0
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ 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
  • 3
    $\begingroup$ 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
2
$\begingroup$

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!

$\endgroup$
0
$\begingroup$

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 $1$.

$\endgroup$
5
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.