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I'm currently working on this problem, and the term "under divisibility" is kind of ambiguous to me. I googled the term "under divisibility" and I couldn't find any definitions.

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The solution also doesn't make sense to me either. It states that all numbers divide 0, but isn't it NOT possible to divide by zero (when talking about natural numbers). Can anyone explain this question/solution in more explicit terms?

This is from this problem set.

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  • $\begingroup$ A partial order $\lt$ is being defined by $x\lt y$ if and only if $x$ divides $y$. $\endgroup$ – André Nicolas Oct 24 '13 at 20:33
  • $\begingroup$ So does that mean "1 divides all natural numbers" mean any number dividing 1 gives another natural number? That's the only interpretation that would make sense to me. $\endgroup$ – user886596 Oct 24 '13 at 20:44
  • $\begingroup$ $a$ divides $b$ means there is a $k$ such that $ak=b$. The phrase "$1$ divides all natural numbers" means that every natural number is divisible by $1$. It definitely does not mean that every natural number divides $1$, that is very false, $2$ does not divide $1$. $\endgroup$ – André Nicolas Oct 24 '13 at 21:05
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If $R$ denotes a weak partial ordering, for example $\subseteq$ ("is a subset of") or $\mid$ ("divides") then the statement "$x$ is minimal under $R$" means that the only $y$ satisfying the statement "$y \mathbin{R} x$" is $x$ itself.

To answer your second question, for any $x$ we have $x \mid 0$ ("$x$ divides zero") because there is a $y$ such that $xy = 0$, namely $y=0$. In other words, $0/x$ is a natural number $y = 0$. So we are not dividing anything by zero, but rather dividing zero by things.

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