TITLE: REVISTED$^2$: Fraction Existence Proof
QUESTION [5 upvotes]: Question 1:
I'm asked to prove that there exists an $n\in\mathbb{N}$ such that
$$\frac{1}{n+1}\leq\frac{a}{b}<\frac{1}{n},$$
where $0<\frac{a}{b}<1$. Here $\frac{a}{b}$ is a fraction in lowest terms.

Q1(Thoughts):
If $\frac{1}{n+1}\leq\frac{a}{b}<\frac{1}{n},$ then with a bit of algebra we get that
$$0\leq\frac{an^2-bn+a}{b}<1,$$
so since $0<\frac{a}{b}<1$ we can set $an^2-bn+a=a$ to get that $an^2-bn=0$, which produces
$$n=\frac{b\pm\sqrt{b^2-4a\cdot 0}}{2a}=\frac{b\pm b}{2a},$$
and since $n\neq 0$, then we get that $n=\frac{b}{a}$.
OK, so that was totally wrong, but I've got that $\frac{a}{b}\in[\frac{1}{2},1)\implies n=1,$ $\frac{a}{b}\in[\frac{1}{3},\frac{1}{2})\implies n=2$, and so on ad infinitum, but I don't know that this is a proof. It looks as though no matter what $\frac{a}{b}\in\bigcup^{\infty}_{n=1}[\frac{1}{n+1},\frac{1}{n}]$.

Question 2:
If $n$ is chosen as in Q1, prove that $\frac{a}{b}-\frac{1}{n+1}$ is a fraction that in lowest terms has a numerator less than $a$.

Q2(Thoughts):
I think this might have something to do with $n<\frac{b}{a}$.

I'll be adding more to this as I go, but my question is essentially something like "Is this right?"

REPLY [3 votes]: Consider the set
$$
\left(\frac12,1\right] \cup \left(\frac13,\frac12\right] \cup \left(\frac14,\frac13\right] \cup \left(\frac15,\frac14\right] \cup \cdots \cup\left(\frac1{b+1},\frac1b\right]
$$
This set certainly contains $\dfrac1b,\dfrac2b,\dfrac3b,\ldots,\dfrac{b-1}b$.  One of those is $\dfrac ab$.