with $Y>1$ and $Z=X/Y$. If there is a factor $B$ of $Y$ such that $B<Y$ and $Z\cdot B=A$ for some factor $A$ of $X$, then $X$ shares a factor with $Y$ other than $1$. If no such $B$ exists, then $X$ and $Y$ are relatively prime.

Proof: Suppose there is a factor $B$ of $Y$ such that $B<Y$ and $Z \cdot B=A$ for some factor $A$ of $X$. Then $(X/Y)\cdot B = A$, so $X= A\cdot (Y/B)$. Note that $Y/B$ is an integer and is greater than $1$, since $B$ is a factor of $Y$ and $B<Y$. So $Y/B$ is a factor of $X$ and a factor of $Y$ which is not $1$.

If no such $B$ exists, let $D=\gcd(X,Y)$ be the greatest common divisor of $X$ and $Y$. We can write $X=D\cdot M$ and $Y=D \cdot N$ for some integers $M$ and $N$ which do not share any factors greater than $1$. Then $Z=X/Y=(D\cdot M)/(D \cdot N)= M/N$. Now $N$ is a factor of $Y$. If $N<Y$, then $N \cdot Z$ is not a factor of $X$ (by assumption). That is, $N \cdot M/N=M$ is not a factor of $X$. Yet $X=D \cdot M$, so $M$ is a factor of $X$. So it must be that in fact $N=Y$ Then $D=1$. Finally, $\gcd(X,Y)=1$ if and only if $X$ and $Y$ are relatively prime.

Claim 2: Suppose for some $1<Y<X$, there exists a factor $B$ of $Y$ such that $B<Y$, and $Z \cdot B = A$ for some factor of $A$, where $Z=X/Y$. Then $X$ is composite. Otherwise, if no such $Y$ exists, $X$ is prime.

Proof: If such a $Y$ exists, then by Claim 1, $X$ and $Y$ share a factor greater than $1$. In particular, they share a prime factor $P$. Then $P \leq Y<X$, so $P$ is a prime divisor of $X$ not equal to $X$, and thus $X$ is composite. If no such $Y$ exists, then by Claim 1, $X$ is relatively prime to every $Y$ such that $1<Y<X$. In particular, $X$ is relatively prime to every prime less than $X$. Being relatively prime to a prime number means precisely that $X$ is not divisible by that prime. So the only possible factors of $X$ are $1$ and $X$, so $X$ is prime.

Ultimately your property of there being a factor of $Y$ less than $Y$ such that $X/Y$ times that factor is a factor of $X$ boils down to saying the $\gcd$ of $X$ and $Y$ is greater than $1$. This is because if $D=\gcd(X,Y)>1$, I can write $X=D\cdot M$ and $Y=D\cdot N$, and choose $N$ to be the factor of $Y$. So you have stumbled upon the fact that if $X$ is relatively prime to everything less than $X$, then $X$ must be prime. Hopefully this shows you why $\gcd$'s are so useful in elementary number theory!

kccu
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  • keccu: good, thoughtful answer. But the line "If no such Bn exists for any 1 < Y < X then X must be prime" makes me think that you and OP are quantifying differently: you are $\forall X\forall Y\text{ s.t.}$... and they are $\forall X\text{ s.t. } \exists Y$... – Eric Stucky May 05 '17 at 02:57
  • I assumed $X$ and $Y$ were fixed since they were defined at the outset, but you may be right. In any case, if you are interpreting OP's claim correctly, then my claim shows that $X$ is relatively prime to $Y$ for every $Y<X$, which certainly implies that $X$ is prime. – kccu May 05 '17 at 03:01
  • Yes, I meant for any Y < X but greater than one so if x is 9 then you have to test 2,3,4,5,6,7,8. Not just 8. I should have said for any value of Y where 1 < Y < X – Samantha Clark May 05 '17 at 03:05
  • I'll edit my answer. – kccu May 05 '17 at 03:06
  • Thank you for taking the time to look at my theorem and try to find the mistakes in it. – Samantha Clark May 05 '17 at 03:14
  • No mistakes! I'm just translating it into more typical mathematical notation and providing a proof. – kccu May 05 '17 at 03:17
  • Oh well still thank you :) So I'm guessing since I only showed something with GCD's there's already a theorem that explains it? Probably better than mine does too lol – Samantha Clark May 05 '17 at 03:24
  • Your theorem boils down to "if $X$ is not divisible by any primes smaller than it, then $X$ is prime." So there is not really a theorem which says all of this, rather that is the very definition of prime, but you have found a way to characterize it in terms of $\gcd$'s. I'm not aware of the name of any theorem which gives this same characterization, but it is certainly known (or, at least if you showed it to a number theorist they would immediately nod their head and say, "yes, that is true"). – kccu May 05 '17 at 03:29
  • Let us continue this discussion in chat. – Samantha Clark May 05 '17 at 13:35