Borel's Other Paradox

Geometry Level 3

You have a ruler, with 0 and 1 marked at either end. You also have a roll of tape, the same width as the ruler.

First you cut a piece of tape of length 1 8 \frac{1}{8} , and stick it with its center at position 1 2 \frac{1}{2} . This tape then covers the ruler from 7 16 \frac{7}{16} to 9 16 \frac{9}{16} .

You then stick two pieces of tape of length 1 27 \frac{1}{27} , centered at positions 1 3 \frac{1}{3} and 2 3 \frac{2}{3} , respectively.

You carry on in this way, sticking pieces of tape of length 1 q 3 \dfrac{1}{q^3} at every fraction p q \dfrac{p}{q} , ( p = 1 , 2 , , q 1 ) (p=1, 2, \ldots, q-1) , lying between 0 0 and 1 1 , regardless of whether there is any tape there already.

By the end of this (infinite) process, are all points of the ruler covered?


Hint: You may wish to use the fact that 1 1 2 + 1 2 2 + 1 3 2 + = π 2 6 . \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots=\frac{\pi^2}{6}.


Relevant wiki: Riemann Zeta function .

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1 solution

Nicholas James
Mar 16, 2017

Let us consider the total length of tape used.

As an example, let us consider the fifths. We use 1 5 3 \frac{1}{5^3} , for each of the fifths between 0 0 and 1 1 : 1 5 \frac{1}{5} , 2 5 \frac{2}{5} , 3 5 \frac{3}{5} and 4 5 \frac{4}{5} . The total length of this tape, L 5 L_5 is:

L 5 = 5 1 5 3 L_5=\frac{5-1}{5^3}

In general, for nths, the length of tape used is L n L_n

L n = n 1 n 3 L_n=\frac{n-1}{n^3}

So the total length of tape used, L t o t a l L_{total} , is

L t o t a l = 1 2 3 + 2 3 3 + 3 4 3 + . . . L_{total}=\frac{1}{2^3}+\frac{2}{3^3}+\frac{3}{4^3}+...

Each of these terms n 1 n 3 \frac{n-1}{n^3} is less than n n 3 = 1 n 2 \frac{n}{n^3}=\frac{1}{n^2} , so,

L t o t a l = 1 2 3 + 2 3 3 + 3 4 3 + . . . 1 2 2 + 1 3 2 + 1 4 2 + . . . L_{total}=\frac{1}{2^3}+\frac{2}{3^3}+\frac{3}{4^3}+...\leq \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...

but,

1 1 2 + 1 2 2 + 1 3 2 + . . . = π 2 6 \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...=\frac{\pi^2}{6}

so,

1 2 2 + 1 3 2 + 1 4 2 + . . . = π 2 6 1 \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...=\frac{\pi^2}{6}-1

We have therefore found that:

L t o t a l π 2 6 1 L_{total} \leq \frac{\pi^2}{6}-1

So, the total length of tape used was less than 1 1 , which means the entire ruler cannot be covered.

The length of the tape comes out to π 2 6 1 k = 2 1 k 3 0.4428 \dfrac{\pi^{2}}{6} - 1 - \displaystyle\sum_{k=2}^{\infty} \dfrac{1}{k^{3}} \approx 0.4428 .

(There's is no closed form for the odd reciprocal powers, so the approximation will have to do.)

I'm not sure how much of the ruler will actually be covered, given the potential for overlap. That would be an interesting calculation to make, given some time.

Brian Charlesworth - 4 years, 2 months ago

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Nice, thanks Brian; I see where that comes from, and it's a nice extension.
It's not a lot of tape, and yet at all real points it's infinitely thick.

The total coverage would be interesting, and far more difficult.

Nicholas James - 4 years, 2 months ago

When I first read the question, I thought we were using pieces of length 1 q 2 \dfrac {1}{q^2} . This gives a pretty interesting result as you continue with the analysis in a similar manner.

Calvin Lin Staff - 4 years, 2 months ago

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