Let $a, b, c \in \mathbb{Z}$, then can every positive fraction less than $1$ be expressed in the form of $$ \frac{1}{a+b} + \frac{1}{a+c} ~? $$

I'm unable to find a contradiction to this statement, but I don't know where to start if I want to prove it to be true. Any ideas?

  • $\begingroup$ Try to express $\dfrac{5}{7}$, $\dfrac{7}{9}$, $\dfrac{7}{11}$... It's enough to use rather small (by $abs$) nonzero denominators, since sum of two numbers with large denominators is pretty small (by $abs$). $\endgroup$ – Oleg567 Jan 24 '20 at 7:28
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    $\begingroup$ What's the point of $a$? Aren't you just asking whether every positive fraction less than $1$ can be expressed in the form $\frac1b+\frac1c$ with $b,c\in\mathbb Z$? $\endgroup$ – joriki Jan 24 '20 at 7:29

I will call $$r := \frac{1}{a+b} + \frac{1}{a+c} \qquad \qquad\qquad\qquad\qquad (1)$$ I see two main cases:

  • First suppose $a+b>0$ and $a+c>0$. In this case, we also have $a+b>1$ and $a+c>1$ since $r≤1$.
    • If $a+b = a+c = 2$, then $r=1$.
    • Else, if $a+b ≥ 2$ and $a+c≥ 3$, then $r≤ \frac{5}{6}$.
    Thus, we see that the rationals in $(\frac{5}{6},1)$ are not covered.
  • If it is not the case, we can suppose $a+b>0$ and $a+c<0$.
    • If $a+b > 1$, then $r≤\frac{1}{a+b}≤ \frac{1}{2}$, so that the rationals in $(\frac{5}{6},1)$ are not covered.
    • Else, if $a+b = 1$, then $r = 1 - \frac{1}{|a+c|}$. The only point of accumulation of such a set is $1$, so that for every $\varepsilon\in(0,\frac{1}{6})$, only a finite number of rationals are covered in $(\frac{5}{6},1-\varepsilon)$.

This proves (if I made not mistakes) that every every positive fraction less than $1$ cannot be expressed in the form $(1)$. There is perhaps a more straightforward proof?


If you are looking for a specific example of number, we see that the numbers of the form $1 - \frac{1}{|a+c|}$ are $\{0,\frac{1}{2}, \frac{2}{3},\frac{3}{4},\frac{4}{5},\frac{5}{6},\frac{6}{7}, ...\}$, all the remaining numbers are smaller than $\frac{6}{7}$. Therefore, all the numbers in $(\frac{5}{6},\frac{6}{7})$ can not be written on the form $(1)$. In particular, the following fraction $$r = \frac{11}{13}$$ is a counterexample!


Here's why it should seem like it shouldn't be true.

First of all the $a+b$ and $a+c$ are irrelevant as any two integers can be written as $n = a+b$ and $m = a+c$ by picking an arbitrary $a$ and setting $b=n-a$ and $c = b-a$.

So this is just saying for any rational $q: 0< q<1$ can be written as $\frac 1m + \frac 1n$.

Now this is handwavey and hard to put into words, but the reason I would first think "Oh, that can't be true" is that each $\frac 1k$ is a distinct distance from $\frac 1{k-1}$ and from $\frac 1{k+1}$ and they aren't close enough to fine tune to any $q$.

Yes, we can get $\frac 1n$ to be as small as we like by taking $n$ to be really big but if we need to. But if we do that then $q$ will have to be very close to some $\frac 1m$ and there are lots of $q$ that won't be close to any $\frac 1m$.

Take for instance: $\frac 12 < q < 1$.

Know we can get $\frac 12 + \frac 13 = \frac 56$ and we can do $\frac 12 + \frac 14= \frac 34$ but suppose $q\ne \frac 56$ nor $q\ne \frac 34$. Suppose $\frac 34 < q < \frac 56$. We cant have $q =\frac 12 + \frac 1m$ because we we have $\frac 14 < q-\frac 12 < \frac 13$ and there are no $\frac 1m$ so that $\frac 14 < \frac 1m < \frac 13$.

And we can't have $q= \frac 1n$ for any $n > 2$ because then $\frac 1m = q-\frac 1n \ge q-\frac 13 \ge \frac 12$ and there is no $m$ so that $\frac 12 < \frac 1m = q-\frac 1n < q < 1$.

And we can't have $n =1$ because them $\frac 1m = q-1$ and $-\frac 13 < q-1< -\frac 14$ and there is no $m$ so that $-\frac 13 < \frac 1m < -\frac 14$.

And since we can't have both $m,n < 0$ (because $q > 0$) we might as well assume that $n$ must be positive.

And that is a legitimate counter-example.

== old answer==

Suppose $0< q< 1$ and $q =\frac 1n + \frac 1m$.

If $n= 1 $ then $\frac 1n=1$ and $q\ge 1$ a contradiction. So $n \ge 2$ and $\frac 1n \le \frac 12$. Likewise $m \ge 2$ and $\frac 1m \le \frac 12$

If $m= n = 2$ then $q = 1$ and that's a contradiction.

So at most one of $m$ or $n$ may be as low as $2$ and the other one must at least $3$.

SO $q = \frac 1n + \frac 1m \le \frac 12 + \frac 13 = \frac 56$.

So what if $\frac 56 < q < 1$? Then there are no $m,n$ so that $q = \frac 1n + \frac 1m$.

And that's that. For any $a, b, c$ we can set $n = a+b$ and $m=a+c$ and there is $\frac 1{a+b} + \frac 1{a+c}$ can never equal any value between $\frac 56$ and $1$.

  • $\begingroup$ Those aren't the only contradictions but they are the simplest $\endgroup$ – fleablood Jan 24 '20 at 7:45
  • $\begingroup$ You still need to treat the case $m<0$ or $n<0$ to complete your contradiction. $\endgroup$ – Medo Jan 24 '20 at 8:08
  • $\begingroup$ @Medo "So at most one of m or n may be as low as 2 and the other one must at least 3." If $m$ or $n$ are negative then they are less than $2$ and $3$ and $\frac 1m < 0 < \frac 12$ and $\frac 1n < 0 < \frac 13$. I do not need to treat those cases. $\endgroup$ – fleablood Jan 24 '20 at 8:17
  • $\begingroup$ Medo has a valid point. You were assuming from the beginning that $m,n$ are positive. This isn't a big hole, but it does allow for the representation of numbers of the form $1-\frac1n$ (which would make your conclusion about $\frac56<q<1$ false, albeit only for a discrete set) $\endgroup$ – Brian Moehring Jan 24 '20 at 17:12
  • $\begingroup$ No, I'm not. I'm checking for the maximum of.... oh. if $n\ge 1$ then ... I assumed $m > 0$. That's true. Okay... you right. But that's easy to figure. "left to the reader as an exercise". $\endgroup$ – fleablood Jan 24 '20 at 19:29

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