If $a0$ then $a\leq b$ Show that if $a,b\in\mathbb{R}$ and $a<b + r$ for every positive rational $r$, then $a\leq b$
I think it might be useful to use the density of $\mathbb{Q}$ in $\mathbb{R}$, but I don't really know how to approach this problem

Suppose that $a>b$. Then we can find a rational number $r$ such that $0 < r < a-b$, and then $a<b+r<a$, which is a contradiction.
