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What is an easy and intuitive explanation to the unique factorisation problem:

If two prime numbers a and b are multiplied, the product (a*b) cannot be divided by a different prime number c.

Kevin Wu
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  • If a prime $p$ divides $ab$ then $p$ divides $a$ or $p$ divides $b$. That "Prime Divisor Property" is as intuitive as it gets. Uniqueness of prime factorizations follows immediately from that by a straightforward induction. If you seek some other type of "intuition" then please clarify what you seek. – Bill Dubuque Feb 01 '17 at 15:52
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    In this answer you can find a list of ten other properties that also imply uniqueness of prime factorizations. You might find the Schreier refinement or Euler Four Number Theorem more intuitive. All shed further light on the essence of uniqueness of prime factorizations. Be aware that there we use the general language, where atoms (or irreducibles) are nontrivial numbers having no nontrivial factorizations, and primes are those satisfying the above Prime Divisor Property = PDP. So for integers the PDP says that Atoms are Prime (= AP property). – Bill Dubuque Feb 01 '17 at 16:12

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I think that in this case contradiction is the most intuitive way to understand this property. So let's consider two prime numbers $p$ and $q$. As you mentioned, if you take the product $pq$ the only two positive divisors, not equal to $1$ are $p$ and $q$.

How can you grok this intuitively?

It was pointed out to me that the following does not provide a necessarily intuitive rationale for why $x$ in this case must divide into $p$ or $q$. I will breakout my intuition further here to show this. Well, what would happen if there was a third prime, call it $x$ that divided into $y$? If $x | y \implies x | pq$. But we know that if some prime number divides into a product of two numbers it must divide either one, or the other, (or perhaps both). This is a relatively straightforward factor, and should make sense based on how division is defined!

A prime number cannot be broken down into further factors, by the definition of primes, and as such $pq$ is the unique way to write this as a product of prime factors.

If it could be written in some other manner [$a, b \neq 0$ are prime], say $(ab)q$ then we would of course have $(ab) = p$ and as such, $p$ would not be prime. If we had that $ab = pq$, then we know that $a$ must divide both sides, as must $q$ so we have that: $\frac{b}{q} = \frac{p}{a}$. But the only two factors that divide $b$ are $b$ and $1$. So then we must have that either $q = b$ or $q = 1$. Similarly, we get that $a = p$ or $a = 1$. Because we know that by definition $a, b, p, q$ are prime we must have that $a = p$ and $b = q$. As such, the factorization is unique.

If the uniqueness of the factorization is not quite enough to fully ensure that the intuition is clear, consider if we had a third prime $x$, which did divide into $pq$. We know from the above logic that $x\times c = pq$ must mean one of the following things:

$x = p \implies x$ is not a third divisor.

$x = q \implies x$ is not a third divisor.

$x = pq \implies x$ is not prime.

$x = 1 \implies x$ is not prime.

The remaining options would be that $x | p$ or $x | q$, since it cannot be that $x$ divides some combination of the factors of $p$ and $q$ since the only factors of $p$ and $q$ are one and themselves! However, for $x | p$ then we know that $x = 1$ or $x = p$ which are options above and for $x | q$ we have the same issues.

So this says that we cannot possibly have a third prime $x$ which divides $pq$ unless $x$ is in fact either $p$ or $q$.

Dylan
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  • Your claim in the second paragraph is false, e.g. $6$ divides $2\cdot 3$ but $6$ neither divides $2$ nor $3$. That property is true only for primes (or $\pm 1$ or $0).,$ See the PDP property in my comments above. – Bill Dubuque Feb 01 '17 at 16:23
  • $x$ is prime by construction. That is the entire point of the proof. The claim holds exactly as stated, but I did make it more clear! Thanks for the note ! – Dylan Feb 01 '17 at 16:39
  • The claim was false as written. Now that you edited it to change "some number" to "some prime number" it is correct. This is important to get correct because some folks do in fact believe that false claim. – Bill Dubuque Feb 01 '17 at 16:44
  • In nay case, this is not a good intuitive explanation The reason that the Prime Divisor Property holds true has little to do with "how division is defined". – Bill Dubuque Feb 01 '17 at 16:53
  • Thanks Bill. When I said it holds "exactly as stated" it was because, in my head, it was clear that the "some number" was in specific reference to the situation I was creating. Upon re-reading I can see why the intuition I was trying to lay out was not necessary intuitive for others beyond myself! I have tried to re-map the argument a little more concretely, which represents the manner in which I came to intuitive appreciate this fact. One of those cases where knowing what I wanted to say meant I said it incorrectly! – Dylan Feb 01 '17 at 18:21