# How much seconds of YouTube video will be played if I will press play/pause button at rational timesteps?

Suppose that I want to watch a continuous YouTube video. I start watching it when my clock shows $$t_0=0$$ seconds. Each time my clock shows rational amount of seconds I instantly press play/pause button. How much video will be played when my clock will show $$t_1$$ seconds?

Intuition tells me that it should be $$\frac{t_1}{2}$$ seconds, but how to prove it?

• Between every two real numbers there are infinitly many ratioanl (and irrational) numbers. So the answer is $\infty$ for every $t_1>0 \in R$ – eminem Jun 6 at 13:41
• There is no thing like next rational number to a given rational number, so I feel you will always be constantly pressing the buttons. – S.Sundara Narasimhan Jun 6 at 13:41
• When after $t_0=0$ do you plan to hit the play/pause button first? – Misha Lavrov Jun 6 at 13:44
• The problem as stated is not well defined. To tell whether the video is playing at irrational time $t$, one would have to decide whether the play/stop button has been pressed an even or odd number of times before. But no matter how small $t$ is, it has been pressed infinitely many times before that. – celtschk Jun 6 at 13:47
• How precise is your clock? – gen-z ready to perish Jun 6 at 15:38

The question does not make sense as this is a supertask. There is no next rational number after $$0$$, so the time you stop is not defined.

Your intuition of half the time seems to assume there are the "same number" of rationals and irrationals. That is not correct. The irrationals are uncountable while the rational are countable. One can prove that the measure of the rationals is zero.

• I don't think you wrote what you meant in the sentence: "One can prove that the measure of the irrationals is zero". – Gae. S. Jun 6 at 13:48
• I understand that measure of the rationals (I think you meant them, not irrationals) is zero. That's why I didn't specify whether video is playing or not at rational timesteps, since it will not affect amount of video played – emptysamurai Jun 6 at 13:48
• Actually the number of irrational numbers doesn't matter here. The problem are the rational numbers (as noted in your first paragraph). – celtschk Jun 6 at 13:48
• @celtschk: I agree that the supertask does not depend on the relative cardinalities, but I was reacting to the intuition that the amount played should be $\frac 12$ of the total time. – Ross Millikan Jun 6 at 13:49
• @RossMillikan: Even for that it doesn't matter. If I start/stop every millisecond, there are still uncountably many irrational numbers in between, but I do indeed get close to $t/2$ of the video played until time $t$, provided $t\gg 1\,\mathrm{ms}$. The expectation is probably by taking the limit of $\Delta t\to 0$ when $\Delta t$ is the time between button clicks. – celtschk Jun 6 at 13:53

We know from measure theory that countable sets have measure zero. https://proofwiki.org/wiki/Countable_Sets_Have_Measure_Zero.

Since you only play the video whenever the time since $$0$$ coincides with a rational number, it follows that the time you spend watching the video equals the measure of the set of rational numbers in your interval, which we know has measure $$0.$$

• Nowhere does the question state that the video is played only during rational numbers. Rather the playing/nonplaying state is supposed to be toggled at those places. If I press the play/pause button every millisecond, I've got a finite number of button presses, so their total time again is a null set. But during every second 1ms interval, the video is playing, so the total play time is clearly not zero. – celtschk Jun 6 at 14:02
• You are right! I Misread the instantly pressing play/pause. Thank you for the clarification. Perhaps we can argue as follows. Since the rationals are dense in the irrationals, we know that there is no fixed time interval such that the video does not play during an irrational time. Hence the total playing time is bounded above by $T$ times the measure of the rationals for some time interval T, which again must be $0$. – Mariuhh Jun 6 at 14:10