Determine the minimum value of $S=p+q$

Question

question

For a fixed natural number $ n \geq 1$ we consider all rational numbers of the form $\frac{p}{q}$, with $p,q \in N*$ and $\frac{n - 1}{n } < \frac{p}{q }< \frac{n}{ n+1}$ Determine the minimum value of the sum $S=p+q$.

my idea

Because $\frac{n - 1}{n }$ and $\frac{n}{ n+1}$ are subunitar we get that

$\frac{p}{q} \in (0,1)$

Which means that $0<p<q$

From here i tried to process the initial inequality, but I have not reached anything.

Hope one of you can help me! Thank you!

Answer 1

It's easier to subtract everything from $1$ first. So write $\frac rq=1-\frac pq$. Now we need $\frac rq\in(\frac1{n+1},\frac1n)$.

Clearly $r=1$ doesn't work, since it would lead to $n<q<n+1$.

Trying $r=2$, we get $\frac1{n+1}<\frac 2q<\frac 1n$, and we can rearrange to get $2n<q<2n+2$. This has a solution, $q=2n+1$, from which we get $p=q-r=2n-1$ and $p+q=4n$.

Now suppose $r\geq 3$. We need $\frac rq<\frac 1n$, so $q>rn$. Since everything's an integer, we have $q\geq rn+1$. Therefore $p+q=2q-r\geq 2rn+2-r\geq 6n-1>4n$. So the solution we found above minimises $p+q$.

Answer 2

the mediant of positive fractions is between them. In this case $$ \frac{2n-1}{2n+1} $$ with sum $4n \; . \; $

Continued fractions are in many number theory books. You want a more specific topic called Farey Series. Let me see what references I can give that discuss mediants.

Good one, Edward Burger, Exploring the Number Jungle: A journey into Diophantine Analysis. Module 2 is Farey Sequences, pages 11-15.

Excellent, Lemma 2.6 on page 13. Type that in

Lemma 2.6. Let $\frac{p}{q} < \frac{r}{s}$ be two rational numbers with $ps-rq = -1$ Suppose that $\frac{a}{b}$ is a rational number satisfying $\frac{p}{q} \leq \frac{a}{b} \leq \frac{r}{s}.$ Then there exist nonnegative integers $\lambda, \mu$ such that $$ a = \lambda p + \mu r \; \; , \; \; \; b = \lambda q + \mu s \; \; . $$

Also in Hardy and Wright, next Niven and Zuckerman

In Niven and Zuckerman, fifth edition (with Montgomery), they put this fact about Farey sequences, Theorem 6.4 on page 298: given consecutive fractions in a Farey sequence, the mediant is the unique fraction with smallest denominator. I would say this is not immediate for your problem as you need to show your fractions are consecutive in the ordered list with denominators no larger than $n+1$