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The set of real numbers $\mathbb{R}$ is closed under division. Does that mean $0$ is also considered?
more specifically, should it be $\mathbb{R}-\{ 0 \}$?
because division by $0$ is not defined.

rohitt
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  • you're right, it makes sense for sets – rohitt Dec 29 '20 at 13:50
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    I frown a bit at asking whether something is "closed under" a certain operation when that operation is undefined. Normally you would already have an operation fully defined on a superset (say, $-$ on $\mathbb R$) and then you would ask for a subset (e.g. $\mathbb N$) whether it is closed under that operation . (In this case, if $x,y\in\mathbb N$, does $x-y\in\mathbb N$?) Thus, wherever this formulation comes from, I think it slightly abuses the words "closed under". –  Dec 29 '20 at 13:57

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The real numbers aren't closed under division unless you don't include zero. So it is like you said: $\mathbb{R}-\{0\}$ is closed under division since we don't encounter that problem.

grumpy
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