The following question is pretty basic, and the underlying idea was used in the "proof" of a statement in this hyperlinked answer to another MSE question.

The question is as follows:

If $a \mid c$ and $b \mid c$ where $a, b, c \in \mathbb{N}$, under what conditions does it follow that $a \mid b$?


Take $c = 20$. $c$ factors as follows: $$c = 20 = 4 \cdot 5 = 2 \cdot 10.$$

Note that we can take $a = 2$, $b = 10$. And also note the counterexample $$a = 4 \nmid 5 = b.$$

So I think a condition under which $$\bigg(a \mid c \text{ and } b \mid c\bigg) \implies a \mid b$$ is when $$\frac{b}{a} \mid c.$$ But that condition is too artificial for my purposes. Are there other more natural conditions?

  • $\begingroup$ What does $\frac ba$ mean if we don't know that $a\,|\,b$? If it is just meant to be rational, then what does it mean to say that it divides $c$? $\endgroup$ – lulu Apr 4 at 9:24
  • $\begingroup$ @lulu, if we know a priori that $\frac{b}{a} \mid c$, then it follows that $\frac{b}{a}$ is an integer. (It forces $a \mid b$.) $\endgroup$ – Jose Arnaldo Bebita-Dris Apr 4 at 9:26
  • $\begingroup$ @lulu, you are merely playing with words. From the context, I meant divisibility in $\mathbb{N}$. $\endgroup$ – Jose Arnaldo Bebita-Dris Apr 4 at 9:28
  • $\begingroup$ So...your test is that if $a$ divides $b$ then $a$ divides $b$? It's obvious that $b\,|\,c$ implies that every factor of $b$ divides $c$. $\endgroup$ – lulu Apr 4 at 9:33
  • $\begingroup$ @lulu, please reconsider the example given in the question. $\endgroup$ – Jose Arnaldo Bebita-Dris Apr 4 at 9:35

$c=1$ is the only possibility.

Indeed, if $c\neq 1$, then $c \text{ } | \text{ }c$ and $1 \text{ } | \text{ } c$, but obviously you don't have $c \text{ } | \text{ } 1$.

  • $\begingroup$ Thank you for your answer, @TheSilverDoe. What about the example I have given? $c = 20 = 2 \cdot 10$, so we can take $a = 2 \mid 10 = b$, but $1 \neq 20 = c$. $\endgroup$ – Jose Arnaldo Bebita-Dris Apr 4 at 9:15
  • $\begingroup$ You give an example where $a \text{ }| \text{ } c$, $b \text{ }| \text{ } c$ and $a \text{ }| \text{ } b$. That does not mean that the implication $(a \text{ }| \text{ } c \text{ and } b \text{ }| \text{ } c) \Rightarrow a \text{ }| \text{ } b$ is true. $\endgroup$ – TheSilverDoe Apr 4 at 9:17
  • 1
    $\begingroup$ Ahh yes! Valid arguments are not always sound! Thank you. Gladly accepting your answer now. $\endgroup$ – Jose Arnaldo Bebita-Dris Apr 4 at 9:19

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