Happy End Problem

As this is the last post of the year, I thought it would be appropriate to end with a post on the "Happy End Problem." I will explain the problem, which is pretty cool in itself, and then talk a little bit about the history behind it.

The initial question was if any five points were placed on a plane with no three of them in a straight line, will four of those points always form a convex quadrilateral? For example, in the following image:

The following convex quadrilateral can be formed:

Will this always work? As usual, I encourage you to grab a scrap piece of paper and try out a few examples. Have fun with it. Get creative! You will end up finding that no matter how you position the five points, you cannot get a combination without a convex quadrilateral.

Why is this true? In fact, there is a very easy way to prove it. Let's analyze three cases.

The first case is the top left one in the red, where the five points form a convex pentagon. In this instance, connecting any four of the points will form a convex quadrilateral by nature.

The second case is the top right one in the blue, where one point is located in between the four outside points. The illustration shows the inside point being included in the quadrilateral, but it could have just as easily been made as just the four outside points. This will continue to work for any combination of this nature by logic.

The third case is the bottom one in the yellow, where two points are enclosed in a triangle. When you draw a line between the two center points, two of the outside points will end up on one side and one will be on the other. Using the two outside points as your third and fourth vertices will form a quadrilateral without flaw.

Now, you might be wondering if one can prove a similar case with a convex pentagon. Could it be done with six? Seven? Eight? Turns out, nine points are required for it to work every time. As you can see below, eight points is just one too few.

What about convex hexagons? Or heptagons? Or octagons? Or chiliagons (1000-sided polygons)? Well, what many mathematicians will do from here is look for a formula to figure out how many points are required for a given n-gon. We know that for a triangle (n = 3), just 3 points are needed (all triangles are convex). For a quadrilateral (n = 4), we proved that 5 points are needed. For a pentagon (n = 5), I mentioned that 9 points are needed. Do you see the pattern?

3, 5, 9, ...

It is not easy to spot at first, but what if I subtract one from each of those terms:

2, 4, 8, ...

They are all now powers of two! This pattern seems to fit the formula A_n = 2^n-2 + 1. Plugging six in for n would give:

A₆ = 2^6-2 + 1
A₆ = 16 + 1
A₆ = 17

This formula predicts that seventeen points would be required for a hexagon. Mathematicians would then work to try to prove that this is the case. Further, they would try to prove that for any value of n, the A_n formula holds true.

George Szekeres (1911-2005; a Hungarian-Austrailian mathematician and analytical chemist) and Esther Klein (1910-2005, another Hungarian-Austrailian mathematician) worked together to prove that all values of n will have a finite A_n output (there will be a number of points that creates the ability for a convex n-gon to be formed), but they could not get this bound down to the formula above. Soon after this proof was published, Szekeres and Klein married each other, which inspired the name "Happy End Problem."

Paul Erdös (1913-1996; a Hungarian mathematician), possibly one of the most influential of the twentieth century), was able to prove successfully that with 71 points, a hexagon can always be drawn.

Sixty years later, Ronald Graham (born 1935; a Californian mathematician) and his wife, Fan Chung, decided to take a swing at the problem. While on a plane ride to a math conference in New Zealand, they were able to lower Erdös's bound to 70 points, which doesn't sound like much, but it brought the problem back into the minds of mathematicians. It was also ironic that another achievement pertaining to the Happy End Problem was from a couple.

Daniel Kleitman (born 1934; an applied mathematician at MIT) and Lior Pachter (born 1972; an Israeli mathematician and molecular biologist at Berkeley College) worked together to lower the upper bound to 65 points. The number was then lowered to 37 points, and has yet to be lowered further.

Although lots of progress has been made on this problem, the overarching proof still has not yet been found. There has been no counterexample to the A_n formula, and there has certainly been no guaranteed formula to generate the future values. People often wonder what a mathematician actually does for his/her job. A big part of it is trying to figure out the answers to these unsolved problems, which can often be understood by the average person. Try playing around with it and you might make a discovery too.

Figure Out The Number Of Digits In Gigantic Numbers

In science and math, you often run across numbers that are too big to be written in standard form. They are usually written in scientific notation, but they are also sometimes written as a number to a certain power. For instance, one might say that there are 2²⁰ outcomes of the flipping of twenty coins rather than saying 1.049x10⁶ ways.

By using that power, your information is likely more accurate. However, this power does not tell you much about the number. Most of us would have no idea if 2²⁰ is in the thousands, millions, billions, etc. at the first glance.

First, lets ask a question. What is the common logarithm of a number, or what can you gather from it? Well, the common logarithm is the power that ten has to be raised to to obtain that number. For instance:

log(100) = 2

log(5000) = 3.69897

log(6283185) = 6.79818

What do you notice about these numbers? It's not clear at first, but count the number of digits in each of the inputs. You will find that the common log is always just a little bit below that number. In fact, to figure out the number of digits in a number, all you have to do is take the common log and round up to the nearest integer.

How can this be used to find the number of digits in a power? Interestingly enough, there is a logarithmic identity stating that the log of a number raised to the power is equal to the power times the log of the number. For example,

log(27) = 3log(3)

ln(32) = 5ln(2)

log(2²⁰) = 20log(2)

Look at the last example there. We just simplified the gigantic 2²⁰ to a reasonable looking 20log(2), which is the formula to figure out the number of digits it has. In other words, the number of digits in 2²⁰ is just 20log(2) rounded to the nearest integer. Plugging this into a calculator tells you that the log is 6.0206, meaning that there are seven digits in the number. If you multiply it out, you will find that 2²⁰ = 1048576, which does indeed have seven digits. 

So whenever a type of problem pops up with a power of this sort, try to determine how many digits it is. Chances are you will gain a much better understanding of the statistic when you perform this quick calculation.

Math in the News: The Influences of Politics

Though mathematics is normally a pretty concrete subject, people's intuition for it is not. Probability and statistics in particular is a very difficult area for us to grasp, as you've seen with the Monty Hall Problem I talked about a few weeks ago.

Here is another example of mathematical aptitude being influenced by an outside source, but this time, it is not just a matter of lack of skill or desire to be correct. It is also influenced sometimes by political views, as Kevin Drum shows in this news article. Check it out!

http://m.motherjones.com/kevin-drum/2013/09/politics-destroys-math-ability

Isosceles Triangle Theorem

The field of geometry is comprised of many different types of questions. Some are construction based, such as “what is the area of this circle?” On the other hand, many are proof based, like “why are these two triangles congruent?” This is slightly different from the proofs I normally discuss; proofs I normally post are very generalized theorems while these questions are more similar to a specific algebra or arithmetic problem.

When writing a geometric proof, many different theorems come into play. One must use the generalized theorems, properties, and postulates to arrive at a conclusion. Let’s try a simple proof just to demonstrate the nature of these theorems. Let’s prove that triangle ABE is congruent to triangle CDE.

First, we would say that it is given that segment AC is parallel to segment BD and that segment AB is parallel to segment CD. We would then say that ABCD is a parallelogram by the definition of a parallelogram (which requires two sets of parallel sides). The definition of a parallelogram also requires AB and CD to be congruent. The vertical angle theorem can be used to say that angle AEB is congruent to angle CED. The alternate interior angle theorem says that angle ABE is congruent to angle DCE. The angle-angle-side triangle congruence postulate then concludes that triangle ABE is congruent to triangle CDE.

As you can see, there are many steps in this proof, and each one uses a different definition, theorem, or postulate. In high school geometry classes, students are told these theorems and postulates, and expected to memorize them for future examples. This makes geometry boring and pointless, when it can be quite fascinating. A way to easily spice up geometric proofs is to actually prove the theorems before they are used in class. If Euclid could do it, then we can do it.

A fun one to prove is the Isosceles Triangle Theorem. This theorem states that when a triangle has two congruent sides, it also has two congruent angles. This can be proven in a similar way as the congruent triangle question I posed earlier. Take an isosceles triangle:

If you were to bisect that top angle, it would create two new triangles. Since the original triangle is isosceles, it is given that the top left segment is congruent to the top right segment. The definition of a bisection (cutting an angle in half) states that the left part of the top angle is congruent to the right part of the top angle. The reflexive property of congruence states that the middle segment is congruent to itself. By the side-angle-side triangle congruence postulate, the left triangle is congruent to the right triangle. And finally, by CPCTC (common parts of congruent triangles are congruent), the bottom left angle is congruent to the bottom right angle.

This sort of geometric proof language sounds extremely long and boring. However, finding uses for it such as proving the Isosceles Triangle Theorem can make it a little more fun. Among many things, I think that schools should teach the reasons behind these theorems to make it more logical and fun to apply them to class.

How YOU Can Memorize 2000 Digits of a Number

Since it is two days after Thanksgiving, and many of us are probably eating leftover pie, I thought it would be appropriate to do a post somewhat relevant to the numerical pi. And I could do something very mathematical, but I just finished my first trimester at Phillips Academy Andover last week, and I needed a break after my nearly impossible MATH-380 final exam. As a result, I thought it could be fun to talk about memorizing numbers, pi and tau in particular.

When people hear about my Tau 2000 event, they often ask me if I have what they call a "photographic memory." This is not at all true. I don't even think the types of photographic memories advertised in pop culture really exist (I'm not an expert on neurology, so for more on that, I'd recommend reading this Scientific American article). The way I memorized 2012 digits of tau was all learned and practiced techniques, similar to my mental math presentations. I was not born with some gift or natural talent, it was just learning the methodology and practicing until I could do it quickly. Just like anyone can do mental math, anyone can be a memory expert as well, ranging from being able to remember 57890 digits of pi to being able to remember your car keys as you leave for work. There are techniques for it all.

First, let me introduce you to the Major System. This is a phonetic code that enables you to turn numbers into words. You store them as words, and later retrieve them as numbers. Basically, each digit is associated with a specific consonant sound.

1 is the t or d sound. It can also be either of the th sounds (see note below).

2 is the n sound.

3 is the m sound.

4 is the r sound.

5 is the l sound.

6 is the j, ch, sh, or zh sound.

7 is the k or g sound.

8 is the f or v sound.

9 is the p or b sound.

0 is the z or s sound.

Note: th (both the th in "that" and the th in "thing") is normally paired with 1, but there are other variations on the system that will put it with 8 or not include it.

This looks hard to memorize on its own, but it is actually not that hard. Here are some mnemonics that can help you.

1. A t or d has 1 downstroke.
2. A n has 2 downstrokes.
3. A m has 3 downstrokes.
4. The number 4 ends in the letter r.
5. If you hold up your hand with 4 fingers up and your thumb at a 90° angle, you will see 5 fingers shaped like an L.
6. A J looks somewhat like a backwards 6.
7. A K can be drawn with two 7s back to back.
8. A lowercase f in cursive looks like an 8.
9. The number 9 is a backwards p or an upside-down b.
10. The word zero begins with the letter z.

You will also notice that the consonants that were paired together sound very similar. Your lip movement and tongue placement are the same in any of the consonant sounds chosen for a number (except for the th sounds, hence the inconsistency of its use).

You might be wondering why there are no vowel sounds on the list. There is also no h, w, or y sound. This is because you can insert these wherever you want between consonants and they mean nothing. With all of this in mind, you can begin turning numbers into words. Let's take the number 15. What words can this become?

Well, one is the t or d sound. Five is the l sound. Insert vowels, and you can get doll. Or tile. Or tail. You can also insert vowels at the beginning or end of the word and make deli, or Adele. You can also insert hs, ws, and ys to get hotel, towel, or yodel. Here are a list of the 66 words that can be made out of  the number 15 (I put the ones that I might use in a mnemonic image in bold print):

Addle, daily, dale, dally, deal, delay, dell, dial, dole, doll, dual, duel, dull, duly, dwell, ethyl, hastily, hostile, hotel, hotly, huddle, ideal, ideally, idle, idly, idol, it'll, italy, oddly, othello, outlaw, outlay, saddle, sadly, seattle, settle, societal, stale, stall, steal, steel, still, stole, stool, style, subtle, subtly, suicidal, sweetly, tail, tale, tall, tally, teal, tel, tell, they'll, tile, till, toil, toll, tool, towel, waddle, widely, yodel

Note that some of the words start with s. Since s is zero, this is referring to the number 015, which is normally still 15. These words do not work if 15 is part of a string of other digits such as in pi or tau.

The ones that I bolded are all nouns that you can create a mental image of in your head. As the Scientific American article that I linked to states, people naturally have a better memory for visuals (the reason why you might remember someone's face, but not be able to place the name). So, you might not be able to remember the number 15, but you can probably picture a doll, or a hotel, or a yodel (for this, I would think of the chocolate pastry, not the verb). If you are trying to remember that it is someone's address or apartment number, picture a relationship between the object and the person. Maybe the person is standing up on a stool shouting to a crowd of confused, awestricken people, or they are on the couch stuffing their face with yodels. The sillier your image, the easier it is to remember.

There are lots of memory experts who will create a list of "peg words," which are essentially 100 words that they will refer to when they are trying to remember a number between 1 and 100. It is certainly not a necessity, but it can often help if you are trying to come up with a word on the fly. Every person has a different list of words that works for them, so this is something that I would encourage you to make on your own. The website www.phoneticmnemonic.com works very well to help create this list.

To memorize shorter strings of digits (something like memorizing 100 digits of pi), the best approach in my opinion is to create sentences out of your words. For instance, take the first five digits of pi: 31415. The only word that can be formed out of this is moderately, which isn't a great start to a sentence. However, it could be turned into "my turtle" or "Madrid law" or "Mother Yodel." The first 24 digits of pi create the sentence:

My turtle Pancho will, my love, pick up my new mover, Ginger.

Say this a few times and you will sadly have it memorized. And since you now know the code, you now have the first 24 digits of pi memorized. If you want to keep going, the next 17 digits are:

My movie monkey plays in a favorite bucket.

The next 19 are:

Ship my puppy Michael to Sullivan's backrubber.

If you want to take it to 100 digits, you can use:

A really open music video cheers Jenny F. Jones.

And my personal favorite:

Have a baby fish knife so Marvin will marinate the goosechick.

This method works great for condensing large quantities of numbers into a small amount of silly, memorable sentences. However, once you get up towards 300, 400, 500 digits, it is really tough to remember the exact prepositions and linking verbs you used, which contribute to the digits. Because of this, the method I used for memorizing 2012 digits of tau is a different variation. Rather than just memorizing plain sentences, I used a technique called the memory palace.

A memory palace is essentially a place that you can mentally visualize that you put the images that you create in. For instance, your drive from your house to work might be a memory palace. Your elementary school campus could be your memory palace. You can even create an imaginary place to be your memory palace. Let's pretend your memory palace is inside of your house. The first ten loci (places to put the images) might be:

1. Your bed (in your bedroom)
2. Your closet
3. Bathroom
4. Hallway
5. Other Bedroom
6. Stairs
7. Living Room
8. Dining Room
9. Kitchen
10. Front Porch

And you might have a grocery list with the following items:

• Grapes
• Carrots
• Corn on the Cob
• Yogurt
• Cheddar Cheese
• Marshmallows
• Cheetos
• Salt
• Pepper
• Ice

All you need to do is mentally "put" each of these items into the corresponding locus in your memory palace. For instance, the first item is grapes. You would put the grapes on your bed. But you wouldn't just put them there, you must do something to make the image stand out. First of all, you must embrace the image. Not only do you see grapes, but you smell the grapes, you taste the grapes. The more of your senses that you alert, the easier the image is to remember. The image also needs to be less dull than just a few grapes sitting on your blanket. Maybe have grapevines growing out of the back of your bed. Maybe visualize the grapes to have legs, and jumping on the bed. As long as it is a silly image that stands out in your mind, you will be able to remember it.

The next item on the list is carrots. The corresponding locus is your closet. Carrots grow out of the ground, so maybe you picture all of the mud that your sneakers have tracked into the closet has carrots growing in it. As long as you pull a carrot out of the mud, you will remember it is carrots. Or maybe there is a snowman inside with a carrot nose, or a carrot shoe-horn. The actual carrot aspect of the image can absolutely be subtle, as long as you can remember the image and this image triggers the thought of carrots in your mind.

Continue through the list, and you will have ten images in your head that will in fact be stuck there until you use other techniques to remove them (yes, there are techniques people use to forget things). Try this out a few times, and I'm sure you will find it very useful. If you have a list of things to do at work, you need to remember when to pick up your kids and bring them to their activities (you may even use the major system for translating times into words - if you need to bring your son to baseball practice at 4:15, you may just picture your son swinging his bat at a "hurdle" (r=4, d=1, l=5) in the appropriate locus), or anything else, the memory palace is a great way to go.

How does this help one memorize the digits of a number, like tau? Well, what the major system does is turns numbers into words, which can then be turned into images. The memory palace then acts as a place holder for those images. For instance, take the digits of tau:

6.28318530717958647692528676655900598...

The first two digits are 62. What words can this form? You can say chain, gin, maybe you know someone named Jane or John. I ended up choosing the word ocean.

The next three digits are 831. This forms the word vomit. Yes, it is disgusting, but it is a word that will create a memorable image.

The next two digits are 85. From this, we can create the word waffle. So the first image will be "an ocean vomiting a waffle." It sounds very silly, but it will be memorable. The smell of the saltwater, the taste of the waffles, the sound of the ocean waves crashing. This all will go into your first locus. My memory palace for tau was my middle school campus, so I remembered this image in the back parking lot of the school.

The next image is comprised of the digits 30717958. This can be turned into "a mask tugging on a bailiff." This was put inside of a staff room that the back parking lot has a door to. It is a very weird image, but still memorable. Picture the bailiff really struggling to get away from this mask, while still fearfully reciting his lines: do you solemnly swear to tell the truth, the whole truth, and nothing but the truth. Make yourself feel scared of this moving mask, and sympathize with the bailiff. The more you relate to and embrace the image, the more memorable it will be. Especially when you are memorizing 2012 digits of tau (which took me 272 images), you need each image to be extremely vivid.

To retrieve the numbers from this memory palace, all you do is go back to the image, find the subject, root verb, and object of it, and translate the consonants back to numbers with the major system. With practice, this becomes easier and easier to do. I strongly recommend practicing at least memorizing grocery lists and to-do lists with the memory palace, and if you want to take it further, learn to convert numbers to words with the major system for more advanced lists and situations. Maybe even memorize your family and friends' phone numbers with the major system and memory palace. These are all great exercises for your mind, and will definitely give you a better memory.

The History of the Monty Hall Problem

One of my very first blog posts was about the Monty Hall Problem. This is an extremely classic example of a probability paradox. Let me quickly describe the problem:

Pretend you are on a game show, and the host gives you three doors to select from. One of these doors has a car behind it, while the other two have goats. Let's say you select door number one. Then, the host (who knows where the car is) opens another door to reveal a goat. Let's say he opens door number three. You are then given the option to either stick to door one or switch to door two. Does either strategy have an advantage?

The common answer would be that it is 50-50, and there is no advantage either way. However, the correct answer is that there is only a 1/3 chance of winning by staying put, and a 2/3 chance of winning by switching. Click here to learn why.

This problem was first posed by Steve Selvin, but it was popularized by Marilyn vos Savant in 1990. Vos Savant is famous for once having the highest IQ in the world, as well as her "Ask Marilyn" column in Parade Magazine.

One week, her column was about the Monty Hall Problem. She posed the question, and then explained her reasoning as to why there is a 2:1 advantage for switching. This created a pandemonium of angry readers who insisted that she was incorrect, and furthermore, accused her of adding to the problem of innumeracy and lack of mathematical intuition in America. Some of these complaints came from a statistician at the National Institutes of Health, the deputy director of the Center for Defense Information, and professors at George Mason University, University of Florida, University of Michigan, Millikin University, Georgetown University, Dickinson State University, Western State College, and more. Even the legendary Paul Erdős couldn't wrap his brain around the paradox.

This problem has continued to baffle everyone it encounters, from average people to accomplished mathematicians. In 2010, Walter Herbranson and Julia Schroeder of Whitman College performed an experiment to see if playing the game multiple times could end up refining the player's strategy. The human test subjects failed to revert to the optimal strategy and switch doors in the experiment. However, when the test was performed on pigeons, with mixed grain as the prize, they were able to pick up on the fact that switching doors gave them the best chance of success. The fact that a pigeon can do better than a human in this situation is fascinating to me.

The Monty Hall Paradox is something that reminds us of how humans are not wired to understand probability and statistics. This is why people can be fooled by mathematical scams and why casinos are packed full of gamblers. If our math curricula put an equal focus on probability and statistics as it did on algebra and calculus, then our world would have much better math minds and critical thinkers in general.

The Mathematics of Ghost

A couple months ago, we were talking about various mathematical games where player two can very easily force a win, such as Chomp and Anti Tic-Tac-Toe. Those games were pretty easy to spot the unfairness, but this is one that is actually commonly played by everyday people. You might even play it yourself. The game is called Ghost.

Ghost is definitely not a scary game, despite the name. In this game, player one starts by saying a letter. Player two must respond with another letter, and they keep taking turns saying letters and forming a word. The goal is to not create a real word. However, you can't call out letters that don't allow a word to be made either. For instance:

Player 1: T

Player 2: O

Player 1: U

Player 2: C

Player 1: A

Player 2: Umm... L

Player 1: I challenge.

Player 2: I didn't have a word in mind.

Player 1 = Winner

If your opponent challenges you, you need to say a word that can be made out of the letters to win. If you can't think of a word, then you lose the round. A better strategy for player 2 in this game might be:

Player 1: T

Player 2: W

Player 1: E

Player 2: A

Player 1: K

Player 2 = Winner

This game seems pretty fair, right? If you play this game with a friend, you will find that the wins are pretty even (unless one of you has a much better vocabulary). However, it is possible for player 2 to force a win.

When all of the words in Scrabble Player's Dictionary are in play, every letter of the alphabet can be followed with another letter that forces an odd lettered word (assuming that player 2 continues perfect play). However, this dictionary has a lot of obscure words in it that might get challenged, and the average player will not know or use.

By keeping it to words that are known to the average player, player 1 does have a chance. If they play H, J, M, or Z, they can force a win if player 2 does not have the dictionary memorized. Here are the words to remember for player 1:

H: Hazard, Haze, Hazily, Hazy, Heterosexual, Hiatus, Hock, Huckster, Hybrid

J: Jazz, Jest, Jilt, Jowl, Just

M: Maverick, Meow, Mizzen, Mnemonic, Mozzarella, Muzzle, Muzzling, Myth

Z: Zaniness, Zany, Zenith, Zigzag, Zombie, Zucchini, Zwieback, Zygote

For any other letter, player 2 can still force a win. Once you know the correct followup letter to use, it is a pretty simple game. Here is the strategy:

Letter	Followup	Possible Words
A	O	Aorta
B	L	Black, Blemish, Blimp, Bloat, Blubber
C	R	Craft, Crepe, Crept, Crick, Crozier, Crucial, Cry
D	W	Dwarf, Dwarves, Dweeb, Dwindle, Dwindling
E	W	Ewe
F	J	Fjord
G	H	Ghastliness, Ghastly, Gherkin, Ghost
I	L	Ilk, Ill
K	H	Khaki
L	L	Llama
N	Y	Nylon, Nymph
O	Z	Ozone
P	N	Pneumonia
Q	U	Quaff, Quest, Quibble, Quibbling, Quondam
R	Y	Rye
S	Q	Squeamish, Squeeze, Squeezing, Squelch
T	W	Twang, Tweak, Twice, Two
U	V	Uvula
V	U	Vulva
W	H	Whack, Where, Whiff, Who, Why
X	Y	Xylem
Y	I	Yield, Yip

By memorizing this list, you can become a perfect Ghost player, and win every time, despite the seemingly fair nature of this game. The other two games were both fun, but with this one, you can really start to impress people with your ability to win.

Math in the News: Teacher Salaries

One of the biggest issues in math education is teacher quality. I have discussed this in both of my TEDx talks about this topic.

We explained it the best we could in our Capstone Research Paper (link is at the top of the page), but I think this New York Times article that just came out describes it fantastically. So, I couldn't resist posting it here.

http://www.nytimes.com/2011/05/01/opinion/01eggers.html?_r=0

Enjoy!

Logarithmic Proofs and Identities

I have talked about logarithms quite a bit on this blog, but they were always being applied to something else, whether it be Benford's Law, the Law of 72, or some other practical use. This week, I would like to show that logarithms are very much a part of pure mathematics as well. Of course they are in mathematical equations just as much as exponents and radicals, but they have some pretty cool features of their own.

First off, let me review what a logarithm is. I have explained it before, but once you understand the notation, you shouldn't need to have done Precalculus to understand this post. They are very easy to understand.

For instance, we know that 10² is 100.

10² = 100

If we take the logarithm of both sides, we are essentially bringing the two out of the exponent. Rather than doing an operation on the 10, we do an operation on the 100 to determine what the exponent is.

log(10²) = log(100)
2 = log(100)

In most situations, it is clear what type of logarithm you are using, especially in this one because ten is a common logarithm to use. However, many people will write a subscript to clarify. For instance:

log₁₀(100) = 2

Logarithms become extremely useful when you need to solve an algebraic equation where the variable is in the exponent. For example:

2^x = 64

As you know, algebra is about doing the inverse operation. If there is addition going on, you subtract. If there is multiplication going on, you divide. Similarly, if there is exponentiation going on, you use a logarithm. In this instance, it would be taking the log₂ of both sides.

c2^x) = log₂(64)
x = log₂(64)
x = 6

In practice, there are three bases that are extremely popular to use in a logarithm. We just used two of them: log₁₀ and log₂, which are also known as the common logarithm and the binary logarithm. The third one is log_e (using the number e that is described here), which is called the natural logarithm, or the natural log. This one is found on most calculators, usually next to the common logarithm.

Though logarithms are not a part of most people's day-to-day life, they do have lots of practical applications. The common logarithm is the basis of the pH system which describes the acidity of water. The natural log is a huge aspect of finance and compound interest (as we saw with the Law of 72). And of course, they are all over nature.

Let's look at an identity of logarithms. Take the following problem:

log₆(24) + log₆(9) =

If you just used a calculator to do this, you would get:

log₆(24) + log₆(9)
1.773705614 + 1.226294386
3

That's odd. Two random numbers happened to have logarithms that summed to three. Let's look closer at this and see if we can figure out why. What number could you find the log₆ of and get 3?

log₆(n) = 3

First of all, let me point out that we just asked an algebraic question. We suddenly got curious about why something happened, so we asked a "what" question, which calls for an unknown quantity, which later becomes an algebraic variable. So when algebra seems like a drag, remember that it is all techniques for answering that "what" question. And "what" is a question asked in all branches of mathematics, science, and engineering.

Anyways, for this equation, we would want to do the inverse operation. We turn both sides into the exponent, and create a base of 6. This gives:

6^log₆(n) = 6³

The left hand side cancels, leaving just n. The right hand side is six cubed, which is 216. So, we end up with:

n = 216

So, this means that the log₆ of 216 makes you end up with 3, or the sum of log₆(24) and log₆(9).

log₆(24) + log₆(9) = log₆(216)

What is the relationship between these three numbers? Well, it shouldn't take to long to determine that 24 x 9 = 216, or:

log₆(24) + log₆(9) = log₆(24 • 9)

In other words, the sum of the logarithms is the logarithm of the product. Wow! That's pretty cool! Is that always the case? Well, let's try to prove that it is for all logarithms.

Let's set a few terms equal to each other and see what happens. Since there are logarithms, we will need a lot of variables.

x = log_a(p)
y = log_a(q)

In other words:

p = a^x
q = a^y

Let's multiply those two equations together. Since they are both equal, multiplying the terms on each side by each other won't make a difference.

p • q = a^x • a^y
pq = a^x+y

The right hand side was simplified using the Law of Exponents, which is explained very well here.

Now, we must take the logarithm of both sides, or specifically, the log_a of both sides.

log_a(pq) = log_a(a^x+y)
log_a(pq) = x + y

But what were x and y? We defined them in terms of a, p, and q earlier. So, let's substitute those values in and see what we get.

log_a(pq) = log_a(p) + log_a(q)

And this creates the identity that we were trying to prove: the sum of the logarithms is the logarithm of the product, and thus, completes our proof. There are other logarithmic identities like this one, but I will save that for another post.

History of Math: Isaac Newton and the Schoolyard Bully

When talking about great mathematicians of the past, many will rank the top three as Archimedes, Gauss, and Newton. I have posted stories about the others, but none about Isaac Newton. So, I think now is a good time.

Isaac Newton

Newton is best known for his work in physics, but he also made huge contributions to calculus, algebra, geometry, and infinite series. Many mathematicians expand their expertise to different diverse branches of math, but Newton stuck to the things that applied the most to his physics and are currently ruling the American school system.

Let me tell you an interesting story about Newton. When he was a young student, he was very shy and not at all the genius that he is known as today. One day at recess, a bully came up to him and punched him in the stomach. Newton chose to fight back, and proceeded to shove his face in the mud. All of his classmates, who did not like this kid, cheered him on as he proved his superiority to the bully.

After this incident, he decided that physical prestige wasn't enough for him, and he wanted mental prestige as well. So, he started working much harder at his schoolwork, and soon after became top of the class, proving to everyone that he was smarter than the bully as well. This motivation could have been what turned him into one of the best scientists and mathematicians of all time.

I think this story shows that anyone who has drive and dedication can become a genius, and it also is a story themed around the negativity of bullying. I also like it because it is an interesting aspect about a mathematician's childhood, which help people get to know who is behind what they are learning and practicing.

A Quick Way to Check Your Work

In school, the teacher is always on top of you for checking your work. When you do a subtraction problem, solve the reversed addition problem and make sure it is right, when you do an algebra problem, make sure you plug your solution back into the original equation. These are all things that are drilled into our heads, but never quite executed.

I did post a year and a half ago about checking your work in algebra problems: plugging the answer into the original equation (click here to see how to do that). But there is also a shortcut for checking work on plain arithmetic problems as well.

Let's take the problem 138 + 253. I would have went smaller, but the method will be easier to demonstrate with larger numbers.

   138
+ 253

If we add that up normally, we would get:


   138
+ 253
   391

How do we know if that is correct? Well, we do something called mod sums. What that means is we add up the digits in the number, and then add up the digits in this sum, and keep going until we find a single digit number. This is called the number's mod sum or digital root.

So, what is the mod sum of 138? Well, we add up the digits.

1 + 3 + 8 = 12

1 + 2 = 3

So, the mod sum or digital root of 138 is 3. Let's find it for 253.

2 + 5 + 3 = 10

1 + 0 = 1

The mod sum of 253 is therefore 1. Let's find the mod sum of the total and see if you notice the pattern.

3 + 9 + 1 = 13

1 + 3 = 4

So, the two addends have mod sums of 3 and 1. The sum has a mod sum of 4. What is the pattern? That's right, the mod sum of the answer is the sum of the mod sums of the addends. What about a subtraction problem?

  924
- 643

The answer to this problem is 281. But how do we confirm it?

The mod sum of 924 is 6 (9+2+4=15 and 1+5=6) and the mod sum of 643 is 4 (6+4+3=13 and 1+3=4). So, the mod sum of the difference must be the difference of the two mod sums. The mod sum of 281 is 2 (2+8+1=11 and 1+1=2), which is the difference of 6 and 4. So, the answer was correct.

What about a multiplication problem? Say 71 x 55. If you do the math, you will find that the answer is 3905. But let's check it with mod sums.

Mod Sum of 71 = 8
Mod Sum of 55 = 1
Mod Sum of 3905 = 8

8 x 1 = 8

So it is correct. There are some glitches in the technique, but this is the basis of it. You might run into scenarios that I didn't quite explain how to deal with, but feel free to comment. I will be happy to respond with some more specific pointers. Have fun actually checking your work now!

Math in the News: Rota's Conjecture is Solved

One of the things that lots of people seem to be oblivious to is that mathematics is developing and innovating just as much as any other discipline, which I allude to in many of my presentations. There are many conjectures, or unsolved problems, out there that mathematicians are working on and trying to prove or solve.

Rota's Conjecture was a problem like this, in the branch of matroid theory. This is a diverse area of mathematics that isn't taught or mentioned in the American school system (another concept I allude to in my presentations). So when I read this article about Geoff Whittle solving the problem, I thought it would make for a great post. Here is the story:

http://phys.org/news/2013-08-rota-conjecture-year-old-math-problem.html

Chomp: A Proof in Game Theory

Game theory and proofs are two of my favorite areas of mathematics; game theory is practical and fun while proofs are interesting and insightful. So, when I learned about this problem that combines the two, I thought that it was definitely worth a post.

This game is called Chomp. It is normally played with just a table of squares, but I find it easier to understand by thinking of a chocolate bar.

The mouth-watering Chomp playing board

Chomp is played where the first player chooses a square on the board, and then takes away everything above and to the right of it (essentially taking a bite out of the top right corner of the chocolate bar). The second player would do the same thing with another remaining square. This process keeps continuing until all that remains is the bottom left square. Whoever is forced to take that square loses.

To better understand how the game works, click here to practice playing it. You will see how easy it is to play and understand.

At this point, any game theorist would be wondering if there is an optimal strategy for this game. From what we saw a couple weeks ago with Anti Tic-Tac-Toe, you might be wondering if symmetry is involved in this game. And yes, you can win this game by playing symmetrical moves in the end game. However, the board is not square, it is a rectangle. So, there cannot be full symmetry.

I do not know what the actual optimal strategy is. But, I do know that one exists that would enable player one to always force a win. I will demonstrate this by an "existence proof" where you prove it exists without finding the actual thing.

Pretend player one just took the top right corner square. This is either a good position or a bad position. If it is a good position, then by definition, player one can continue to play perfectly and force a win. If it is a bad position, then player two must have a responding move that will force them to win.

But, this responding move must be a square that player one could have hit on their first move. Since the top right square really doesn't have an effect on the rest of the board, this would not be a problem. So, player one could have played this strategy, which would allow them to force a win as well.

In either of these situations, player one wins. So, there is our proof. I find these existence proofs really interesting because you don't always think you can know if a statement is true without being able to see an example, but with mathematics, it can be done.

Carl Friedrich Gauss: The Child Prodigy

One of the most famous mathematicians of all time is the German mathematician Carl Friedrich Gauss (1777 - 1855). Gauss was one of the leading number theorists of all time, as well as a contributor to algebra, statistics, geometry, analysis, and applied mathematics.

There is a piece of mathematical folklore (which may or may not be 100% accurate) that involved a child Gauss. It is a great story, highlights a great point, and shows the intelligence of a great mathematician.

A fifth grade teacher is teaching a class, and started to get frustrated with the students. So, in an attempt to punish them, she demanded that they add up all of the numbers from 1 to 100. This is a daunting task for the average person. She expected to have the students start working on the problem, and she could leave and take a break.

As she was about to walk out the door, the young Gauss raised his hand and declared that the answer is 5050. The teacher was stunned. After checking his work, they found that 5050 was the correct answer.

How did he do it? Well, he visualized a horizontal line with all 100 numbers:

1   2   3   4   5   ...   96   97   98   99   100

And then he took the second half of that line (51 - 100) and flipped it around underneath to look like so:

    1     2     3     4     5   ...   46   47   48   49   50

100   99   98   97   96   ...   55   54   53   52   51

Each of these vertical columns is its own addition problem. And in all fifty columns, the sum is 101. So, the sum of the numbers from one to one-hundred is the same as fifty 101's, or 50 x 101. Since 50 ends in a zero, it is a pretty quick computation: 50 x 101 = 5050. And there is the answer.

I think this is a great story when it comes to historical mathematicians, regardless of how true it is. Gauss did go on to study triangular numbers, which are the sums of consecutive integers up to a point. And since triangular numbers are absolutely fascinating, this story is a great way to begin an endeavor in that topic.

Anti Tic-Tac-Toe

We've all heard of the game tic-tac-toe. It's easy to set up, simple to understand, and fun to play. All you have to do is get three in a row.

There is some mathematics behind correct play of tic-tac-toe, but the real goal is to see where the other player has a chance of forcing a win and make sure that doesn't happen.

However, there is an interesting mathematical fact in the game "anti tic-tac-toe." This is where you have to avoid getting three in a row.

As X, what would be a good starting move in this game? In regular tic-tac-toe, most people start in the center because it is part of 4 of the 8 winning possibilities. So in this instance, it is part of 4 of the 8 losing possibilities. So, I don't think the center would be an appealing move to most players.

Let's start in the left corner square and see what happens.

X

Since O wants to see X get three in a row, they will get as out of the way as possible.

X
		O

Now, where are X's safest moves? Well, if X moves in any of the squares with asterisks, then it would give them two in a row, which would put them closer to a loss. So, X's best move is in the bottom side square.

X	*	*
*	*	O
*	X	*

Now, O would probably go in the top side square to keep out of having 2 in a row.

X	O
		O
	X

Now, X's best move is in the top right corner, since all other squares would have an asterisk.

X	O	X
		O
	X

In this instance, the only technically "safe" move for O is in the bottom left corner, but you can see that this move eliminates 3 possible three in a rows for X. Since the bottom right corner is pretty safe too, O's best move would be there.

X	O	X
		O
	X	O

Now, X is forced to go in an unsafe square. In fact, O can force a win wherever X goes. If X goes in the left side square, O will go in the center. If X goes in the center square, O will go in the left side square. If X goes in the bottom left corner square, O can go in either square. There is no way for X to win.

You might notice that in this game, X has a huge disadvantage. O has one less letter to put down, so it is impossible for X to win against a rational player. However, X can force a tie.

Eight of the squares on the board are a guaranteed loss for X. If X moves there on the first turn, O should be able to succeed. However, this ninth square is a square that X can move to and O will not be able to force three in a row.

Which square is it? It is the one that would never be suspected: the center square. Earlier, we said that the center square has so many losing possibilities that nobody would consider it. But, it is actually the right move. Let's look at it.

	X

Currently, there is no specific square that O would have an advantage in. Let's say they went in the top left corner square.

O
	X

Then, X would play the bottom right corner square. This would be the symmetrical move.

O
	X
		X

X would continue playing the symmetrical move through the whole game. With any logical player, this would result in a draw.

Though anti tic-tac-toe wouldn't be a popular game to play, there are much more fun games that use symmetric properties to force ties/wins like nim or cram. Here is another game with this type of strategy called napkin chess that I learned on the show Scam School which can also be turned into a fun game with friends:

Math in the News: Giving Students More Independence

About a month ago, I was speaking at a TEDx conference that was themed around education. During these TEDx conferences, there are always live speakers as well as videos chosen by the organizers that fit well with the occasion. One of the videos we watched really got us thinking. Here it is:


This talk is not directly mathematical. It isn't meant to be a math education talk. But, these points apply to math education as well.
For example, the mathematician Euclid had nothing. He pretty much was starting from scratch, just like the kids described in the talk. So what did he do? Well, after creating five axioms (foundational ideas that can be concluded with basic logic), he started asking questions, and answering them with mathematical proofs. He would then think about other questions, and find ways to answer them using everything he had discovered. This was the basis of his book series called Elements.
Students should be able to approach math in a similar way. A teacher could lay out five axioms (or develop them with the students), and then back off. He or she might also provide terminology (line, triangle, square, circle, angle, bisect, trisect, etc.), but the students can discover the rest. By learning math this way, they will understand everything they are doing so much better because they created it. For more on that topic, check out my Capstone Research Paper (link is on the top of the page), and scroll to "Chronological Cognition."

Law of Sines

A few months ago, I posted the proof of the Law of Cosines, which is an extremely important aspect of trigonometry. If you do not know what sines and cosines are, it is a very easy concept. Click here to view the post discussing that (which is also the post with the proof of the Law of Cosines).

But this law only works if you are given all three sides of the triangle or two sides and the enclosed angle. What if you are given two sides and a non-enclosed angle, or two angles and one side (three angles isn't enough information to generate side lengths)? How could you approach this problem?

This is where you use the Law of Sines. This law goes as follows:

This was taught to me last year in school, and I immediately wondered what the proof was. Though the law of cosines one was a bit clunky, I found that this proof was quite simple and elegant. So, I thought that it would be great to share.

Since it would require many diagrams, I thought it would be easier to just watch a video of it. It is pretty short, and explains the proof well.

A big part of the reason why most of the cool stuff I post isn't taught in school is that it is not mandated in the curriculum or test standards. Of course, I do believe there are changes that need to be made to these (click here for my Capstone research paper explaining those). However, the Law of Sines is something already taught in school. Same with the Law of Cosines.

These proofs, especially the sine one, fit right into the curriculum. The Law of Sines is already being taught, so why not take an extra 5 minutes to explain the proof? Or even better, explain the basic thought process behind the proof and have the students generate the formula (which works really well for the Quadratic Formula as well). This increases the students' ability to understand and apply the concept, as well as making it fun and interesting. On top of that, the common core standards do want students able to "construct viable arguments," which is the whole purpose of proofs. I think that this proof is not only interesting, but shows that cool math stuff can be integrated into the classroom while keeping it relevant and obedient to the standards.

Don't Borrow Tens Ever Again in a Subtraction Problem!

When the traditional method of subtraction is taught in school, it can often be quite confusing, especially when the number you are subtracting from has lots of zeros. It is a pain to go borrow the tens from other numbers, especially when they are ones or zeros. But, there is an easier way to approach this that isn't taught in class.

In geometry, the word "complimentary" is used to describe the relationship between two angles that sum to 90°. For instance, 36 and 54 are complimentary angles. Similarly, the compliment of 36 would be 54 when talking about geometry.

A geometric application of complimentary angles (and a funny picture)

With arithmetic on the other hand, a number's compliment is its difference from the power of ten above it. So, the compliment of 36 would be 64 (100 - 36 = 64), or the compliment of 473 would be 527 (1000 - 473 = 527). This is more practical for solving arithmetic problems than the distance from 90.

One-hundred or one-thousand are very difficult numbers to subtract from. These are ones where you need to keep turning zeros into nines until you reach that first digit, and then figure out the problem from there. This can be quite the pain. But, there is a shortcut for these examples.

All you have to do is subtract the last digit of the number from ten. This is the last digit of the compliment. Then, subtract all of the rest of the digits from nine, and place them in their corresponding place value. For instance, let's find the compliment of 36.

10 - 6 = 4
9 - 3 = 6

100 - 36 = 64

Pretty easy, right? Let's try it with 473.

10 - 3 = 7
9 - 7 = 2
9 - 4 = 5

1000 - 473 = 527

This can easily be extended to numbers in the thousands, millions, billions, and more as long as you can keep track of the digits. For example, 5647823 would have a compliment of:

10 - 3 = 7
9 - 2 = 7
9 - 8 = 1
9 - 7 = 2
9 - 4 = 5
9 - 6 = 3
9 - 5 = 4

10000000 - 5647823 = 4352177

Couldn't be easier! This method is basically just remembering the fact that no matter what the situation, you will have borrowed a ten from each zero, making them all nines except for the very last one which will remain a ten. Try a few subtraction problems yourself and you will see why it works.

This can be applied to geometry in a way as well. To subtract quickly from ninety, you still subtract the last digit from ten, and the first one from eight (one less than the nine). So, the compliment of 29° is:

10 - 9 = 1
8 - 2 = 6

90° - 29° = 61°

For supplements of angles (the angle's difference from 180), you can do a similar technique as well. Subtract the last digit from ten and the first from seventeen. For instance, the supplement of 48° is:

10 - 8 = 2
17 - 4 = 13

180° - 48° = 132°

In fact, any number ending in zero(s) can be subtracted from just by altering this method. I have found this extremely helpful when performing mental math (three-digit and four-digit squaring requires you to quickly identify how far you are from the nearest hundred/thousand, which often needs compliments). It is also very practical. When you give the cashier a hundred dollar bill, they are usually impressed when you tell them the change before they have time to punch the bill into the register. I'd recommend practicing this technique because it is useful and also quite a bit of fun.

Noam Elkies: Summing the Fourth Powers

A few weekends ago, I went to the MOVES (Mathematics of Various Entertaining Subjects) conference held by the Museum of Mathematics in New York City. I gave a workshop on the Dynamic World of Mathematics, and got to see some very interesting presentations and people.

Meeting Noam Elkies at MOVES

One of the people who I met was Noam Elkies. Noam Elkies is currently a professor at Harvard, and is the youngest full tenured professor in the history of the school.

In 1772, Leonhard Euler conjectured that there was no way to solve the following equation with whole numbers:

a⁴ = b⁴ + c⁴ + d⁴

For over 200 years, mathematicians were trying to prove this conjecture, but to no avail. Finally, in 1986, a twenty-year-old Noam Elkies offered a different type of answer.

In most posts, I will explain how someone proved a conjecture true. But in this instance, Noam Elkies proved the conjecture false. He found values of a, b, c, and d that made the equation true, and thereby showing that the equation is possible.

a = 20,615,673
b = 18,796,760
c = 15,365,649
d = 2,682,440

In addition, he proved that there are an infinite number of solutions to this equation. That completely went against what Euler had thought! This shows that you can never be 100% sure if something is true until you have proof. That philosophy applies to math, science, critical thinking in general, and it is also a great story to show the importance of mathematical proofs.

Ironically, my TEDx talk from this last June just went up a week and a half ago, and I told that exact story in my talk. So, I thought this would be an appropriate post to share the video in.

Time to Double Your Money!

Last year, I did a post about a number called e. This number is around 2.71828, and has many applications to calculus and finance. In the other post, I discussed its application to finding compound interest. Click here to see that post.

You might remember that the formula for finding the amount of money you have when your interest gets compounded continuously is:

Pe^rt

P = amount of money originally deposited
r = interest rate
t = time (years)

This is a prettier formula, but a more practical formula, which also has variable n for the number of times the money was compounded in the year is:

P(1 + r/n)^nt

For instance, if you deposited 1000 dollars in the bank, and your money got compounded every year with a 3% interest rate, after 30 years, you would have:

P = 1000
r = 3% = .03
t = 30
n = 1

P(1 + r/n)^nt
1000(1 + .03/1)¹⁽³⁰⁾
1000(1 + .03)³⁰
1000(1.03)³⁰
1000(2.42726)
2427.26

So, after thirty years, you would have about $2427.26, which means you were able to more than double your money! This might be a little surprising that it is possible for your money to double if you leave it alone for long enough.

You might be wondering how long it will take to double. How many years does it have to sit there? In other words, what value of t makes that equation equal to 2P? We deposited P, so to make it double, we must get 2P.

2P = P(1 + r/n)^nt

In our scenario, we deposited $1000 with an annually compounding interest rate of 3%. So, plug all of this in and we get:

2P = P(1 + r/n)^nt
2(1000) = 1000(1 + .03/1)^1t
2000 = 1000(1.03)^t
2000/1000 = (1000(1.03)^t)/1000
2 = (1.03)^t
ln(2) = ln((1.03)^t)
ln(2) = t • ln(1.03)
ln(2)/ln(1.03) = t
24 ≈ t

Note: this computation required something called logarithms. They look weird, but are very easy to understand. I explained them in my post on Benford's Law. Click here to read it.

So, it will take 24 years to double. So, let's set a rule to this - figure out a simple formula where you can figure out how long it will take for your interest to double, assuming it is compounded annually.

2P = P(1 + r/1)^1t
2P/P = (P(1 + r)^t)/P
2 = (1 + r)^t
ln(2) = ln((1 + r)^t)
ln(2) = t • ln(1 + r)
ln(2)/ln(1 + r) = t

This is as simplified as the equation will get without using calculus. And this doesn't look very simple anyways. But, let's look at a graph of it. It is easier to see as a picture than a messy jumble of logarithms.

This is a logarithmic function. However, it looks very much like a rational function, or a function that is the quotient of two polynomials (for example, 1/x is a rational function). So, let's try to find a rational function that fits this blue curve.

You can play around with it on your graphing calculator if you want, but I will just tell you that the function that fits it best is 72/(100r). Here are the two graphs:

As you can see, the two graphs are practically touching. In fact, all the way up through 0.5, they are very close together, as you can see here:

Since no bank on this planet offers 50% interest rates (if anyone has heard otherwise, please contact me), the 72/(100r) should be a good approximation for any of our purposes. Even to see the difference in the ones and tens, I had to set the graph below two. So, they are very close together.

This 72/(100r) equation can look even better. You might remember that r is currently in decimal form. To turn it into a percentage, you must move the decimal over twice, or multiply it by 100. So, dividing the interest percentage into 72 will give the same approximation.

Let's try it out on the original example. We said that it was a 3% interest rate. 72 ÷ 3 = 24, and we did conclude that it would take about 24 years. For a 6% interest rate, it would take about 72 ÷ 6 = 12 years. For a 4.5% interest rate, it would take about 72 ÷ 4.5 = 16 years. Couldn't be easier!

This rule is normally called the Law of 72. I find it very cool because you don't even need to know the amount of money you deposited to figure out this time. It doesn't matter. You don't even need to convert the interest rate into a decimal. I think this is a really intriguing and practical formula.