# Prove or disprove every positive fraction less than 1 can be expressed with $\frac{1}{a+b} + \frac{1}{a+c}$

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?

• 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 $abs$). – Oleg567 Jan 24 '20 at 7:28
• 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 $b,c\in\mathbb Z$? – 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.

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 $$1$$.

• Those aren't the only contradictions but they are the simplest – fleablood Jan 24 '20 at 7:45
• You still need to treat the case $m<0$ or $n<0$ to complete your contradiction. – Medo Jan 24 '20 at 8:08
• @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. – fleablood Jan 24 '20 at 8:17
• 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) – Brian Moehring Jan 24 '20 at 17:12
• 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". – fleablood Jan 24 '20 at 19:29