In ring theory, is every real number divisible by any other nonzero real number, precisely because nonzero division is closed in $\mathbb{R}$?
And so in general this idea of "closure under division" can be used to describe divisibility in a ring? E.g. one might say:
$\mathbb{R}^*$ is closed under division, hence each one of its elements is divisible by any other one. (Which for example, in $\mathbb{Z}$, remains false.)
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I don't think it is wise to talk about the division operator being closed in the ring minus the $0$ element, but rather that every nonzero element has a inverse element such that $x\cdot x^{-1}=1$. Every real number "is divisible" by every nonzero real number because every nonzero real number has a multiplicative inverse.
I want to make this seemingly pedantic distinction because in a more general ring it is not clear what "division" should be for elements that don't already have a defined inverse. The best we can do is build the field of fractions, but again it is all defined in terms of the multiplication operation and the idea of what a multiplicative inverse is.
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Well, yes, it is true that if $d\in\mathbb R \setminus \{0\}$ and $a\in \mathbb R$, then $d|a$, since this means $$\exists k\in \mathbb R, \;a=dk,$$ (and is enough to take $k=\frac ad$). But this is equivalent to have a multiplicative inverse for every non-zero element of the ring, which implies that this is in fact a field (or a division ring, if it were not commutative).
I would not say that $\mathbb R^*$ is "closed under division", since for a ring $R$ the set $R^*$ is by definition the set of elements of $R$ with an inverse, so this is kind of the same. For instance, $$\mathbb Z^*=\{-1,1\},$$ and so $\mathbb Z^*$ is also "closed under division", since $$1|1, \quad 1|-1, \quad -1|1 \quad \text{and}\quad -1|-1.$$
The answer to the first question is unequivocally yes. This is why it's important that divisibility (and hence primality) is only meaningful relative to a base ring.
Mainly answered to say that since the previous answers weren't being sufficiently definite about it. That said:
As to the second question, one would normally just say that $\mathbb{R}$ is a field. An element that every element is divisible by is called a unit, so you equivalently say that every nonzero element is a unit. The reverse condition, being divisible by every nonzero element, doesn't have a name I'm aware of, but is a nontrivial condition in certain rings.
People are objecting to the term "closure" on the grounds that it doesn't make sense except in the context of an operation defined on a larger set. This is a valid objection which also applies to a common way you'll see group theory taught: it's fine to say "a subgroup is a subset closed under the group operations," but it's not okay to say that "a group is a set closed under multiplication such that..." because what would it even mean for it not to be closed?