If $a0$ then $a\leq b$ Show that if $a,b\in\mathbb{R}$ and $a&lt;b + r$ 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 $a&gt;b$. Then we can find a rational number $r$ such that $0 &lt; r &lt; a-b$, and then $a&lt;b+r&lt;a$, which is a contradiction.
another idea is to take $r=\frac{1}{n}$ so $$a&lt;b+\frac 1n$$ how to read inequality when $n$ tends to $+\infty$ ?