Proving density of two sets in $\mathbb R$

Question

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 $\mathbb{R}$.

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?

Answer 1

First fix one element $b\in A$. Since $A$ is dense in $\mathbb{R}$, for any $x,y\in \mathbb{R}$ and $x<y$，one can find $a\in A$, such that $x+b<a<y+b$.

By the definion, $a-b\in B$. That is, for any $x,y\in \mathbb{R}$ and $x<y$, we find one element $c=a-b\in B$, such that $x<c<y$. So $B$ is dense in $\mathbb{R}$.

For your approach 1, if you can show all elements in $B$ are in $\mathbb{N}$, then $B$ is not dense in $\mathbb{R}$. But this is impossible. For approach 2, from $x<a<y $ and $x'<a'<y'$, we can't get $x-x'<a-a'<y-y'$. For example, $1<2<3$ and $-1<2<2.5$. But $1-(-1)<2-2<3-2.5$ is totally wrong.

Answer 2

Lemma. If the subset $X\subseteq\mathbb{R}$ is dense and $r\in\mathbb{R}$, then also $X-r=\{x-r:x\in X\}$ is dense.

Once you have proved the lemma, you have $$ B=\bigcup_{a'\in A}(A-a') $$ Since $A$ is not empty, $B$ contains a dense set.

Your first approach can't work, sorry. Even if you find $a,a'\in A$ such that $a-a'$ is integer, what would this lead to?

Approach two is more promising, but you're overcomplicating things.