History of Mathematics: Charles Dodgson

Mainly, I talk about proofs, patterns, and ways to apply mathematics to the real world. However, there are other ways that mathematics is cool too. Not only is the actual content interesting, but the stories are as well.

Today, I will tell the story of a mathematician named Charles Lutwidge Dodgson. Dodgson was born in 1832 in Cheshire, England. He graduated from Christ Church College at Oxford, and then began his career as a mathematician, lecturing and tutoring at Oxford.

Dodgson was not the type of mathematician who often made breakthroughs and discoveries. However, he did find some interesting things in mathematics and logic. One of them, I actually talked about in a previous post, but never mentioned that Dodgson was the mind behind it. Click here to see it.

He is also known for a method of election he developed. In America, we just vote for the candidate of our choosing, which is called simple plurality. Yet, we run into problems in elections such as the Bush-Gore-Nader election of 2000. In this election, most Nader supporters preferred Gore to Bush and most Bush supporters preferred Gore to Nader, even if it was just by a little bit. So, Gore was never the least favorite of anybody, while Bush was the least favorite of most Nader supporters and probably many Gore supporters. Using a ranking method rather than a plurality method, we can find a winner that the most people are satisfied with.

Yet, this method can fail in elections such as the Obama-Romney election of 2012. The leading third party candidate was Gary Johnson, but was supported by only a small percentage of the population. But, Obama fans would rank Johnson as their second favorite in order to give as little support as possible to Romney, and vice versa. This may have led to a win for Gary Johnson, which very few people would be satisfied with.

So, mathematicians in the field of game theory are always struggling to find a perfect election method. There are dozens of ideas out there, one of which is created by Charles Dodgson. His method makes use of finding a Condorcet winner. If on every ballot, a certain candidate is ranked higher than another one, then this beaten candidate is eliminated. For example, if all Americans preferred Barack Obama to Jill Stein, the Green Party candidate, then Stein would be eliminated. If you can eliminate every candidate in this way, the candidate still standing is the Condorcet winner.

Since it is near impossible to have a Condorcet winner, this method is impractical. However, Dodgson extended it by saying that after eliminating everyone possible, you begin swapping rankings on people’s ballots until you are able to have a Condorcet winner. The candidate that requires the least swaps wins. This method will eliminate weak third party candidates, but still take them into strong consideration.

On a different note, Dodgson was also known for his ability to write. He wrote dozens of famous math textbooks, and compiled texts for undergraduate students. But he also enjoyed implementing his mathematical knowledge into fantasy writing. As a man who loved children, he would tell stories that had some mathematics and logic infused in them.

One of his favorite children was named Alice Liddell, whose father was the Dean of the school where Dodgson taught. He began telling her stories about a girl named Alice, which were always continued every time they saw each other. Years later, he gave her a written manuscript of the story for Christmas.

In the story, there were many mentions of mathematics. For instance, there is one point in the story where Alice is three inches tall. She finds out that she must eat from a mushroom to grow back to normal, but half of the mushroom stretches her neck and half shrinks her torso. She must find the correct proportions to grow properly.

This is the foundation of Algebra. In fact, the word “Algebra” comes from an Arabic algebra book whose title translates to “Restoration and Reduction.” And she must use the concepts in this book, by finding how much of each side of the mushroom is necessary to make her neck proportions equivalent to her torso proportions. This resembles an algebraic equation.

Dodgson even used his ability to play with words in the story, by describing the branches of arithmetic as ambition, distraction, uglification, and derision.

These manuscripts ended up getting published, and eventually found their way to the queen of England: Queen Victoria. After reading them, she demanded that she receive every book written by this man. To her surprise, she ended up with a huge stack of mathematics textbooks.

This book that the queen loved, that has many mathematical references, that is written by a mathematician from Oxford, is titled Alice and Wonderland. And you might know Charles Dodgson better by his pen name, Lewis Carroll.

Bonus: Lots of these historical mathematicians have a few funny stories in their successful careers as well. Here is one about Charles Dodgson.

Because of his friendliness to children, Dodgson was a popular guest at parties. One time when he was invited to a party in London, he decided to crawl into the room as a surprise to the kids. However, he crawled into the wrong household, where a group of adults were also having a party.

Game Theory and Soccer

When I gave my TEDx talk in India this past December, one of the things I talked about was an application of game theory to soccer. Since I think it is pretty cool, I thought I would share it.

You may know that a penalty kick is when a player gets a free shot at goal from a point about 12 yards (or 11 meters) from it with only the goalkeeper allowed to block the shot. Since this is such a close distance, the goalie just has to take a guess as to whether the kicker will shoot to the left or the right. The kicker also has to choose to kick to his left or his right, since kicking to the center is kicking right to the goalkeeper.

So, we know what each player's strategies are. Let's create a grid to represent it, like we do in most game theory examples.

	Dive Left	Dive Right
Kick Left
Kick Right

Next, we need to figure out what percent of the time the kicker scores in these four outcomes. So, a professor from the London School of Economics named Ignacio Palacios-Huerta figured these statistics out by taking data from over 1400 penalty kicks. This was his result:

	Dive Left	Dive Right
Kick Left	.58, .42	.95, .05
Kick Right	.93, .07	.7, .3

Now, we will use some game theory techniques to figure out what the optimal strategy of each player is. In this case, it would be best to find the mixed-strategy equilibrium

Goalkeeper's Optimal Strategy (Diving Left)
.58x + .95(1 - x) = .93x + .7(1 - x)
.58x + .95 - .95x = .93x + .7 - .7x
.95 - .37x = .23x + .7
.25 = .6x
.42 ≈ x

Kicker's Optimal Strategy (Kicking Left)
.42x + .07(1 - x) = .05x + .3(1 - x)
.42x + .07 - .07x = .05x + .3 - .3x
.35x + .07 = .3 - .25x
.6x = .23
x ≈ .39

So, the goalie's optimal strategy is to dive to the left 42% of the time and the kicker's optimal strategy is to kick to the left 39% of the time.

This alone is pretty cool, that we can determine the best way for a soccer player to handle this situation. However, we don't know if this actually works. So, Ignacio Palacios-Huerta took data from the best kickers and goalies in the world to see how their strategies matched up with the math.

Surprisingly enough, they were using the exact same strategy. The fact that even though these soccer players don't know game theory, but happened to stumble upon this perfect strategy was really impressive.

Bonus: While on the topic of game theory in soccer, I thought I would mention a famous soccer game that involved some game theory. In the 1994 Caribbean Cup, Granada faced Barbados. Because of some unique rules, a very interesting thing happened. Click here to read the Wikipedia article about the match.

A Dumbing Down of the Riemann Hypothesis

Today is my first post on math in the news. I recently came across this article on the Riemann Hypothesis, which I had planned to talk about in India, but didn't get a chance to. Let me give a brief background and then I will share the article.

Back in 2000, Clay Mathematics Institute of Providence, Rhode Island announced the Millennium Prizes, which consisted of seven problems that had been stumping mathematicians for a long time. They set aside a million US dollars for any person who solved one of the problems.

I find it interesting just on its own that you can become wealthy as a mathematician. Other than the Nobel Economics Prize, math has its own way of getting a million dollars.

One of the more popular of these problems is the Riemann Hypothesis, which I wanted to talk about today. I will try to explain here what the Riemann Hypothesis is (it is a difficult concept, but online sources complicate it drastically), and then show the article.

First off, you might remember the number i, which is the square root of -1. This is not a normal variable that can just replace anything you want; it is always the square root of -1. You may have heard in math class the term "real number." A number that has just an i in it are imaginary numbers, like 2i or -5i.

A little over a year ago, I did a post about the complex plane. This takes our horizontal number line from first grade and makes it our x-axis. It then takes these imaginary numbers and makes those the intervals of the y-axis.

A point on the x-axis is a real number, represented with the letter a. A point on the y-axis is an imaginary number, represented with the term bi. A point that is just floating around somewhere not on one of these lines is a complex number. You can write it with the expression a + bi, with a being the number it lines up with on the x-axis and bi being the number it lines up with on the y-axis.

Say you had to take the equation y = x^3 - 2 and start plugging in values for x (replacing the x with a number and then figuring out what it equals). Most people would start plugging in real numbers like 0, 1, 2, 3, -1, -2, -3, and so on. However, this Riemann Hypothesis requires us to open up our minds a little bit. Rather than just plugging real numbers into equations, we have to plug complex numbers into equations.

The Riemann Zeta Function is the equation we are plugging these numbers into (a little side-note: the Riemann Zeta Function has a 2π in it for any tauists). For the Riemann Hypothesis, it is concerned to find when this equation equals zero. This is called the zero of the equation.

The real numbers that are zeros of this equation are called the trivial zeros. They are equal to -2, -4, -6, -8, and so on. The complex numbers that are zeros of this equation are called the non-trivial zeros. As far as we know, these non-trivial zeros all have different b values in our a + bi, but the a value always seems to be 1/2.

The Riemann Hypothesis is simply asking the question is there a non-trivial zero of the Riemann Zeta Function whose a value is not equal to 1/2. Imagine winning a million dollars after submitting a hundred page paper that answers just a yes or no question.

It seems like proving either side would be extremely difficult. This article gives a nice explanation of how they are going about proving the yes side of it.

http://www.rdmag.com/news/2012/11/supercomputing-solve-superproblem-mathematics

The article brings up a very good way to do it. If you find just one time where the a value is not 1/2 and it is complex, then the statement is proven. So, Yuri Matiyasevich decided to turn the supercomputers on and start cranking out values. They have not found any values without the 1/2, but they also cannot mathematically prove that it always is the 1/2, and that is where the dispute lies.

Let me finish by saying what I find cool about the Riemann Hypothesis. So what if some weird function might have a consistency with its x-intercepts? But there is a very practical and interesting connection.

You may notice how with the prime numbers, there really isn't any relation between them. I mean, the Fibonacci numbers are the sum of the two before it, the powers of two are one more than the sum of all the ones before it, the triangulars are the sum of the natural numbers, the squares are the sum of the odd natural numbers, but the primes have no relation like that. I have always wondered why that is, or if there was one.

If the Riemann Hypothesis gets solved, it will shine a light on the distribution of prime numbers. We will be able to see if they do have a pattern or if there is no pattern. Some might find the technological, algebraic, or financial parts of this problem interesting, but I think the practical aspect is really cool.

Turning Numbers Rational

Today, I wanted to do something that I haven't done in a while. I wanted to prove something, but with no patterns involved. Just take a question and prove the answer.

What I want to prove is that an irrational number to an irrational power can be rational. Like, could you raise π to a power and get a whole number, or a simple fraction?

π is a very hard number to work with since there is not a clean way of deriving it (click here to see how you can derive it), so let's take a different irrational number and use that for the example. Say, the square root of 2. We proved it irrational in November of 2011 (click here for that), so we can use that for this example.

Let's take (√2)^(√2). What does that equal?

Most people, myself included, would just say that they don't know. That is a good answer. Let's stick to two possibilities.

1. It is a rational number (it can be written as a fraction, so it is a terminating or repeating decimal)
2. It is an irrational number (it cannot be written as a fraction)

If it is a rational number, we are done. We proved that an irrational number to an irrational power is rational. That would be easy!

What do we do if it's irrational? Since we know it's irrational, we can use it as the irrational number being raised to the irrational power. Let's just see what happens if we use [(√2)^(√2)] as the base and (√2) as the exponent.

[(√2)^(√2)]^(√2)

The law of exponents says that (a^b)^c = a^bc, so we can use that to our advantage.

[(√2)^(√2)]^(√2)
(√2)^(√2)(√2)
(√2)^2
2

So, we ended up with a rational number. This means that if it were irrational, we have proven the original statement true as well. For either of the two possibilities, we have a proof.

How to Divide Any Number by 91

Since this is the last post of 2012, I thought I'd finish the year with a mental math technique. Before I do that though, I want to show you a pattern that is closely related.

Take any three digit number. For this, I will use 123.

123

Multiply that by 13. You should get 1599.

Now, multiply that by 11. You should get 17589. You might have done that in your head using the multiplication by eleven trick I taught last October.

Now, multiply that by 7. This will give you 123123. And if you'll notice, we started with the number 123 and finished with two 123s.

Why did this happen? It is very simple, and just relies on the fact that 13 x 11 x 7 is 1001. So, by multiplying by these seemingly random numbers is really multiplying by 1001. And any number times 1001 is just itself repeated twice.

Division by 1001 results in a similar answer. For instance, 123 ÷ 1001 = 0.122877122877...

If you notice, the first three digits are 122, which is just 123 - 1. The next three are 877, and 122 + 877 = 999.

This pattern continues as well. I am not sure how to prove that, but please comment if you do know.

So, to divide by 1001, you just subtract one for the first three digits, and subtract the first three digits from 999 for the next three digits.

However, telling someone you can divide any number by 1001 doesn't sound that impressive. Since 1001 is right next to 1000, people will get very suspicious.

That is why I showed you the multiplication pattern. Dividing by 1001 is basically dividing by 13, then 7, then 11.

123 ÷ 1001 = 123 ÷ (13 x 7 x 11)

So, instead of going all the way to 1001, let's just get part way there by dividing by the 13 x 7, which is 91. We will use 33 as the number we are dividing by, or the dividend.

33
91

Our goal is to make it a number divided by 1001, since we know how to do that. That means that the first step is multiplying the 91 by eleven. But, if we multiply the denominator by eleven, we must also multiply the numerator by eleven.

33 x 11
91 x 11

We know 91 x 11 is 1001, so we have the problem we are looking for. What is 33 x 11 though? If you go to the multiplication by eleven post, you will see that it is just 363 (add the 3 + 3, and stick it in the middle). This gives us the problem 363 ÷ 1001.

363 - 1 is 362 and 999 - 362 is 637, so the answer is 0.362637362637...

So, to divide by 91, you just multiply the number by 11, subtract one, and subtract that from 999 to get the answer.

Though it is a lot tougher, you can also divide by 77 (11 x 7) and 143 (13 x 11) with the same principle. You are just multiplying the top number by 13 or 7 instead of the easy 11.

The 1001 pattern is really cool, and something you can turn into a really cool trick. The multiplication by eleven is also a really cool effect. But putting them together into the division by 91 is a really impressive feat of mental math that will fool all of your friends.