I have a non-empty set $A\subseteq\mathbb{R}$ and let $B:=\{a-a':a,a'\in A\}$.

> 
> I want to prove that if $A$ is dense in $\mathbb{R}$, then $B$ is also
> dense in $\mathbb{R}$.
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> 
> 

I have thought of the following two approaches but I am not confident in either of them.

Approach 1: find values $a,a'$ such that $a-a'\in\mathbb{N}$ and state that since $\mathbb{N}$ is not dense in $\mathbb{R}$, then $B$ is not dense in $\mathbb{R}$. I was told this would not hold as a proof- why not?
Approach 2: if $A$ is dense in $\mathbb{R}$, then $\exists x<a<y$ and $\exists x'<a'<y'$, so if $B=\{a-a':a,a'\in A\}$, then I can say that $x-x'<a-a'<y-y'$ pertains to set $B$, which therefore is dense in $\mathbb{R}$. This doesn't seem right either. Are any of these in the right direction, and if not, how can I solve this?