Recently while at my tutoring I had a question that said: "Aladin has a pair of magic scissors that can cut things in to tiny pieces. If he cuts a carpet in half, cuts the half into half and continues forever will he eventually reach nothing?".
I answered no. I said that he would eventually reach a piece that his scissors wont be able to cut (eg. atoms).

The teacher accepted my answer but told me that the "right" one was Yes. He then explained to me how $\bigg(\dfrac12\bigg)^{\infty} = 0$. I can't understand how that makes sense.

I entered the equation into wolframalpha and received the same answer. Can someone please explain how this works?


  • $\begingroup$ http://en.wikipedia.org/wiki/Limit_(mathematics) $\endgroup$ – blue Mar 24 '14 at 8:56
  • $\begingroup$ He's right in the sense that in math numbers are infinitely divisible and there are no atoms, but he's wrong because Aladin never reaches infinity, not even eventually. $\endgroup$ – RemcoGerlich Mar 24 '14 at 10:01
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    $\begingroup$ Indeed. There are infinitely many natural numbers, but all of them are finite. $\endgroup$ – celtschk Mar 24 '14 at 10:19

I object to your teacher's answer (even if we remove physical obstructions, such as indivisibility of some particle or another): after any finite amount of time, Aladdin will have only divided it into pieces of size $(1/2)^n$ for some finite $n$; this is still positive. However, the limit is zero, which is what is meant by $(1/2)^\infty$. As $n$ gets bigger, $(1/2)^n$ gets as small as you want, so we say that its limit as $n \rightarrow \infty$ is zero.

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    $\begingroup$ Thanks for your answer. I thought that was the case but he pulled out the calculator show me. I'm a little surprised that I got down-voted for my question, I guess it must have been 'noobish'. $\endgroup$ – C1D Mar 24 '14 at 9:11
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    $\begingroup$ @Mike: "after any finite amount of time..." Not if every cut is twice as quick as the previous one. ;-) $\endgroup$ – Hans Lundmark Mar 24 '14 at 10:02
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    $\begingroup$ @C1D The calculator showed zero (upon successively halving) simply because it is a finite-precision machine that cannot represent arbitrarily small quantities. It ran out of decimal places to store the number, so it returned zero. $\endgroup$ – AmadeusDrZaius Mar 26 '14 at 17:59
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    $\begingroup$ @AmadeusDrZaius that was what I was thinking. $\endgroup$ – C1D Mar 27 '14 at 11:24

What your teacher said was not very mathematically exact. You have to observe the infinite sequence of powers of one half,


And the proper term is that the sequence tends to zero (it converges to its limit which is $0$).

The fact that this goes to zero shouldn't surprise you. But you need an infinite number of cuts.

Of course this is assuming that he's throwing the rest of the carpet away. Otherwise you just get a heap of differently sized pieces :)


I would also object to your teacher's answer. The advantage of writing $\left(\frac{1}{2}\right)^{\infty}=0$ is that you don't have to answer the question "what do you actually end up with when you cut a carpet in half an infinite amount of times?".

Your teacher was probably trying to explain the idea of a limit, which is more like asking the question "If I cut the carpet into a half, and then that half into a half, and so on, am I always going to get closer to having nothing?"

A limit is like a prediction that gets better the more data you use for your prediction. The more terms of $\left(\frac{1}{2}\right)^n$ you calculate, the closer it gets to 0. If your prediction always gets better the more terms you take, then you can write $\lim_{n\rightarrow\infty} \left(\frac{1}{2}\right)^n=0$.

So Aladdin is always getting closer to having nothing, but will he ever actually end up with nothing? That question is actually not being answered on purpose. He's always going to be getting closer to having nothing the longer he keeps on cutting though: $\lim_{\text{cutting in half}\rightarrow \infty} \text{Aladdin's amount of carpet }=0$.


Short answer, limit is not the same as value, so it means expression goes closer and closer to zero.

What is more to mention, if you reach infinity (is not physically possible) you will get zero (it is also physically impossible).

  • $\begingroup$ If you actually reach infinity, you are working in an algebra or field which contains infinity in one way or another. And then I'd expect the value to depend on the exact algebra or field you use. For example, I'd be surprised if in the hyperreal numbers this gives zero instead of an infinitesimal number. $\endgroup$ – celtschk Mar 24 '14 at 10:15

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