# Where the hell is 1/3, or Introduction to Chaos

*This article is translation of our Turkish article with the same name.*

*For another article about Mandelbrot and Julia Set, refer to our other article (to be translated soon). *

Chaos is a relatively new branch of mathematics. The verb chaos means “disarray” in Greek.

Let me try to expand the chaos concept.

Now, have a loud, intense cough. You have just made a tiny ripple in the atmosphere. You inhaled and exhaled some air. You might not observe this small effect on the atmosphere today or tomorrow. However your intense cough might yield a typhoon that devastates the houses in Bangladesh after one year. Maybe… Who knows… You cannot even predict the effect of a butterfly after ten years.

Thus, the meteorology discipline can only make one week prediction, not one year.

The very close initial conditions can yield very distant situations after a long time.

Now, the chaos theory just deals just with this problem.

Is there such chaos/disarray in mathematics? Indeed there is, so much so there is a whole theory named after it!

In this article, I will give an example from chaos in mathematics.

Have you ever wondered where exactly is 1/3? We will try to find where is 1/3.

We know that 1/3 is a number between zero and one, meaning in [0, 1] interval.

But knowing this information does not mean that we know where the hell is 1/3. Let us try to understand better where is 1/3.

If we divide the interval [0, 1] to exact half, where does the 1/3 fall? Right? Left? That is, is 1/3 smaller or bigger than 1/2? Of course 1/3 < 1/2 inequality is valid and 1/3 falls to the interval on the left.

In the figure above, we divided the [0, 1] interval by half and obtained [0, 1/2] ve [1/2, 1] intervals. We can see that 1/3 is in [0, 1/2] intervals, so those inequalities involving 0 1/3 1/2 are valid. Now divide the [0, 1/2] interval by half of which 1/3 sits: we obtained

[0, 1/4] and [1/4, 2/4]

intervals. Of which of these intervals does 1/3 falls into? Right or left? Since 1/4 1/3 2/4 inequalities are valid, 1/3 falls into second interval, that is, [1/4, 2/4] interval on the right.

Now divide [1/4, 2/4] interval by half. We obtained

[2/8, 3/8] and [3/8, 4/8]

intervals. Since 2/8 < 1/3 < 3/8 inequalities are valid, 1/3 falls into the first interval on the left.

What if you divide [2/8, 3/8] interval by half? And continue to divide the intervals like this?

In general, if you divide the interval [0, 1] to 2^{n} equal length intervals, to which interval does the 1/3 fall into?

Let me give you the answer: the number 1/3 falls into alternating left and right of those equally divided intervals. Meaning that, it falls into left of 1/2, right of 1/4, left of 3/8, right of 6/16…

But, where is the chaos here?

We know the equality 1/3 = 0.333333… 0.333333 number is very close to 1/3, but is not 1/3. Like 1/3, 0.333333 falls into alternating left and right, but its behavior diverges completely from that of 1/3, as if it is completely independent of 1/3. Now, *that* is the chaos. Two similar numbers behave likewise initially, but one gives no clue about the another after a while.

Let’s turn back to our search for 1/3. Below we present these identities, but it is not difficult to check if the identities that we skipped here are correct or not:

for any value of* n*, there is an integer *a*_{n} such that,

*a*_{n}/2^{n} ≤ 1/3 ≤ (*a*_{n}+1)/2^{n} (1)

inequalities are valid. As you can infer from above,

*a*_{1}= 0

*a*_{2}= 1

*a*_{3}= 2

*a*_{4}= 5

*a*_{5}= 10

*a*_{6}= 21

*a*_{7}= 42

expressions are valid. It is not hard to find the subsequent numbers. Well, the array goes like this: 0, 1, 2, 5, 10, 21, 42, 85, 170, 341, 682… The reader can predict what the next elements of this array. However, a prediction is only a, well, prediction. This predictions needs to be proven.

Now, we will prove the statements:

If* n* is odd, *a*_{n }is even

If* n* is even, *a*_{n} is odd.

Before diving into the calculations, we have to prove that there cannot be any equality in the inequalities of (1) above. Indeed, for an integer *a,* if *a*/2^{n} = 1/3 , 3*a*= 2^{n} is also true, thus we can deduce that 3 divides 2^{n}, which is incorrect. Therefore we can write the inequalities at (1) in a more precise way,

*a*_{n}/2^{n} < 1/3 < (*a*_{n}+1)/2^{n} .

Let’s save those inequalities from their denominators and multiply those numbers by 3 *x* 2^{n} . We obtain:

3*a*_{n} < 2^{n} < 3*a*_{n}+3

inequalities. Therefore the number 2^{n} is bigger than 3*a*_{n}, but smaller than 3*a*_{n}+3. There are two numbers that are bigger than 3*a*_{n }and smaller than 3*a*_{n}+3: 3*a*_{n}+1 ve 3*a*_{n}+2. Consequently the number 2^{n} is equal to one of those numbers:

Either 2^{n}= 3*a*_{n}+1 or 2^{n}= 3*a*_{n}+2. (2)

Which one of these is true? The answer depends on if *n *is odd or even. If *n* is even 2^{n}= 3*a*_{n}+1, where is if odd 2^{n}= 3*a*_{n}+2. Why? I will tell you why in a minute. First we need some preliminaries.

This needs to be known: 2^{n}-2, 2^{n}-1 and 2^{n} are three consecutive numbers. This means that one of these has to be divided by three. Since 2^{n} cannot be divided by 3, we can understand that one and only one of the 2^{n}-2 and 2^{n}-1 numbers can be divided by three.

**Claim**: If *n *is even, 2^{n}-1 is divided by three. If *n *is odd, 2^{n}-2 is divided by three.

**Proof**: First assume that *n *is even and write the *n *as 2^{m}:

*n*= 2^{m}.

Since 2^{m}-1, 2^{m} and 2^{m}+1 are consecutive numbers, one of them is divided by three. But of course 2^{m} is not the one. It means that one of these 2^{m}-1 and 2^{m}+1 numbers is divided by three. Keep this in mind, it will be required in the next paragraph.

Now let’s carry out this simple calculation:

2^{n}-1 = 2^{2m}-1 = (2^{m})^{2}-1 = (2^{m}-1)(2^{m}+1).

We have already learned that one of the numbers 2^{m}-1 ve 2^{m}+1 on the right hand side of the equation can be divided by three. So their product, 2^{n}-1, is divided by three. We have proven the first part of our claim.

Now assume that *n *is an odd number and write *n*= 2^{m}+1. Let’s calculate this again:

2^{n}-2 = 2^{2m}+1-2 = 2(2^{2m}-1).

Hmm, did not we learn that 2^{2m}-1 is divided by three? Thus the numbers in the equation above can be divided by three, 2^{n}-2 is divided by three.

Our claim is proven.

Now, let’s go back to (2).

If 2^{n}= 3*a*_{n}+1 equality is true, so is 3*a*_{n}= 2^{n}-1. This means 2^{n}-1 is divided by three (and 2^{n}-2 is not divisible by three). From the claim above, it is clear that *n *is even.

If 2^{n}= 3*a*_{n}+2 equality is true, 3*a*_{n}= 2^{n}-2 is also true, meaning that 2^{n}-2 is divided by three (and 2^{n}-1 is not divisible by three). And from the claim above, we understand that *n *is odd.

This means we have proven,

*If n* is even, 2^{n}= 3*a*_{n}+1

*If n* is odd, 2^{n}= 3*a*_{n}+2

propositions. Namely we have proven,

*If n* is even, *a*_{n}= (2^{n}-1)/3

*If n* is odd, *a*_{n}= (2^{n}-2)/3

propositions. Yaay!

Now let’s investigate if *a*_{n }is even or odd.

If *n* is even, meaning if *a*_{n}= (2^{n}-1)/3, *a*_{n} cannot be divided by two, since 2^{n}-1 cannot be divided by two. This means *a*_{n} is odd.

If *n* is odd, meaning if *a*_{n}= (2^{n}-2)/3 = 2(2^{n-}^{1}-1)/2, of course *a*_{n} is divided by two.

All these hassle means that 1/3 falls into alternating left and right of the intervals.

Lastly, let’s prove the following beautiful equation:

1/3 = 1/2-1/4+1/8-1/16+1/32-1/64+… (3)

The summation on the right hand side of the equation is an infinite summation. Let’s call the infinite sum, *x* :

*x*= 1/2-1/4+1/8-1/16+1/32-1/64+…

Now multiply the *x *by two :

2*x *= 2(1/2-1/4+1/8-1/16+1/32-1/64+1/128-1/256+…)

= 1-1/2+1/4 –1/8+1/16-1/32+1/64 –1/128+…

= 1-(1/2 –1/4+1/8-1/16+1/32-1/64+1/128-…)

= 1-*x.*

So 2*x *= 1-*x*. And we easily find that *x*= 1/3 !

The equation (3) is proven.^{1}

If this proof did not convince you for the validity of equation (3), try it yourself:

1/2 = 0,5

1/2–1/4 = 0,25

1/2–1/4+1/8 = 0,375

1/2–1/4+1/8–1/16 = 0,3125

1/2–1/4+1/8–1/16+1/32 = 0,3475

1/2–1/4+1/8–1/16+1/32–1/64 = 0,328125

1/2–1/4+1/8–1/16+1/32–1/64+1/128 = 0,3359375

1/2–1/4+1/8–1/16+1/32–1/64+1/128–1/256 = 0,33203125

1/2–1/4+1/8–1/16+1/32–1/64+1/128–1/256+1/512 = 0,333984375

1/2–1/4+1/8–1/16+1/32–1/64+1/128–1/256+1/512–1/1024 = 0,3330078125

1/2–1/4+1/8–1/16+1/32–1/64+1/128–1/256+1/512–1/1024+1/2048 = 0,33349609375

1/2–1/4+1/8–1/16+1/32–1/64+1/128–1/256+1/512–1/1024+1/2048–1/4096= 0,333251953125

1/2–1/4+1/8–1/16+1/32–1/64+1/128–1/256+1/512–1/1024+1/2048–1/4096+1/8192= 0,3333740234375.

As you can see, we converge to 0.33333… or 1/3, as we carry out the calculations. If we had infinite time, we could exactly find 1/3. Even if we cannot carry out the calculations infinitely, we can employ mathematics and find that the equation is just equal to 1/3.

The answer to the question in the first part of this article can be found from equation (3).

The curious reader can answer “WHERE THE HELL IS 1/5?”.^{2}

1 : I slightly lied here. This proof is true if the infinite sum converges to a finite number.

2 : I would like to thank Tayfun Akgül for teaching me that this article is relevant to chaos.

Translated by Bilgecan Dede

Source : Matematik ve Sonsuz (Mathematics and Infinity)

#### Yazar: Ali Nesin (tümünü gör)

- Where the hell is 1/3, or Introduction to Chaos - 13 December 2018