Can a set of real numbers be closed under division but not under addition, multiplication and subtraction? I have recently learned a bit about the concept of closure under an operation, and I'm curious to know if a set can be closed only under division (when considering the four basic arithmetic operations). It is clear to me that for a set $\Bbb{X}\subset\Bbb{R}$ to be closed under division:
$$1\in\Bbb{X},0\notin\Bbb{X}$$
which means that, for any real $k$:
$$k\in\Bbb{X}\iff\frac{1}{k}\in\Bbb{X}$$
And if $k\in\Bbb{X}\Rightarrow k^2=1$, then $\Bbb{X}$ is closed under division but also under multiplication, so if $\Bbb{X}$ is to be closed only under division:
$$\neg\forall k(k\in\Bbb{X}\Rightarrow k\in\Bbb{Z})$$
Which is to say, it is not the case that all members of $\Bbb{X}$ are integers.
This is about as far as I've gotten with restricting the possibilities. Where do I go from here?  Is the answer completely obvious?
Thanks for your time.
 A: If $k, l\in\mathbb{X}$, then as you point out $\frac{1}{l}\in\mathbb{X}$, so that $\frac{k}{1/l} = kl\in \mathbb{X}$. So $\mathbb{X}$ must be closed under multiplication.
A: Hope this helps you.
I think your statement that $k\in\mathbb{X}$ iff $\frac{1}{k}\in\mathbb{X}$ doesn't mean that the set $\mathbb{X}$ is closed under division. 
The reason why is because division is a binary operetion and the operation you've mentioned is a uniary operation
To be closed under an operation (in this case under division) is understood as $\frac{a}{b}\in\mathbb{X}$ every time that $a, b$ lie in $\mathbb{X}$.
Therefore, as @rogerl mentioned, let $\mathbb{X}$ to be a non-empty subset from real numbers that is closed under division. Since $\mathbb{X}$ is not empty, take $k\in\mathbb{X}$ and as a result from $\mathbb{X}$ to be closed under division $1=\frac{k}{k}\in\mathbb{X}$ then is closed under multiplication. Think of for instance in the set of rational positive numbers $\mathbb{Q}^{+}$ which is closed under division.
Remember that an operation over a set $X$ is just a function $f:X\times X\rightarrow X$.
