A Paradox in an Ellipsis (Tuklas Vol. 15, No. 1 - October 19, 2013)

A PARADOX IN AN ELLIPSIS

Are you familiar with infinite decimals? Perhaps you have already encountered a number like this on your calculator screen:  $$0.333333\ldots .$$ Recall that to express this infinite decimal as a fraction, we simply represent this value as $x$ and do the following computations:  $$\begin{align*} x &= 0.333333\ldots \\ 10x &= 3.333333\ldots \\ 10x - x &= 3.333\ldots - 0.333\ldots \\ 9x &= 3 \\ x &= \frac{1}{3}. \end{align*}$$ Another method for this would be to consider the geometric series  $$a_1 = \frac{3}{10}, r = \frac{1}{10},$$  which yields  $$0.333\ldots = 0.3 + 0.03 + 0.003 + \ldots = \sum_{i=1}^{\infty}{\frac{3}{10^i}} = \frac{3/10}{1-1/10} = \frac{1}{3}.$$ The problematic part of the number $0.333\ldots$ lies in the fact that it is composed of a sequence of $3$'s repeated infinitely, and yet we somehow paradoxically resolve this as a finite value. Recall that a paradox is a statement that is seemingly contradictory by nature but might actually be true. In our context, our problem is that we are dealing with infinity (which is, well, infinite) as a complete object (that is, finite). Let us see how this tricky concept was addressed, and gain some perspective on the importance of paradoxes in mathematics.

We begin with the ancient Greek philosopher Parmenides. He believed that there was only one indivisible unchanging reality. To that end, he said that plurality and change were illusions and our senses were lying to us. In particular, he said that motion was an illusion. One of his students, Zeno of Elea, proposed a series of famous paradoxes that demonstrated Parmenides' claims. We discuss one of them below.

Suppose that Achilles and the tortoise are in a 100-meter footrace, where both racers run at a constant speed. Suppose further that Achilles is faster and gives the tortoise a head start, say fifty meters. Zeno argues that by the time Achilles covers the distance from his starting position to the starting position of the tortoise (which we call $P_1$), the tortoise will have moved ahead to a new position $P_2$. When Achilles reaches the position $P_2$, the tortoise will have moved to a new position further ahead, say $P_3$. This will go on forever; in effect, Achilles will never catch up to the tortoise. However, if we compute the time it takes for each racer to finish the race, it is clear that Achilles will win.

Let us plug some numbers into the example to make the comparison more obvious. Say Achilles runs at a speed of 10 meters/minute and the tortoise runs at a speed of 1 meter/minute. From there, we can see that Achilles completes the race in ten minutes whereas the tortoise completes it in 100 minutes. More explicitly, we can use algebra to compute that Achilles and the tortoise meet $5\frac{5}{9}$ minutes after the race begins at the $55\frac{5}{9}$-meter position. Since Achilles always runs faster, this means that after $5\frac{5}{9}$ minutes he will pass the tortoise and eventually finish the race ahead of the tortoise.

What is the problem here? Remember, Parmenides (and many of his contemporaries) said that reality was unchanging and indivisible. The computations above yielded the values $55\frac{5}{9} = 5.555\ldots$ meters and $5\frac{5}{9} = 5.555\ldots$ minutes. In a sense, what we have done is slice our observations of distance and time into smaller and smaller segments under the assumption that these things are infinitely divisible. This process is made explicit in our geometric series representation, where we keep adding a new addend that is a fraction of the previous value. The problem is that there seems to be a disconnect from one side of the equation (the fraction) to the other side (the infinite decimal). 

How do we solve this problem? Practically speaking, it is absurd to think of taking the sum of an infinite number of terms. Look back to the number $0.333\ldots$ and its geometric series representation. We are applying a finite process (taking the sum) an infinite number of times. The best that we can do is attempt to approximate the value of $0.333\ldots$ by adding the value $\frac{3}{10^k}$ to our approximation $\widehat{x}$, which will always be slightly smaller than $0.333\ldots$. That is, we have  $$\begin{eqnarray*} k = 1, & \widehat{x} = 0.3 &< 0.33\ldots \\ k = 2, & \widehat{x} = 0.33 &< 0.333\ldots \\ k = 3, & \widehat{x} = 0.333 &< 0.3333\ldots \\ \vdots \\ \end{eqnarray*}$$ We now make the observation that as we continue this process, the values seemingly get closer and closer to the value $\frac{1}{3}$. However, the formal argument hinges largely on a particular assumption of a property of real numbers (called an axiom) which says that every increasing sequence of (real) numbers that is bounded above must approximate a (unique) real number. Yet we see that the equation does make sense, if viewed in this light.

Thus, we are able to combine two paradoxical concepts under one simple equation. It becomes clear that the summation does not only signify a single process but an infinite continued process that is bounded by a natural property of the numbers used. Hence, it is natural that the other side could be written as a finite value. While some philosophical arguments still debate the merits of Zeno's paradox, this particular resolution by mathematics leads to many beautiful results. In particular, it is this construction of infinite decimals (and by extension, the real numbers) from which one can intuit the notion of a limit. The limit is the central concept of calculus, and leads into further mathematics that surprisingly enough finds applications in fields as varied as engineering and economics.

We now see how the paradox, far from highlighting the pitfalls of mathematics, in fact forces it to grow and enhance its potential. Furthermore, it is able to do so in a beautiful manner, one that makes sense to the point that when we look at it afterwards, we cannot imagine how it could have been otherwise.

ABOUT THE AUTHOR:
Jesus Lemuel L. Martin, Jr. is an Instructor at the Ateneo de Manila University. He obtained his B.S. in Applied Mathematics major in Computational Science at the Ateneo de Manila University in 2010 and his M.S. in Mathematics at the same university in 2013.

REFERENCES:

[1] Byers, William. How Mathematicians Think.
[2] Rademacher, H., Toeplitz, O. The Enjoyment of Mathematics.

OLYMPIAD CORNER

from the 22nd Irish Mathematical Olympiad, 2009

Problem: Let $ABCD$ be a square. The line segment $AB$ is divided internally at $H$ so that $$\left|AB\right| \left| BH\right| = \left| AH\right| ^{2}.$$ Let $E$ be the midpoint of $AD$ and $X$ be the midpoint of $AH$. Let $Y$ be a point on $EB$ such that $XY$ is perpendicular to $BE$. Prove that $\left|XY\right| = \left| XH\right|$.

Solution: 

Let the square be of length $2k$ for some $k>0$, and $\left| AH\right| =x$. Then
$$\begin{eqnarray*} \left| AB\right| \left| BH\right|  &=&\left| AH\right| ^{2} \\ 2k\left( 2k-x\right)  &=&x^{2} \\ x^{2}+2kx-4k^{2} &=&0. \end{eqnarray*}$$ Solving for $x$, we have $$\begin{eqnarray*} x &=&\frac{-2k\pm \sqrt{4k^{2}+16k^{2}}}{2} \\ &=&\frac{-2k\pm 2\sqrt{5}k}{2} \\ &=&\left( \pm \sqrt{5}-1\right) k \end{eqnarray*}$$ But since $x>0$, then $x=\left( \sqrt{5}-1\right) k$.

Note that $$\begin{eqnarray*} \left| BE\right|  &=&\sqrt{\left( 2k\right) ^{2}+a^{2}}=k, \\ \left| BH\right| &=&2a-\left( \sqrt{5}-1\right) k=\left( 3-\sqrt{5}\right) k, \\ \left| XH\right|  &=&\frac{\left( \sqrt{5}-1\right) k}{2}. \end{eqnarray*}$$ Moreover, $\Delta BXY$ and $\Delta BEA$ are similar. It then follows that $$\begin{equation*} \frac{\left| XY\right| }{\left| BX\right| }=\frac{\left| EA\right| }{\left|BE\right| }\Rightarrow \frac{\left| XY\right| }{\left| BX\right| }=\frac{k}{\sqrt{5}k}\Rightarrow \left| XY\right| =\frac{\left| BX\right| }{\sqrt{5}}\end{equation*}$$ and $$\begin{eqnarray*} \left| XY\right| &=&\frac{\left| BH\right| +\left| XH\right| }{\sqrt{5}} \\ &=&\frac{\left( 3-\sqrt{5}\right) k+\frac{1}{2}\left( \sqrt{5}-1\right) k}{\sqrt{5}} \\ &=&\frac{\left( 5-\sqrt{5}\right) k}{2\sqrt{5}}=\frac{\left( \sqrt{5}-1\right) k}{2}. \end{eqnarray*}$$ Therefore, $\left| XH\right| =\left| XY\right|$.

PROBLEMS

1. Show that if $n$ is a perfect square, then the sum of the divisors of $n$ is odd.
2. Let $A$ be any set of $12$ distinct integers chosen from the arithmetic progression $9, 13, 17 \ldots, 77$. Prove that there must be two distinct integers in $A$ whose sum is $90$.
3. Let $p\left(x\right)$ be a polynomial with integer coefficients satisfying $$p\left(0\right) = p\left(1\right) = 2011.$$ Show that $p$ has no integer zeros.
4. Evaluate $$\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{n+1}\right)\left(\frac{1}{n+2}\right).$$
5. Let $ABC$ be an equilateral triangle and $P$ a point in its interior. Consider $XYZ$, the triangle with $XY=PC$, $YZ=PA$, and $ZX=PB$, and $M$ a point in its interior such that $\angle XMY=\angle YMZ=\angle ZMX=120^{\circ}$. Prove that $XM+YM+ZM=AB$.
6. An interior point $P$ is chosen in the rectangle (ABCD) such that $\angle APD+\angle BPC=180^{\circ}$. Find the sum of the angles $\angle DAP$ and $\angle BCP$.

We welcome readers to submit solutions to the problems posed below for publication consideration. Solutions may be submitted to the PEM facilitators on the deadline date or online via vantonio1992@gmail.com. Solutions must be preceded by the solver's name, school affiliation and year level. The deadline for submission is 12:30 PM November 9, 2013.

PEM session for October 12, 2013 cancelled

In light of the current weather, tomorrow's PEM session (October 12, 2013) will be cancelled. Stay safe everyone! :]