2

A rational number is one that can be written as $a/b$ where $a$ and $b$ are integers, $b\gt0$ ($a$ can take care of negative rationals), and I suppose $\gcd(a,b) = 1$.

Given some $n\in\mathbb{Q}$ where $n=a/b$, what is the next rational number?

At first, I naively thought that it was $(a+1)/b$ but of course that is absurd. Consider $n=1/2$. The next rational number is obviously not $1$.

I decided that one must make the "granularity" finer. For instance $n=1/2=2/4$, so by the above idea, the next one is $3/4$, which is better.

Extending that, is it correct that the next rational number is:

$$ \frac{1+\prod_{n\in\mathbb{Z^+},n\neq b}n}{\prod_{n\in\mathbb{Z^+}}n} $$

(We can factor in the sign of $a$ if we wanted to make this more correct)

Michael Hardy
  • 1
pushkin
  • 313
  • 2
    Do you mean the next where they are linearly ordered? In that case there is no next because if $n_i$ and $n_{i+1}$ existed you could take the average which would contradict that $n_{i+1}$ was the next. If you mean the next as an enumeration you can get that from the Cantor diagonal enumeration. – John Douma Oct 07 '15 at 02:05
  • I meant the next where they are linearly ordered. – pushkin Oct 07 '15 at 02:09
  • I see, it wouldn't be enough to just multiply all integers together. You could multiply the denominator by another $2$ to get a rational in between the two supposedly sequential rationals. – pushkin Oct 07 '15 at 02:11

1 Answers1

5

The point of this exercise is to make you realize that there is no such thing as "the next rational number". You did well to realize $\frac{a+1}b$ does not work. But the problem is more general: suppose you do decide that some number $\frac c d$ is the "next" one after $\frac a b$. But ask yourself: what is the average of $\frac a b$ and $\frac c d$? And where is it located?

Paul Sinclair
  • 43,643