The answer is $C = \frac{1}{2}$. For construction, take the five numbers $\varepsilon$, $1$, $2-\varepsilon$, $2$, $2+\varepsilon$ for arbitrarily small $\varepsilon$. We now show selecting the five numbers is always possible. Assuming the numbers are $a < b < c < d < e$, consider the five fractions \[ a/c, \quad b/d, \quad c/e, \quad a/d, \quad b/e \]and arrange them in a pentagon. These fractions are less than $1$, so three of them must be on the same side of $\frac{1}{2}$; two are adjacent on the pentagon and give the desired. Remark: Finding the answer is a large part of the difficulty of this problem. The way I motivated is solving the sub-problem with four numbers $a < b < c < d$, the minimum fraction is $|\frac{a}{c} - \frac{b}{d}|$ (check this). Thus given five distinct numbers $a < b < c < d < e$, the five values in consideration are: $a/c - b/d$ $a/c - b/e$ $a/d - b/e$ $a/d - c/e$ $b/d - c/e$ One can then arrange these five fractions (less than $1$) around the vertices of a pentagon, connecting fractions with no common term; trying to maximize the minimal difference naturally gives the optimal value of $1/2$.

mathwizard888 wrote: Find the smallest constant $C > 0$ for which the following statement holds: among any five distinct positive real numbers, one can label four different ones as $p, q, r, s$ such that \[ \left| \frac{p}{q} - \frac {r}{s} \right| \le C. \] In the official said that (not necessarily distinct).

This was also problem 5 at Bosnia Herzegovina TST 2017.

Also German TSTST #4 and my favorite problem of 2017. We'll prove $C=\frac{1}{2}$. Take for instance $1,1,1,2,N$ with an arbitrarily large $N$ to show that $C \geq \frac12$. Let us now assume that there exists a choice of numbers, such that $C>\frac12$ to eventually prove $C \leq \frac12$. We will need a Lemma. Lemma. If $x$ and $y$ are both larger or both smaller than $\frac12$, then $|x-y| \leq \frac12$. Look at the case $\frac{a_1}{a_5} \geq \frac12$. The Lemma gives us $\frac{a_2}{a_3}, \frac{a_2}{a_4} \leq \frac12$ but a repeated use of the Lemma on those numbers give us $\frac{a_4}{a_5}, \frac{a_1}{a_3} \geq \frac12$. Now \[ \left| \frac{a_4}{a_5} -\frac{a_1}{a_3} \right| \leq \frac12, \]a contradiction. The case $\frac{a_1}{a_5} \leq \tfrac12$ can be done in the same way, just reverse the inequality signs. Thus, we are done.

Easy enough as a TST problem ! $C=1/2$ ; For $a>b>c>d>e$ consider the 5 fraction $a/b,b/c,c/d,d/e,e/a$ . Since they are in $(0,1]$ so atleast 3 of them are in same side of $1/2$ . So done ! For equality consider the 5 tupple ${1,2,x,1-2x,1+2x}$ Actually the main part of the problem which is to find $C$ is get easy when one notes the fractions at that order and notes that any 3 of them contains 4 elements of the given 5 tupple and thinks about PHP which is trivial !

Solution. I will prove that $C=\frac{1}2$ works. For a construction, consider $\varepsilon, \frac{1}2, 1,1,1$; where $\varepsilon \to 0$ So it suffices to show $C \leq \frac{1}2$. WLOG let $a_1 \leq a_2 \leq a_3 \leq a_4 \leq a_5$, and assume $C > \frac{1}2$. Clearly, $\left \{ \frac{a_1}{a_2}, \frac{a_1}{a_5}, \frac{a_2}{a_5} \right \}$ and $\left \{\frac{a_3}{a_4} \right \}$ must not lie on same side of $\frac{1}2$. Also, $\frac{a_1}{a_2} \geq \frac{a_1}{a_4} \geq \frac{a_1}{a_5} \implies\frac{a_1}{a_4},\frac{a_2}{a_5}$ lie on same side of $\frac{1}2$. Therefore, $\left| \frac{a_1}{a_4} - \frac {a_2}{a_5} \right| \leq \frac{1}2$, contradicting our assumption. Hence, $C=\frac{1}2$ is indeed the solution. $\square$

This was actually pretty hard The answer is $C = \frac{1}{2}$. A construction is $\varepsilon, \frac{1}{2}, 1, 1, 1$ where $\varepsilon \to 0$. WLOG $a_1 \le a_2 \le a_3 \le a_4 \le a_5$. We will only consider the values $\frac{a_{i}}{a_{j}}$ less than $1$, since we seek to minimize $C$. Assume FTSOC that each pair of ratios (with different indices) differ by $> \frac{1}{2}$. That is, $\left| \frac{a_1}{a_2} - \frac {a_3}{a_4} \right| > \frac{1}{2}$, $\left| \frac{a_1}{a_2} - \frac {a_3}{a_5} \right| > \frac{1}{2}$, and $\left| \frac{a_1}{a_2} - \frac {a_4}{a_5} \right| > \frac{1}{2}$, and so on. Suppose $\frac{a_i}{a_j} = x$ if $\frac{a_i}{a_j} < \frac{1}{2}$ and $\frac{a_i}{a_j} = y$ if $\frac{a_i}{a_j} \ge \frac{1}{2}$. If $\frac{a_1}{a_2} = x$, then we have $\frac{a_3}{a_5} = \frac{a_4}{a_5} = \frac{a_3}{a_4} = y$, but this implies $\frac{a_1}{a_4} = \frac{a_2}{a_4} = \frac{a_1}{a_3} = \frac{a_2}{a_3} = \frac{a_1}{a_5} = \frac{a_2}{a_5} = x$, which is a contradiction because then $\left| \frac{a_1}{a_4} - \frac {a_2}{a_5} \right| <\frac{1}{2}$ . $\blacksquare$

Generalization to more than 5 variables looks pretty hard. I think the answer in the 6-variable case is $C=1/4$. A construction is $1,1,1,2,4,N$ with an arbitrarily large $ N$

Here's my solution: Let $a \leq b \leq c \leq d \leq e$ denote the $5$ numbers. We claim the answer is $C=\frac{1}{2}$. Taking the $5$ numbers as $(1,1,1,2,x)$ with $x \rightarrow \infty$ gives that $C \geq \frac{1}{2}$, since the minimum value of the given expression can reach arbitrarily close to $\frac{1}{2}$. Now we show that this is indeed the minimum. Assume not. Then there exists some sequence $(a,b,c,d,e)$ such that for any $4$ distinct reals $w,x,y,z$ out of these five numbers, we have $$\left | \frac{w}{x}-\frac{y}{z} \right| > \frac{1}{2}$$Note that $\frac{c}{d} \geq \frac{b}{e}$, which gives $$1 \geq \frac{c}{d}>\frac{b}{e}+\frac{1}{2} \Rightarrow \frac{c}{d}>\frac{1}{2}> \frac{b}{e}$$If $\frac{a}{b} \geq \frac{c}{d}$, then we have $$\frac{a}{b}>\frac{1}{2}+\frac{c}{d}>1 \Rightarrow a>b$$which is not according to our assumption that $a \leq b$. So we get $$\frac{c}{d}>\frac{a}{b} \Rightarrow \frac{b}{d}>\frac{a}{c} \Rightarrow 1 \geq \frac{b}{d}>\frac{a}{c}+\frac{1}{2} \Rightarrow \frac{a}{c}<\frac{1}{2}$$In a similar fashion, if $\frac{b}{e} \geq \frac{a}{d}$, then $$\frac{b}{e}-\frac{a}{d}>\frac{1}{2} \Rightarrow \frac{b}{e}>\frac{1}{2}$$contrary to the previous statement. Else $$\frac{b}{e}<\frac{a}{d} \Rightarrow \frac{a}{d}>\frac{b}{e}+\frac{1}{2}>\frac{1}{2} \Rightarrow \frac{a}{c} \geq \frac{a}{d}>\frac{1}{2}$$which is again a contradiction to what we had discovered earlier. Thus, no such sequence exists, as desired. $\blacksquare$ REMARKS: To be honest, I didn't like this problem much. After getting the answer, it was more of a "keep playing with fractions till you find something" sort of a thing, which made it a bit boring. Although, looking at other solutions now (especially that pentagon/PHP idea), the problem doesn't seem so bad for an A2

Also Croatia TST.

Solved together with @Blastoor The answer is $C=\frac{1}{2}$. For construction take numbers $1,1,1, 2, N$, where $N$ is large enough number. Assume that there are given positive real numbers $a \ge b \ge c \ge d \ge e$. We split problem in two cases: 1) $\frac{b}{a} \ge \frac{1}{2}$. If at least one of the numbers $\frac{d}{c}, \frac{e}{d}, \frac{e}{c}$ is greater or equal to $\frac{1}{2}$ we are done. Otherwise $\frac{e}{a} \le \frac{e}{d} \le \frac{1}{2}$. So we pick numbers $\frac{e}{a}$ and $\frac{d}{c}$, because their absolute difference will be less or equal to $\frac{1}{2}$, since their are both less or equal to $\frac{1}{2}$ 2) $\frac{b}{a} \le \frac{1}{2}$. If at least one of the numbers $\frac{d}{c}, \frac{e}{d}, \frac{e}{c}$ is less or equal to $\frac{1}{2}$ we are done. Otherwise consider numbers $\frac{c}{b}, \frac{d}{b}, \frac{e}{b}$. If at least one of the numbers $\frac{c}{b}, \frac{d}{b}, \frac{e}{b}$ is greater or equal to $\frac{1}{2}$, then we can easily pick appropriate number from $\frac{d}{c}, \frac{e}{d}, \frac{e}{c}$, since they are all greater or equal to $\frac{1}{2}$. Otherwise $\frac{e}{a} \le \frac{b}{a} \le \frac{1}{2}$ and $\frac{c}{b} \le \frac{1}{2}$. So we pick numbers $\frac{e}{a}$ and $\frac{c}{b}$ and we are done.

v_Enhance wrote: The answer is $C = \frac{1}{2}$. For construction, take the five numbers $\varepsilon$, $1$, $2-\varepsilon$, $2$, $2+\varepsilon$ for arbitrarily small $\varepsilon$. We now show selecting the five numbers is always possible. Assuming the numbers are $a < b < c < d < e$, consider the five fractions \[ a/c, \quad b/d, \quad c/e, \quad a/d, \quad b/e \]and arrange them in a pentagon. These fractions are less than $1$, so three of them must be on the same side of $\frac{1}{2}$; two are adjacent on the pentagon and give the desired. Remark: Finding the answer is a large part of the difficulty of this problem. The way I motivated is solving the sub-problem with four numbers $a < b < c < d$, the minimum fraction is $|\frac{a}{c} - \frac{b}{d}|$ (check this). Thus given five distinct numbers $a < b < c < d < e$, the five values in consideration are: $a/c - b/d$ $a/c - b/e$ $a/d - b/e$ $a/d - c/e$ $b/d - c/e$ One can then arrange these five fractions (less than $1$) around the vertices of a pentagon, connecting fractions with no common term; trying to maximize the minimal difference naturally gives the optimal value of $1/2$. but how can we know that those two(which are adjacent) has distinct subscripts? I am asking the “two are adjacent on the pentagon and give the desired” part.

Hello @goodgood, the pentagonal arrangement that v_Enhance is talking about is $$\frac{a}{c} \rightarrow \frac{b}{d} \rightarrow \frac{c}{e} \rightarrow \frac{a}{d} \rightarrow \frac{b}{e} \rightarrow$$In this way, all adjacent fractions have distinct subscripts.

Replace $a_1,a_2,a_3,a_4,a_5$ with $a,b,c,d,e$. The answer is $C=\frac{1}{2}$. As a construction, take $1,1,1,2,N$ for extremely large $N$. We have $\left| \frac{1}{2} - \frac {1}{N} \right| \leq C$, but making $C$ any smaller fails, which can easily be checked by considering a few cases. Now we will show that all sequences work with this value of $C$. First note that scaling the entire sequence doesn't change anything, so WLOG let $a=1$. Also suppose that $a \leq b \leq c \leq d \leq e$. Suppose FTSOC that some sequence $a,b,c,d,e$ does not work, and any $i,j,k,l$ satisfies $\left| \frac{a_i}{a_j} - \frac {a_k}{a_l} \right| \leq \frac{1}{2}$. First note that $d>2$, since if $d \leq 2$ then $\left| \frac{1}{d} - \frac {b}{c} \right| \leq \frac{1}{2}$. Then note that $\frac{b}{c}\geq \frac{1}{2}$, otherwise $\left| \frac{1}{d} - \frac {b}{c} \right| \leq \frac{1}{2}$. This means that $\frac{d}{e}<\frac{1}{2}$, otherwise $\left| \frac{b}{c} - \frac {d}{e} \right| \leq \frac{1}{2}$. But then $\frac{1}{b}\leq \frac{1}{2}$, otherwise $\left| \frac{d}{e} - \frac {1}{b} \right| \leq \frac{1}{2}$. Then $\frac{c}{d}<\frac{1}{2}$, otherwise $\left| \frac{1}{b} - \frac {c}{d} \right| \leq \frac{1}{2}$. Finally, $\frac{1}{e}\geq \frac{1}{2}$, otherwise $\left| \frac{c}{d} - \frac {1}{e} \right| \leq \frac{1}{2}$. This implies that $e\leq 2$, but we defined $e \geq d$, and $d>2$, so this is absurd. Therefore all such sequences work, so we are done. $\blacksquare$

Proof that two disjoint pairs of numbers from $a_1,\cdots ,a_5$ will lie on the same intervals between $(0,0.5]$ or $(0.5,1]$. There are $10$ ways to choose a pair of numbers. We have to put each pair into one of the intervals. Let a pair be $(a_x,a_y)$ and let the sets of pairs in the given intervals be $A$ and $B$. ( for example $\frac{a_1}{a_2} \in (0,0.5] \Rightarrow (a_1,a_2) \in A$ WLOG $(a_1,a_2) \in A \Rightarrow (a_3,a_4),(a_4,a_5),(a_3,a_5) \in B$. $(a_4,a_5) \in B \Rightarrow (a_1,a_3), (a_2,a_3) \in A$. We can't put $(a_1,a_5)$ in neither $A$ or $B$, because $(a_2,a_3) \in A$ and $(a_3,a_4) \in B$, which is a desired contradiction.

$C=\frac12$ is the best constant. We can't do any better because of the counterexample 1, 1, 1, 2, $N$, where $N$ is arbitrarily large. Now show that $C=\frac12$ indeed works. WLOG, let $a_1\leq a_2\leq\dots\leq a_5$ and consider only ratios $\frac{a_i}{a_j}$ with $i<j$, so that everyhing is at most 1. If a ratio is exactly $\frac12$, then choose any other ratio. Otherwise, split the ratios into "large" ones ($>\frac12$) and "small" ones ($<\frac12$). The ratios $\frac{a_1}{a_4}$ and $\frac{a_2}{a_5}$ cannot be both small, so at least one of them is large. But then either both $\frac{a_1}{a_2}$ and $\frac{a_3}{a_4}$ or $\frac{a_2}{a_3}$ and $\frac{a_4}{a_5}$ are both large, with which we are done.

Also Peru TST 2017

dame dame

Let the 5 numbers be $a<b<c<d<e$. If $C< \frac12$, then for some $\varepsilon$, take \[a=\varepsilon,b=1,c=2-\varepsilon,d=2,e=2+\varepsilon,\]Then, all fractions get arbitrarily close to half-integers, so the minimum difference gets arbitrarily close to $\frac12$, so such $C$ fail. If $C=\frac12$. Consider the cycle of $\left(\frac{a}{c},\frac{b}{e},\frac{a}{d},\frac{c}{e},\frac{b}{d}\right)$ all of which are disjoint from the adjacent fraction. Since 5 is odd, there exists some series of $0<f_1<f_2<f_3<1$, so there must exist some $|f_i-f_j|\leq \frac12$, so $C=\frac12$ works.

Oh, the pentagon solution is much nicer. The answer is $C=\frac{1}{2}$. Necessity is shown by considering the five numbers $1,2,2,2,X$ for arbitrarily large $X$. Now we prove sufficiency. Suppose there existed five real numbers $x_1\le x_2\le\ldots\le x_5$ such that $C=\frac{1}{2}$ failed and let $r_i=\frac{x_i}{x_{i+1}}$ for $1\le i\le 4$. Note that all $r_i$ are between $0$ and $1$. Consider the following three values: Value 1. $\left| \frac{a_1}{a_3} - \frac {a_2}{a_4} \right| = r_2|r_1-r_3|$. Value 2. $\left| \frac{a_1}{a_4} - \frac {a_2}{a_5} \right| = r_2r_3|r_1-r_4|$. Value 3. $\left| \frac{a_2}{a_4} - \frac {a_3}{a_5} \right| = r_3|r_2-r_4|$. By assumption, all three values are greater than $\frac{1}{2}$. Note that $r_2$ and $r_3$ must be greater than $\frac{1}{2}$, otherwise Value 2 must be less than or equal to $\frac{1}{2}$. Using the values in a similar manner, $|r_1-r_3|, |r_1-r_4|, |r_2-r_4| > \frac{1}{2}$. Since $|r_1-r_4|>\frac{1}{2}$ and the $r_i$ are between $0$ and $1$ we must have one of them (WLOG $r_1$) be greater than $\frac{1}{2}$. But then since $|r_1-r_3|>\frac{1}{2}$, we have that $r_3<\frac{1}{2}$, contradiction.

The answer is $C=\frac{1}{2}$; we can use the construction $x,\frac{1}{2},1,1,1$ for arbitrarily small $x$. Now suppose that $C>\frac{1}{2}$, and WLOG $a_1\le a_2\le a_3\le a_4\le a_5$. Then since $\frac{a_3}{a_4}\ge \frac{a_2}{a_5}$ we clearly also have $\frac{a_3}{a_4}\ge \frac{a_1}{a_2}$ so that $\frac{a_1}{a_2}<\frac{1}{2}$. Similarly $\frac{a_4}{a_5}<\frac{1}{2}$ which is a contradiction. $\blacksquare$

WLOG $a_1 \le a_2 \le a_3 \le a_4 \le a_5$. I claim that $C= \frac{1}{2}$ is the smallest constant possible. Claim 1. $C \ge \frac{1}{2}$ Simply take numbers $1,1,1,2,x$ where $x$ is arbitrarly larger real number. Considering all the possible values, we can see that indeed $C \ge \frac{1}{2}$. $\square$ Lemma 1. If $0 \le x,y \le \frac{1}{2}$ or $ \frac{1}{2} \le x,y \le 1$ , then $|x-y| \le \frac{1}{2}$. By definition of modulo, $|a|$ means distance between $0$ and $a$, if $M\le a \le N$, then distance can't be greater than $N-M$, hence conclusion follows. $\square$ Claim 2. for any real numbers $a_1 \le a_2 \le a_3 \le a_4 \le a_5$, the set $A= \{ \frac{a_i}{a_{i+j}} \ : i,j \in \mathbb{N} \}$ contains at least two elements lying on interval $(0, \frac{1}{2}]$ or $[\frac{1}{2},1]$ Let $x_i \in A$. Note that for any $i$, $x_i \le 1$ by WLOG. Since $|A| \ge 5$, by PHP there are at least $2$ elements lying on the same interval, hence by Lemma 1, we are finished!

We claim that the answer is $C=\frac 12.$ From the set of numbers $1,1,1,\frac 12,\epsilon$ with arbitrary $\epsilon$ we have $C\geq \frac 12 -\epsilon \implies C\geq \frac 12.$ To see that value $C=\frac 12$ suffices assume WLOG $a_i\leq a_{i+1}$ and consider five fractions $\frac{a_1}{a_2},\frac{a_3}{a_4},\frac{a_2}{a_5},\frac{a_1}{a_3},\frac{a_4}{a_5}$ in a cycle. They all don't exceed $1$ and by PHP there are three of them on one side about $\frac 12.$ Two of these are adjacent in a cycle, and the absolute difference between them doesn't exceed $\frac 12$, so it satisfies.

I don't know what I am doing with my life. The answer is $C = \frac{1}{2}$, WLOG assume $a \leq b \leq c \leq d \leq e$, then consider these 5 fractions $a/c, b/d, c/e, b/e$, 3 of them are either $\geq 1/2$ or $\leq 1/2$, two of them will have to be disjoint, consider their absolute difference which will be $\leq \frac{1}{2}$, the construction is the same as evan chen $(a, 2,2,2,1)$, where $a$ is arbitrarily small.

The answer is $C=\tfrac{1}{2}$. First, we will show that it works. WLOG assume $a_1<a_2<a_3<a_4<a_5$. Then, from \[\frac{a_1}{a_2}, \frac{a_3}{a_4},\frac{a_1}{a_5},\frac{a_2}{a_3},\frac{a_4}{a_5}\]note that all of them are less than $1$. If there are two of them that lie on the same side of $\tfrac12$ and have all different subscripts then we are done. Fortunately, we have by pigeonhole principle that there will be two cyclically adjacent fractions in our above list that will lie on the same side of $\tfrac12$ which finishes the proof. Now, consider $x,1,2,2,2$ where $x$ is small. Note that the only ratios present are either small, $\tfrac{1}{2}$, $1$, $2$, or large. Therefore, $C$ cannot go any lower than $\tfrac12$.

The answer is $\boxed{\frac{1}{2}}$. To see that $C \ge \frac{1}{2}$, we consider $1,1,2,2,N$ for some arbitrarily large positive integer $N$. Now we show that $C = \frac{1}{2}$ works. WLOG that $a_1\le a_2\le a_3\le a_4\le a_5$. Suppose FTSOC that for all distinct subscripts $i,j,k,l$, \[ \left| \frac{a_i}{a_j} - \frac {a_k}{a_l} \right| > \frac{1}{2}. \]Therefore if $i < j$ and $k < l$, then $\frac{a_i}{a_j}$ and $\frac{a_k}{a_l}$ are on different sides of $\frac{1}{2}$ (note they are both in the interval $(0,1]$) Now, we get that $S = \left\{\frac{a_3}{a_4}, \frac{a_4}{a_5}, \frac{a_3}{a_5}\right \}$ are all on a different side of $\frac{1}{2}$ from $\frac{a_1}{a_2}$, so they themselves must be on the same side of $\frac{1}{2}$. Similarly, we have $T = \left\{\frac{a_2}{a_3}, \frac{a_3}{a_4}, \frac{a_2}{a_4} \right\} $ are all on the same side of $\frac{1}{2}$, since they are on the opposite side from $\frac{a_1}{a_5}$. However, since $S$ and $T$ have an element in common, they must both be on the same side of $\frac{1}{2}$, hence \[\left | \frac{a_2}{a_3} - \frac{a_4}{a_5} \right| \le \frac{1}{2} ,\]absurd. Therefore $C = \frac{1}{2}$ works.

First, consider the $4$ variable case for $a < b < c < d$. By switching opposite numerator and denominators, it only remains to consider the fractions \[ \frac{a}{c} - \frac{b}{d}, \frac{b}{c} - \frac{a}{d}, \frac{c}{b} - \frac{a}{d} \]Since $|ad - bc| < |bd - ac| < |cd - ab|$, it follows that $\left|\frac{a}{c} - \frac{b}{d}\right|$ is minimal. Now, suppose that $a < b < c < d < e$. We only need to consider \[ \frac{a}{c} - \frac{b}{d}, \frac{a}{c} - \frac{b}{e}, \frac{a}{d} - \frac{b}{e}, \frac{a}{d} - \frac{c}{e}, \frac{b}{d} - \frac{c}{e} \]Arrange these values in a circle, $\frac{a}{c}, \frac{b}{d}, \frac{c}{e}, \frac{a}{d}, \frac{b}{e}$. One of the differences must have magnitude less than $\frac{1}{2}$, or else contradiction follows by considering buckets $(0, \frac{1}{2})$ and $(\frac{1}{2}, 1)$. Then by taking $a = 1, b = x, c = 2x - 2, d = 2x - 1, e = 2x$ and taking $x \to \infty$ shows that $C = \frac{1}{2}$ is tight.

$\color{magenta} \boxed {\textbf{SOLUTION A2}}$ $\color{green} \textbf{Claim 1 :}$ $C \ge \frac{1}{2}$ $\textbf{proof :}$ Just Consider $1,1,1,2,a$ where $a \rightarrow \infty \square$ $\color{green} \textbf{Claim 2 :}$ $C=\frac{1}{2}$ $\textbf{proof :}$ Let $a \ge b \ge c \ge d \ge d$ Consider the fractions $$\frac{e}{a}, \frac{e}{d},\frac{d}{c},\frac{c}{b},\frac{b}{a}$$ Now divide them into to intervals, $0 < f_j \le \frac{1}{2}$ and $\frac{1}{2} < f_k \le 1$ by $\textbf{PHP}$ atleast $3$ of the fractions lie in same intervals. Among any $3$ fractions of the above $5$ farctions we will always find $2$ fractions with $4$ different numbers $\blacksquare$

A selection of $(1,1,1,2,m)$ with $m \rightarrow \infty$ ensures that $C\geq 1/2$ . Without loss of generality,$a\leq b\leq c\leq d\leq e$ FTSOC, assume that for any distinct $a,b,c,d$ chosen from the ${a,b,c,d,e}$ we have $|\frac{a}{b}-\frac{c}{d}|>1/2$ CLAIM:- Any inequality of the form $\frac{a_i}{a_j} \leq \frac{a_k}{a_l}\leq1$ where $a_i's$ are distinct and chosen from $a,b,c,d,e$ is basically $\frac{a_i}{a_j} <1/2< \frac{a_k}{a_l}\leq 1$ PROOF:- $\frac{a_i}{a_j} < \frac{a_k}{a_l}\leq 1$ implies that $\frac{a_i}{a_j} +\frac{1}{2} < \frac{a_k}{a_l}\leq 1$ which basically proves the above claim. As $\frac{b}{e} \leq \frac{c}{d}$, therefore $\frac{b}{e} < 1/2 < \frac{c}{d}$ . Similarly as $\frac{a}{b} \leq \frac{c}{d}$ and $\frac{b}{e}\leq \frac{a}{d}$ ,we get $\frac{a}{c} < 1/2< \frac{b}{d}$ and $\frac{b}{e} <1/2< \frac{a}{d}$. We can conclude from all these that $ \frac{a}{c} \geq \frac{a}{d}> 1/2 $ which is a clear contradiction to $\frac{a}{c} < 1/2< \frac{b}{d}$. Hence, $C=1/2$ is indeed the minimum.

From walkthroughs. We first solve the four variable case. Say we are given $a < b < c < d$. Say we wish to minimize $\left| \frac{w}{x} - \frac{y}{z}\right|$ for some permutation $(w, x, y, z)$ of $(a, b, c, d)$. Note that the only fractions we really need to consider are $\frac{a}{c} - \frac{b}{d}$, $\frac{b}{c} - \frac{a}{d}$ and $\frac{c}{b} - \frac{a}{d}$. Of these it is easily shown the first is minimal. Now we move on to the five variable case. Assume without loss of generality that we have $a < b < c < d < e$. Then it suffices to consider the five permutations, \begin{align*} \frac{a}{c} - \frac{b}{d}, \frac{a}{c} - \frac{b}{e}, \frac{a}{d} - \frac{b}{e}, \frac{a}{d} - \frac{c}{e}, \frac{b}{d} - \frac{c}{e} \end{align*}from the four variable case. Namely, these five differences arise from the $\binom{5}{4}$ ways of choosing numbers from $\{a, b, c, d, e\}$. Now consider the fractions $\frac{a}{c}$ , $\frac{b}{d}$ , $\frac{b}{e}$ , $\frac{c}{e}$ and $\frac{a}{d}$. Note that all of them are less than $1$. Then place them in the intervals $(0, \frac{1}{2}]$ and $[\frac{1}{2}, 1)$. Clearly at least three fractions lie in the same region, and for any three such fractions it can be observed that we may find two fractions involving four different variables. Thus we have attained a bound of $1 - \frac{1}{2} = \frac{1}{2} - 0 = \boxed{\frac{1}{2}}$. Now to show this is acheivable consider the constructions $(\epsilon, 1, 2 - \epsilon, 2, 2+ \epsilon)$.