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What's the "simplest" equation with only rational coefficients that produces a graph with no rational coordinates? Obviously I haven't precisely defined "simple" so any answer will have to either specifically define it or not be entirely rigorous. However, to give you an idea of what I mean by simple:

I consider higher degree polynomials to be less simple, so $y = 3$ would be more simple than $y = 2x^2 + 4$. Functions defined by infinite polynomials would then be more complicated than every polynomial functions, so $sin(x)$ is less simple than $x^{2789} + 3x^{37} + 14$. I would also consider infinite products more complicated than infinite sums, and infinite powers more complicated still. Also I'm not super familiar with the territory so this might not support natural way to define simple, but I think it makes sense to consider any function only defined by the analytic continuation of another function to be more complex than an infinite polynomial & similar.

I suppose something like y = x/0 might work, as it's undefined everywhere, but what I'm looking for is something with a non-zero and not arbitrarily small domain and range.

My original idea only related to 2D graphs, but higher dimensions may also be interesting.

As a final note, I'm only super familiar with math up to a highschool level, though I do have knowledge beyond that level, so putting things in simpler/intuitive terms would always be helpful if at all possible.

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    $\begingroup$ Notice that the accepted answer has no points where both coordinates are rational. While I think this is what you want, your discussion of $y=f(x)$ reads like even a single rational coordinate is prohibited. $\endgroup$ Commented Jun 15, 2023 at 20:42
  • $\begingroup$ @GregMartin indeed. My original thinking was that neither coordinate would be rational, but I didn't think of the implications, and was really thinking of a function where not both of the coordinates were rational - I've updated the question accordingly. $\endgroup$
    – MilesZew
    Commented Jun 15, 2023 at 21:58
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    $\begingroup$ If only one coordinate needs to be irrational, can't you just have something like the graph y=√2 ? $\endgroup$
    – Showsni
    Commented Jun 16, 2023 at 12:35
  • $\begingroup$ An example of an equation that can be seen as the graph of a function is $y=x^{\sqrt 2}$ for $x\ge 0$. I think by Gelfond–Schneider theorem (clearly much "deeper" than Robert Israel's proof) it passes through exactly two rational points, $(0,0)$ and $(1,1)$. But then if you add square root two to get $y=x^{\sqrt 2}+\sqrt 2$, $x\ge 0$, I suppose you can use Gelfond–Schneider theorem to conclude it passes through no rational points. Edit: Oh, forgot you said rational "coefficients". $\endgroup$ Commented Jun 16, 2023 at 13:45
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    $\begingroup$ if I added some additional natural constraints, the accepted $x^2 + y^2 = 3$ could probably be the simplest, but as it stands things like $y^2 = 2$ are indeed simpler $\endgroup$
    – MilesZew
    Commented Jun 16, 2023 at 17:54

3 Answers 3

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You can't get much simpler than the equation of a circle of radius $\sqrt{3}$ centred at the origin: $$x^2 + y^2 = 3$$

Proof: Suppose we have a solution with $x$ and $y$ rational. Taking a common denominator, we can write the equation as $X^2 + Y^2 = 3 Z^2$ where $X, Y, Z$ are integers and not all even. But this is impossible, as we see by considering the possibilities mod $4$.

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  • $\begingroup$ ah, very nice, very simple - I was expecting things to be a bit more complicated. I wonder what additional constraints could be added to force more exotic/esoteric solutions $\endgroup$
    – MilesZew
    Commented Jun 15, 2023 at 19:40
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    $\begingroup$ Hmm, for example this breaks down if we allow x and y to be complex: $i^2 + 2^2 = 3$ $\endgroup$
    – MilesZew
    Commented Jun 15, 2023 at 20:08
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    $\begingroup$ As a comparison, the unit circle does not only have the rational points $(\pm1,0)$ and $(0,\pm1)$, but actually an infinite number of rational points. $\endgroup$
    – md2perpe
    Commented Jun 16, 2023 at 6:55
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    $\begingroup$ Hi. The Devil's advocate was here and asked me to tell you that $y^2 = 2$ is more simple than $x^2 + y^2 = 3$. $\endgroup$
    – Stef
    Commented Jun 16, 2023 at 16:17
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    $\begingroup$ @Devil: I did write "You can't get much simpler", not "you can't get simpler". $\endgroup$ Commented Jun 16, 2023 at 19:08
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The square root of two is irrational; in other words $x^2-2=0$ has no rational solutions.

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    $\begingroup$ It's not clear to me what you're getting at here. If you mean to consider (say) $y = x^2-2$ then it is a non-example, since it has infinitely many rational points. $\endgroup$ Commented Jun 16, 2023 at 12:40
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    $\begingroup$ The onedimensional equation $x^2-2=0$ was maybe what the asker meant to exclude by not arbitrarily small domain and range. It has two solutions. The asker likely thought of "curves" having a continuum of solutions, like the circle from the earlier answer. $\endgroup$ Commented Jun 16, 2023 at 13:25
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    $\begingroup$ @preferred_anon I'd interpret it as $x^2 - 2 + 0\cdot y = 0$. $\endgroup$ Commented Jun 16, 2023 at 16:52
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    $\begingroup$ @preferred_anon $x^2 - 2 = 0$ in x,y coordinates is the equation of a vertical line through $x = \sqrt{2}$. It passes through an infinity of y values, some of them rational — but the x coordinate is irrational for all of them. It is, of course, a boring example. $\endgroup$
    – hobbs
    Commented Jun 16, 2023 at 17:31
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    $\begingroup$ @hobbs Well, more precisely, $x^2 - 2 = 0$ is the union of two vertical lines, which does not invalidate the rest of your comment. $\endgroup$ Commented Jun 16, 2023 at 21:10
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If one wants y = f(x) with not both x, y rational, where dom(f)=R (set of reals) and range(f) is at least infinite, how about this?

f(x) = x * chi(Q') + pi * chi(Q) where Q={rationals}, Q'={irrationals}, chi(set)=characteristic function of set (which returns 1 on elements of the set and 0 otherwise). And, pi = 3.14159... is what you think it is.

Ergo, for x irrational, f(x)=x, and for x rational, f(x)=pi*x is irrational.

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