I was solving an inequality today and I proved it for rational numbers (it was easier because I was able to "strengthen" by doing things like "$\frac{p}{q}>\frac{r}{s}\implies ps\ge qr+1$ since $ps,qr\in \mathbb{Z}^+$". I then noted that "since it is proven for rationals, it must be true for reals, since every real can be approximated by rationals arbitrarily". Is this wrong? I wasn't sure of this step.
Thanks!
EDIT: Problem: If $a,b>0$, prove that:
$$\frac{1}{a+b+1}-\frac{1}{(a+1)(b+1)}<\frac{1}{11}$$
I plugged $a=p/q$ and $b=r/s$ and proved it to be true in the case where $q\ge p$, i.e. $a\le 1$. I then plugged $a\rightarrow 1/a$ and showed it to be true in this case, so it must be true for all positive $a,b$ that are rational. I then said it follows for reals, but this is the step I wasn't sure.
EDIT#2: My question is basically, can you show me how to rigorous prove that if this inequality holds for positive rationals, then it must hold for positive reals?