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There is a problem in my calculus textbook that wants me to draw/look at the following function $f:\Bbb{R}\to\Bbb{R}$

$$f(x)=\begin{cases} 1 & : x \in \Bbb{Q} \\ 0 & : x \in \Bbb{P} \\ \end{cases}$$

The rational numbers are given by:

$\Bbb{Q}=${$\frac{a}{b} | a,b\in \Bbb{Z}, b\neq0$}

But I am choosing to only look at the positive numbers since the negative numbers will include the same patterns but mirrored on the negative x-axis.

I thought you could just list every combination of a and b as fractions to end up with all the rational numbers. So if you just look at the first 4 positive integers: $a,b\in${$0, 1, 2, 3$} you get the rational numbers: $\frac{a}{b}\in${$\frac{0}{0}, \frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \frac{2}{1}, \frac{2}{3}, \frac{3}{1}, \frac{3}{2}$} after removing all combinations of a and b that leads to the same integer or by-zero-division ($\frac{1}{0}, \frac{0}{3}, \frac{2}{2}$ and so on). All other points on the x-axis would be irrational. When marking the rational numbers it would look like this: drawn points

It seemed from the drawing that over half of the rational numbers will always be in the interval [0, 1]. It seems logical from the sample above because all the fractions in the range (0, 1) make up the set: {$\frac{1}{2}, \frac{1}{3}, \frac{2}{3}$} and all the other fractions excluding the special cases $\frac{0}{0}$ and $\frac{1}{1}$ make up the set {$\frac{2}{1}, \frac{3}{1}, \frac{3}{2}$} which are just the reciprocals of other fractions. If you extend the set to: $a,b\in${$0, 1, 2, 3, 4$} you get the extra combinations: {$\frac{1}{4}, \frac{2}{4}, \frac{3}{4}$} and {$\frac{4}{1}, \frac{4}{2}, \frac{4}{3}$} where the fractions $\frac{2}{4}$ and $\frac{4}{2}$ already are points on the x-axis. This means that over half of the rational numbers are still in the interval [0, 1] when you include the fractions $\frac{0}{0}$ and $\frac{1}{1}$.

I want to prove that this pattern holds but I am not able to. One thing I am struggling with proving is that if a new combination $\frac{a}{b}$ is already accounted for, then $\frac{b}{a}$ will always also be accounted for like in the example above with $\frac{4}{2}$ and $\frac{2}{4}$.

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  • $\begingroup$ Over half of the rational numbers in your particular sequence lie in $[0,1]$. But this is just an artefact of the way you have chosen to list them. $\endgroup$
    – TonyK
    Commented May 2 at 15:54
  • $\begingroup$ Well uhm technically $\forall x≥1,x\in \mathbb{Q} \exists 0<\frac{1}{x}≤1$. But you can't call that "half". It's uncountable, infinite $\endgroup$
    – Gwen
    Commented May 2 at 16:00
  • $\begingroup$ It’s countably infinite @Gwen $\endgroup$
    – Malady
    Commented May 2 at 17:28
  • $\begingroup$ @TonyK is the way I have listed them wrong? As in would I have missed any rational numbers when continuing to use this listing method for ever? $\endgroup$
    – volticus
    Commented May 2 at 17:51
  • $\begingroup$ No, but there are many correct ways of listing the rationals. (By the way, $0/0$ is undefined!) $\endgroup$
    – TonyK
    Commented May 2 at 17:59

1 Answer 1

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There are infinitely many positive rational numbers. It does not formally make sense to talk about 'half' of them. The way you chose your method, half are lying there. But we can choose other sequences in different ways where it won't happen. I recommend reading up some more on infinities.

Also think about this: using similar argument as you, I can also show more than half of them lie in $[0,\frac{1}{2}]$, then again you can show that for $[0,\frac{1}{4}]$ and so on. Now you can see that will lead to some problems if you keep going like this.

Hope this helps :)

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  • $\begingroup$ I am not able to see how you can argue that half of the rational numbers lie in [0, $\frac{1}{2}$]. Could you explain? $\endgroup$
    – volticus
    Commented May 2 at 17:43
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    $\begingroup$ @volticus: just take your sequence and divide every element by $2$. $\endgroup$
    – TonyK
    Commented May 2 at 18:00

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