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Recently I was wondering: Why does pi have an irrational value as it is simply the ratio of diameter to circumference of a circle? As the value of diameter is rational then the irrationality must come from the circumference.

Then I used calculus to calculate the arc length of various functions with curved graphs (between two rational points) and found the arc length two be irrational again.

Do all curved paths have irrational lengths?

My logic is that while calculating the arc length (calculus) we assume that the arc is composed of infinitely small line segments and we are never close the real value and unlike the area under a curve, there do not exist an upper and lower limit which converges to the same value.

If yes, are these the reasons irrational values exist in the first place?

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    $\begingroup$ "As the value of diameter is rational then the irrationality must come from the circumference." No, you could just as easily say that your circumference is rational. Then your diameter would necessarily be irrational. Or your diameter and circumference could BOTH be irrational. $\endgroup$ – James Jul 11 at 18:08
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    $\begingroup$ Your title asks about all arc lengths, but the question text asks about all curved paths. They're not the same thing at all. $\endgroup$ – Barmar Jul 11 at 19:17
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    $\begingroup$ Maybe what you meant to ask is "Is the ratio between the secant length and the arc length always irrational?" $\endgroup$ – Barmar Jul 11 at 19:18
  • $\begingroup$ Also, consider that if irrational values existed only because of curves, that would imply that no straight line segment could have an irrational length, but of course it is easy to think of segments that do, e.g. the diagonal of a unit square. $\endgroup$ – Tom Zych Jul 12 at 22:32
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    $\begingroup$ The OP's concept of "arc" is too general for this to have any hope of being true, but the question is perhaps more interesting under better constraint of what "arc" means. $\endgroup$ – R.. Jul 14 at 14:57
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An example of a curve with rational arc lengths between at least some pairs of rational points is a cardioid.

Down to scaling and rotation, a cardioid may be rendered in polar coordinates by the equation

$$r=1-\cos\theta$$

with arc length differential

$$ds=\left(\sqrt{r^2+(dr/d\theta)^2}\right)d\theta=\sqrt{2-2\cos\theta}~d\theta=2\sin(\theta/2)d\theta$$

Integrating this from $\theta=0$ to an arbitrary value of $\theta$ gives the arc length function

$$s=4(1-\cos(\theta/2))$$

Thus the arc length from the origin to $(-2,0)$ ($\theta=\pi$) is $4$. Moreover, suppose we select $\theta=2\cos^{-1}(a/c)$ where $a^2+b^2=c^2$ is a Pythagorean triple. Then we have

$$\cos\theta=2(a^2/c^2)-1$$

$$\sin\theta=2(b/c)(a/c)=2ab/c^2$$

Clearly giving rational values for the Cartesian coordinates $x=(1-\cos\theta)\cos\theta$ and $y=(1-\cos\theta)\sin\theta$. The arc length from the origin is then the rational quantity

$$s=4(1-\cos(\theta/2))=4(1-a/c)$$

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Obviously, a straight line between two rational points can have rational length $-$ just take $(0,0)$ and $(1,0)$ as your rational points.

But a curved line can also have rational length. Consider parabolas of the form $y=\lambda x(1-x)$, which all pass through the rational points $(0,0)$ and $(1,0)$. If $\lambda=0$, then we get a straight line, with arc length $1$. And if $\lambda=4$, then the curve passes through $(\frac12,1)$, so the arc length is greater than $2$.

Now let $\lambda$ vary smoothly from $0$ to $4$. The arc length also varies smoothly, from $1$ to some value greater than $2$; so for some value of $\lambda$, the arc length must be $2$, which is a rational number.

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  • $\begingroup$ But how do we know that the value of $\lambda$ is not itself irrational?. As there can be infinite irrational numbers between two rational numbers. $\endgroup$ – Rupanshu Yadav Jul 11 at 10:37
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    $\begingroup$ @RupanshuYadav: I didn’t claim that $\lambda$ had to be rational. $\endgroup$ – TonyK Jul 11 at 10:42
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    $\begingroup$ I like this answer since it implicitly shows something much stronger. As $\lambda$ varies, it will take on infinitely many values where the length is rational, and infinitely many values where the length is irrational. Furthermore, there is nothing about this argument that really relies on the points having rational coordinates. Between any two points there will be infinitely many curves of rational length, and infinitely many of irrational length. $\endgroup$ – John Coleman Jul 12 at 14:42
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    $\begingroup$ How do you know the arc length also varies smoothly? And that the arc length must be 2 somewhere? The former requires continuity at least. The latter from intermediate value theorem and the real line. $\endgroup$ – qwr Jul 12 at 20:49
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    $\begingroup$ @qwr: if you want, you can take the formula for arc length yourself, and apply it to this case. You will find that for all $\varepsilon>0$, there exists $\delta>0$ such that if $1\le\lambda_1<\lambda_2\le 4$, and $\lambda_2-\lambda_1<\delta$, then the arc lengths of the two curves $y=\lambda_1 x(1-x)$ and $y=\lambda_2 x(1-x)$ differ by less than $\varepsilon$. Hence the arc length varies smoothly, and $-$ as you say $-$ the Intermediate Value Theorem does the rest. $\endgroup$ – TonyK Jul 12 at 21:21
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So, my question is that do all curved path have irrational lengths?

Of course not. A circle with radius $\frac{1}{2\pi}$ is a curved path and has length $1$ which is a rational number. If you put the center of the circle to $(-\frac1{2\pi}, 0)$, then $(0,0)$, a "rational" point, is on the circle, and the circle can be seen as a path from $(0,0)$ to $(0,0)$.

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    $\begingroup$ That's kind of cheating! $\endgroup$ – TonyK Jul 11 at 9:17
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    $\begingroup$ @TonyK It's more like showing that it's important to ask specific well defined questions... $\endgroup$ – 5xum Jul 11 at 9:23
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    $\begingroup$ How does $(0,0)$ count as "two rational points"? $\endgroup$ – TonyK Jul 11 at 9:38
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    $\begingroup$ If you consider the semicircle it is not defined between two rational points. $\endgroup$ – Rupanshu Yadav Jul 11 at 10:26
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    $\begingroup$ @RupanshuYadav Sure. But I am not considering the semicircle. I am considering the circle. Is that somehow not allowed? $\endgroup$ – 5xum Jul 11 at 10:56
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Consider the two points $(-\frac12,0)$ and $(\frac12,0)$. For any real value of $y_0$, we can draw a circular arc between these two points which is centered at $(0,y_0)$ and which lies entirely in the upper half-plane. As $y \to - \infty$, the length of this arc approaches 1 (since the arc approaches a straight line); as $y \to +\infty$, the arc length approaches $\infty$. Since the arc length varies continuously with $y_0$, it must be the case that the arc length can be any real number greater than 1, including all rational lengths greater than 1.

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No.

Take any smooth curve between two rational points and deform it to change its length by a finite amount. During deformation, you will cross infinitely many rational lengths.

A simple example is a polynomial having two rational roots, times a variable factor.


Now consider the curve of parametric equations

$$\begin{cases}x=\dfrac{t^3}3-t,\\y=t^2\end{cases}$$

(a modified Tschirnhausen cubic).

We have

$$s=\int_a^b\sqrt{(t^2-1)+4t^2}\,dt=\int_a^b(t^2+1)\,dt=\frac{b^3-a^3}3+b-a,$$

so that the length between two rational $t$ (giving rational endpoints) is always rational.

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Contributing another simple counterexample, let $f(x)=-\cos x$ with $x\in[0,\pi]$. Then, the length of $f$ between $A(0,-1)$ and $B(\pi,1)$ is given by: $$\ell(f)=\int_0^\pi|f'(t)|dt=\int_0^\pi\sin tdt=[-\cos t]_0^\pi=2.$$ Note, also, that the ratio between the curve and the "diameter" of it, $AB$ is: $$\frac{\ell(f)}{(AB)}=\frac{2}{\sqrt{\pi^2+2}},$$ which is, again, irrational.

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