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Dummit and Foote says the following about integer division.

If $a,b \in \mathbb{Z}$ with $a \neq 0$, we say $a$ divides $b$ if there is an element $c \in \mathbb{Z}$ such that $b = ac$. In this case we write $a \mid b$; if $a$ does not divide $b$, we write $a \not \mid b$.

My question, though perhaps a bit silly, is why we require that $a \neq 0$. If $a = 0$, then the only possible value of $b$ would be $0$. We could then pick any value of $c$.

Is it correct that say that $a = 0$ divides $b = 0$? Is there another reason we exclude $a = 0$?

J. W. Tanner
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John P.
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  • Might be useful https://math.stackexchange.com/questions/666103/why-would-some-elementary-number-theory-notes-exclude-00 – aangulog Aug 30 '20 at 01:57
  • Cf. Wikipedia – J. W. Tanner Aug 30 '20 at 01:57
  • This is just one way to define divisibility. Although yes, using the definition: "for $a,b \in \mathbb Z$ we say that $a \mid b$ if and only if there exists $c \in \mathbb Z$ such that $b = ac$", then $0 \mid 0$. – azif00 Aug 30 '20 at 01:59
  • Personally, I find it nice to allow $a=0$. It lets me think of the "divides" relation as another ordering on (all of) the integers, with zero being the largest element, and one being the smallest. – Brian Shin Aug 30 '20 at 02:03
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    Most definitions I've seen require that $a$ be nonzero. If I had to take a stab at why, it's because the division algorithm gives a unique $c$ such that $b=ac$, and if $a=0$, then $c$ can be any real number. – Stephen Goree Aug 30 '20 at 03:45

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