# An interesting problem from Real Analysis-Abbot

This is taken from "Undestanding Analysis"- Abbot,

Exercise 4.5.8: Imagine a clock where the hour hand and the minute hand are indistinguishable from each other. Assuming the hands move continuously around the face of the clock, and assuming their positions can be measured with perfect accuracy, is it always possible to determine the time?

From my point of view, I think we CAN tell the time because if you observe for, say, 5 minutes, you can see that the minute hand moves faster than the hour hand. Having that, you can determine which one is the hour hand and which one is the minute hand, and hence, you can easily deduct the exact time. However, this is not very mathematical, and although there is a solution for this question, I do not want to look at it so it would be better that I come up with my own original solution.

So, please help me, did I answer this question correctly? How do I move up my argument to a solid proof statement? I thank you very much for your help.

• Well, if you observe long enough time, it is easy to tell. But the more interesting question is to ask whether one can tell immediately the time. For example, if at some moment one hand is pointing exactly at 12 and the other one exactly at 6. Then it has to be six o'clock (It can't be 12:30, as the hour hand will point in between 12 and 1 instead). One solution to this problem uses some topology of $\mathbb S^1 \times \mathbb S^1$. Not sure if it was posted on SE. – user99914 Feb 10 '15 at 9:44
• Aha, so we are assuming continuous movements. But then, if one hand is pointing between 12 and 1 and the other at 6 it could well be 12h 30min but also 6h 2,5 min because the difference in the hour hand would be unobservable to our eyes. So I guess the hypothesis that we can measure their position with perfect accuracy must be used. Oh! this problem would be much easier with a digital clock with numbers of equal size :D – Martingalo Feb 10 '15 at 10:32

Take $[0,1]$ as an interval of 12 hours. Then it is easy to see that at the time $t \in [0,1]$ the hour hand has an angle of $2\pi t$ (in radians), while the minute hand has an angle of $24\pi t$.

So we can consider a model of a clock with the following function $$f: [0, 1) \longrightarrow S^1 \times S^1$$ $$t \mapsto (e^{2i\pi t}, e^{24i\pi t})$$

If the two hands were distinguishable, then clearly we would be able to say what is the time just as seeing the clock, since the function $f$ is injective.

Now, your problem is equivalent on asking if the following system has a solution $$\left\{ \begin{matrix} e^{2i \pi t} = e^{24i \pi s} \\ e^{2i \pi s} = e^{24i \pi t} \end{matrix} \right.$$ where $s \neq t$ are two distinct times in the interval $[0,1)$. This system is equivalent on asking that $t-12s, s-12 t$ are both integers. By trial and error, I found that at least one solution exists, precisely $$(t,s) = \left( \frac{12}{143} , \frac{1}{143}\right)$$ is the solution of $$\left\{ \begin{matrix} t -12s &=& 0 \\ s - 12 t &=& -1 \end{matrix} \right.$$

so the answer is that there exist distinct times corresponding to equivalent representation of the clock, however it is easy to see that these are only a finite number.

• No need for trial and error. Set $u=e^{2i\pi t}$ and $v=e^{2i\pi s}$. Then $u=v^{12}$ and $v=u^{12}$, and so $v=v^{144}$ and $v^{143}=1$, which is your solution. – lhf Feb 10 '15 at 12:50
• @Crostul is i in the exponential function a complex number? Wow, why do you use exponential function? – SON TO Feb 10 '15 at 20:12
• $e^{i \alpha }$ is just a compact way to write $(\cos \alpha, \sin \alpha ) \in \mathbb{R}^2$. – Crostul Feb 10 '15 at 22:28
• @Crostul so, $e^{2i \pi t}$ means that $(cos(2 \pi t),sin (2 \pi t))$, does it not ? – SON TO Feb 10 '15 at 22:35
• Exactly. There is a good reason to write it in this way, one of them is the fact that the rule $e^{i\alpha}\cdot e^{i \beta} = e^{i (\alpha + \beta)}$. I suggest you to see at en.wikipedia.org/wiki/Euler%27s_formula – Crostul Feb 10 '15 at 22:47