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.

Math in the News: Playing with Pentagonal Tilings

This last weekend, I attended MathFest 2013, which is held by the Mathematical Association of America in Hartford, Connecticut. I gave a talk on magic squares, and got to see many other fascinating speakers. Many of my future posts will be content I learned at this convention.

One of the speakers was Frank Morgan, who talked about tiling the plane with different polygons, primarily pentagons. During the presentation, he mentioned that he wrote a news article about it in the Huffington Post. Since the presentation was so interesting, I thought this article would be a perfect "Math in the News" post.

http://www.huffingtonpost.com/frank-morgan/bees-honeycomb-mathematics_b_1318258.html

Triangular Day: More Figurative Families

Two weeks ago, we talked about the relationships between figurative families. We looked at these in a more basic light, by just analyzing the quick and easy patterns that are noticed. This week, I would like to go a little deeper.

Let's find the explicit formula for the different sequences. We'll start with triangular numbers.

T_n = n(n + 1)/2

We have used this formula in many of the previous posts about triangular numbers, so that one didn't take as much work. What about the square numbers?

S_n = n²

This formula is pretty obvious, considering that the definition of square numbers is a natural number squared.

What about pentagonal numbers? This might take some more work, but it can be found by solving a system of equations. At the end, you would get:

P_n = n(3n - 1)/2

What about hexagonal numbers? Again, this one would take some work. Let's see what it ends up with:

H_n = 2n² + n

There isn't an obvious pattern right now, but let's rewrite each thing in the terms of n(an - b)/2 with a and b being coefficients and constants in the formula.

T_n = n(1n - (-1))/2
S_n = n(2n - 0)/2
P_n = n(3n - 1)/2
H_n = n(4n - 2)/2

Now the pattern is pretty clear. Each formula is just adding one to a and b. Algebraically, a will always be equal to the number of sides on the figure minus 2, and b will be equal to the number of sides minus 4. I found this pattern to be pretty cool, considering how different the different shapes are.

Bonus:
I just found out that my TEDx talk in India was posted on YouTube. Here is the video of it.

I will also be speaking at TEDxBushnellPark in a week, and I will make sure to post that video when it is available as well.

Triangular Day: Are the Figurative Families that Related?

I don't know if you noticed, but today is a triangular day. It is the first of June, and 1 is indeed a triangular number.

1 is also a special number because it is a square number as well. In fact, if you look at all of the regular polygonal figurative families, one is the first number. It is the first pentagonal number, hexagonal number, and so on. In fact, a dot can represent whatever figure you want it to, which shows the creativity of mathematics.

Because of this specialness to the number one, I thought we should analyze all of the regular polygonal figurative families. Let's look at the first fifteen numbers in each family.

Name	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15
Triangular	1	3	6	10	15	21	28	36	45	55	66	78	91	105	120
Square	1	4	9	16	25	36	49	64	81	100	121	144	169	196	225
Pentagonal	1	5	12	22	35	51	70	92	117	145	176	210	247	287	330
Hexagonal	1	6	15	28	45	66	91	120	153	190	231	276	325	378	435
Heptagonal	1	7	18	34	55	81	112	148	189	235	286	342	403	469	540
Octagonal	1	8	21	40	65	96	133	176	225	280	341	408	481	560	645
Nonagonal	1	9	24	46	75	111	154	204	261	325	396	474	559	651	750
Decagonal	1	10	27	52	85	126	175	232	297	370	451	540	637	742	855

This table just looks like a regular table of numbers. However, there are tons of patterns buried inside of it.

For example, look at each row. If you take the common differences of each one, what do you find?

The very top row has a constant difference of one, as you would expect. The next row, which contains the triangular numbers, has a difference of all natural numbers. The 1 and 3 have a difference of 2, the 3 and 6 have a difference of 3, the 6 and 10 have a difference of 4, and so on. In other words, rows n-1 and n have a difference of n. This is the definition of triangular numbers, so that point is obvious.

The next row, which contains the square numbers, has a difference of all odd numbers. The 1 and 4 have a difference of 3, the 4 and 9 have a difference of 5, the 9 and 16 have a difference of 7, and so on. In other words, rows n-1 and n have a difference of 2n-1. If you want to see a quick proof of it, it isn't too hard. It just requires some algebra. Just do n squared minus n-1 squared, and you should receive 2n-1 to prove it.

n² - (n - 1)² = 2n - 1

n²  - (n² - 2n + 1) = 2n - 1
n² - n² + 2n - 1 = 2n - 1
2n - 1 = 2n - 1

If you look at the pentagonal numbers, it isn't as obvious what the pattern is. However, the differences are 4, 7, 10, 13, 16, and so on. The differences all differ by three, making the equation 3n-2. This can be proven using similar logic to above.

So, the first row has a difference of 1, which can be rewritten as 0n - (-1). The second row has difference of n, or 1n - 0. The third row has 2n - 1, the fourth row has 3n - 2. Do you see the pattern?

0n - (-1)
1n - 0
2n - 1
3n - 2

The fifth row of hexagonal numbers should have a difference of 4n - 3 if this pattern continues, right? Let's see if it worked.

1 - 0 = 1 = 4(1) - 3
6 - 1 = 5 = 4(2) - 3
15 - 6 = 9 = 4(3) - 3
28 - 15 = 13 = 4(4) - 3
45 - 28 = 17 = 4(5) - 3

The pattern indeed continued. For heptagonal numbers, the difference is 5n - 4, for octagonal numbers, the difference is 6n - 5, and so on.

Instead of looking at the rows this time, let's look at the columns and their differences. The first column has a difference of zero obviously. That is the nature of why one is in every figurative family. The second column has a difference of one. Since this number is the number defining the family itself, that isn't too surprising.

The third column has a difference of three. This is a little strange, but we can keep going. The fourth column has a difference of six. The fifth column has a difference of ten. Now do you see the pattern?

0, 1, 3, 6, 10

These are all triangular numbers! In fact, the triangular numbers have come back in the whole figurative family.

I find this table really cool to analyze. On June 15th (15 is a triangular number), we will revisit this table and look at some other neat identities within it.

Triangular Day: The Oblong Numbers

I don't know if you noticed, but today is a triangular day. It is April 6th, and six is the third triangular number.

You might remember from December when I brought up figurative families. The triangular numbers are a figurative family because each number can be represented by a triangular array of dots. Each array forms a geometric figure.

I also talked about square numbers as being a figurative family. Each number in that sequence can be represented by a square array, which is another geometric figure.

What about a rectangle? Considering that any number can be turned into a rectangular array, it would not be as interesting of a family. However, what about a rectangle whose array has just one more row than it does columns. For instance:

•  •  •  •
•  •  •  •
•  •  •  •

This has four rows and three columns, making it fit into this family. These numbers are called the "oblong numbers."

The oblong numbers go as follows:

2, 6, 12, 20, 30, 42, 56, 72, 90, 110, 132...

Here is a little property of oblong numbers. Divide each number in that sequence by two. You get:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66...

You end up with the triangular numbers. Why is this? Remember the formula used to find any triangular number?

T_n = n(n + 1)/2

We said that a number is oblong if its array had one more row than it had columns. So, an oblong number must be factorable into two numbers that are just one apart. We can denote these two factors as n and n+1. This makes the oblong number equal to:

O_n = n(n + 1)

After dividing by two, we ended up with the triangular numbers.

I enjoy trying to find ways to combine these different families together. For instance, take the first triangular number, first square number, and first oblong number. This would be 1, 1, and 2 respectively. We will add the first two together, and then subtract the last one.

1 + 1 - 2 = 0

Let's try that with some more:

T₁ + S₁ - O₁ =     1 + 1 - 2         = 0
T₂ + S₂ - O₂ =     3 + 4 - 6         = 1
T₃ + S₃ - O₃ =     6 + 9 - 12       = 3
T₄ + S₄ - O₄ =     10 + 16 - 20   = 6
T₅ + S₅ - O₅ =     15 + 25 - 30   = 10
T₆ + S₆ - O₆ =     21 + 36 - 42   = 15

Do you see the pattern? Each of the answers is a triangular number. In fact, it is one less than the triangular number used in the example.

Why is this? When I saw it, I first thought that we could check by using algebra. Take the explicit formula for each sequence:

T_n = n(n + 1)/2
_Sn = n²
_On = n(n + 1)

Our goal is to end with T_n-1, so we will use n(n - 1)/2 to denote that.

[n(n + 1)/2] + [n²] - [n(n + 1)] = [n(n - 1)/2]

Let's simplify all of the brackets so we don't have parentheses to work with.

[n(n + 1)/2] + [n²] - [n(n + 1)] = [n(n - 1)/2]
[1/2n² + 1/2n] + [n²] - [n² + n] = [1/2n² - 1/2n]

Now, we will combine like terms.

[1/2n² + n² - n²] + [1/2n - n] = 1/2n² - 1/2n
1/2n² - 1/2n = 1/2n² - 1/2n

And there is our proof. Though oblong numbers are not the family of a regular polygon, it is still an interesting sequence to look at.

Handshake Problem

Let me start by giving a little math problem. n people are in a room, and each one has to shake every other person's hand. How many handshakes will it take for all n people to have shaken every other hand?

Let's first try a few numbers. For two people, it would obviously just take one handshake. For three, it would take three handshakes. For four, it would take six handshakes. For five, it would take ten handshakes.

Do you see the pattern? 1, 3, 6, and 10 are the triangular numbers. In fact, this pattern always continues. 

If you look at it logically, you will see why. Person 1 has a number of hands to shake. Person 2 would have to shake all of the hands except for person 1's (it was already counted). Each person has to shake one less than the one before until there is just one left.

Adding up these handshakes will be a sum of the first n natural numbers, which is the definition of triangular numbers. 

I found it cool that a famous number sequence could be applied to a practical problem like this. 

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.

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.

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.

Triangular Day: A Second Figurative Family

Happy December, and happy triangular day! It is the first of December, and one is the first triangular number. One is also a square number, which is a good reason to bring up this topic today.

The triangular numbers are a type of "figurative family," which are sequences of numbers whose elements are numbers that if you draw that many dots, they can form an equilateral polygon (a shape with no curves or openings whose sides and angles are all equal in measure), which is the most common type of figure. For instance, the triangular numbers are a sequence of numbers whose elements form a equilateral or equiangular triangle. This is the first of the figurative families (you cannot draw a one or two sided polygon).

Today, we will turn to the next figurative family, which would be the family whose elements form a equilateral quadrilateral.

Think for just a moment about that. Do you ever hear kids in math class finding the area or equilateral quadrilaterals, or equilateral quadrangles, or equiangular quadrilaterals or quadrangles? I would be surprised if they do, because there is a very simple name for that shape. I defined an equilateral polygon earlier as a shape with no curves or openings whose sides and angles are all equal in measure, and a quadrilateral is simply a four-sided polygon. So, a equilateral quadrilateral is basically:

A four-sided shape with no curves or openings whose sides and angles are all equal in measure.

Do you recognize this definition from math class? It is the exact definition of a square. So, a equilateral quadrilateral can simply be called a square. Because of that, instead of calling the next figurative family the equilateral quadrangular numbers or something, we can call them the square numbers.

Wait a second! The name "square number" is already taken by the numbers whose square root is a integer. The square numbers are 1, 4, 9, 16, 25, and so on. We can't have two sets of square numbers.

But what are the elements of this new figurative family? You can form a 1x1 square with 1 dot, a 2x2 square with 4 dots, a 3x3 square with 9 dots, and so on. We are forming square numbers!

Believe it or not, the square numbers that we normally think of are the next figurative family. And they all form a square, whose side length is the square root of that number.

Before I go into a pattern with square numbers, I want to mention one quick thing that will be important when we study square numbers further. The explicit formula for triangular numbers was essential again and again. So, I want you to think for a moment about what the explicit formula for square numbers would be.

Don't think too hard! It is simply n^2. That is the definition of square numbers, so we can use it as its formula. We may have to complicate it in a later post with something like 2n(n+0)/2, but for now, it is perfect to leave it as n^2.

Okay, let's look at a pattern now. Take two random numbers, like 2 and 3. Find the sum of the square of each of those numbers.

2^2 + 3^2 =
4 + 9 =
13

Now, double that.

13 • 2 = 26

Can you find a sum of two square numbers that equals 26? This is a pretty easy one, namely the sum of 25 and 1, or 5^2 and 1^2.

Let's try a harder one, the sum of the squares of 4 and 6.

4^2 + 6^2 =
16 + 36 =
52

Double 52 and you will get 104, which is the sum of 4 and 10, or 2^2 and 10^2.

Let's try a big one. The sum of the squares of 27 and 41.

27^2 + 41^2 =
729 + 1681 =
2410 =

Double 2410 and you will get 4820, which is the sum of 4624 and 196, or 68^2 and 14^2.

I found it amazing that you can square two numbers, double their sum, and find a sum of two new square numbers. I was almost skeptical about a proof for it, but it actually isn't that bad.

In each of the triangular number proofs (except for the induction one), we found a correlation between the number(s) we started out with and number(s) we finished with. This was essential to turn it into a format that can be worked with.

In these three examples, the original numbers that were being squared did have a correlation with the square roots at the end. Those square roots were actually the sum and the difference of the two starting numbers. So, we can now label everything.

Let a = the bigger original number
Let b = the smaller original number

We can then conclude that the square roots are a+b and a-b.

Let's write this as an equation. We are doubling the sum of the squares of a and b to get the sum of the squares of a+b and a-b.

2(a^2 + b^2) = (a + b)^2 + (a - b)^2

Similar to the triangular number proofs, we will multiply each side out and see what we end up with.

2(a^2 + b^2) = (a + b)^2 + (a - b)^2
2a^2 + 2b^2 = (a + b)(a + b) + (a - b)(a - b)
2a^2 + 2b^2 = a^2 + 2ab + b^2 + a^2 - 2ab + b^2
2a^2 + 2b^2 = a^2 + a^2 + 2ab - 2ab + b^2 + b^2

2a^2 + 2b^2 = 2a^2 + 0ab + 2b^2

2a^2 + 2b^2 = 2a^2 + 2b^2

And we ended up with the two sides being equal! What I find kind of cool is that you can now easily create a brain teaser with this, asking something like taking the number 123 and 456 and asking people what two numbers can be squared and then added together to get double the sum of the squares of those two numbers. With this proof, you can easily determine that the answer is 579 and 333. And if you remember from last December, 333^2 is a pretty easy one to figure out.

Triangular Day: Sum of the Cubes is the Square of the Sum

Just like last week, today is a triangular day! It is the tenth of November and 10 is the fourth triangular number. Here are the triangular numbers just to remind you.

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, ...

Most of the proofs I've done with triangular numbers just involved some simple algebra. We just would perform the necessary operations to the explicit formula for triangular numbers n(n+1)/2 and it would simplify to the answer we wanted.

Today, I will be doing something very different that can't be proved with this method. Rather than looking at triangular numbers, let's look at cube numbers. Here are the first few:

1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, ...

Let's just add up the cubes. We will start by adding the first one, then the first two, the first three, and so on.

1 = 1
1 + 8 = 9
1 + 8 + 27 = 36
1 + 8 + 27 + 64 = 100
1 + 8 + 27 + 64 + 125 = 225
1 + 8 + 27 + 64 + 125 + 216 = 441

Do you notice any pattern with the sums? If you look closely, you'll see that they are all square numbers. But not any square numbers. Look again.


1 = 1 = (1)^2
1 + 8 = 9 = (3)^2
1 + 8 + 27 = 36 = (6)^2
1 + 8 + 27 + 64 = 100 = (10)^2
1 + 8 + 27 + 64 + 125 = 225 = (15)^2
1 + 8 + 27 + 64 + 125 + 216 = 441 = (21)^2

They are squares of the triangular numbers. When I first saw this, I was stunned that this happened, and thought there was no way this could continue to work forever. But in fact, we can prove it.

You could try doing it algebraically, but we can't represent cube numbers in a form that can be simplified. For this, we have to use something that I've mentioned before called proof by induction.

We already know it works for the first number. We know that:

1^3 = 1^2
or
1^3 = (1(1+1)/2)^2

Remember that triangular numbers can be found with the formula n(n+1)/2. Since we are squaring it, let's just simplify that.

(n(n + 1)/2)^2
((n(n + 1))^2)/((2)^2)
n^2(n + 1)^2/4

We know it works at some point, that is when n = 1, we can state our induction hypothesis. This is saying that for some number k, we know it will work.

1^3 + 2^3 + ... + (k - 1)^3 + k^3 = k^2(k + 1)^2/4

Now, we want to find out if we continue forward, it will keep working. In other words, we want to know if it holds true for k+1.

1^3 + 2^3 + ... + (k - 1)^3 + k^3 + (k + 1)^3 = (k + 1)^2(k + 2)^2/4

Based on our induction hypothesis, we know what the sum of the first k cubes is. So, we can substitute that into the new equation to get:

1^3 + 2^3 + ... + (k - 1)^3 + k^3 + (k + 1)^3 = (k + 1)^2(k + 2)^2/4
k^2(k + 1)^2/4 + (k + 1)^3 = (k + 1)^2(k + 2)^2/4

If you look on the left side, you will see that both terms have a factor of (k+1)^2. The distributive property says that we can "factor out" or restructure the equation so that the (k+1)^2 is multiplied at a different time. This gives us:

k^2(k + 1)^2/4 + (k + 1)^3 = (k + 1)^2(k + 2)^2/4
(k + 1)^2 • (k^2/4) + (k + 1)^2 • (k + 1) = (k + 1)^2(k + 2)^2/4
((k + 1)^2)(k^2/4 + k + 1) = (k + 1)^2(k + 2)^2/4

Now, rather than having a denominator on just one term, let's create a common denominator of 4.

((k + 1)^2)(k^2/4 + k + 1) = (k + 1)^2(k + 2)^2/4
(k + 1)^2(k^2 + 4k + 4)/4 = (k + 1)^2(k + 2)^2/4

The two sides are almost identical. The only thing that is different is the (k + 2)^2. But we can simplify that.

(k + 2)^2
(k + 2)(k + 2)
(k)(k) + (k)(2) + (2)(k) + (2)(2)
k^2 + 2k + 2k + 4
k^2 + 4k + 4

Let's substitute that in for (k + 2)^2.


(k + 1)^2(k^2 + 4k + 4)/4 = (k + 1)^2(k + 2)^2/4

(k + 1)^2(k^2 + 4k + 4)/4 = (k + 1)^2(k^2 + 4k + 4)/4

And now the two sides are equal. So, when all of the dust settled, it showed that the sum of the first k+1 cubes equalled the square of the k+1st triangular number, or the sum of the cubes is the square of the sum. In other words, the sequence continued to work from k to k+1 meaning that whatever value k is, the next value will hold true as well.

This proof is very complicated, with factoring, substituting and simplifying. But if you look at it a couple of times, it makes perfect sense and makes the fact even cooler.

Triangular Day... times eight plus one

I don't know if you noticed, but today is a triangular day! It is the third of November and three is a triangular number. Next week is also a triangular day, and I will be mentioning probably my favorite thing about them.

Let me quickly list some triangular numbers.

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120,...

They are basically the sum of consecutive natural numbers. So, 1 = 1, 3 = 1+2, 6 = 1+2+3, and so on.

This week, I wanted to try something. In August, we tried doing a triangular number times nine plus one. That resulted in other triangular numbers, which was fairly cool.

Today, let's try taking that down a notch. We will multiply by eight and add one.

1 x 8 + 1 = 9
3 x 8 + 1 = 25
6 x 8 + 1 = 49
10 x 8 + 1 = 81
15 x 8 + 1 = 121
21 x 8 + 1 = 169
28 x 8 + 1 = 225

Do you see any pattern? This is a little bit harder than the rest, but look closely.

They are all square numbers! It is fascinating to me that triangular numbers generate squares on their own, but it is true.


1 x 8 + 1 = 9 = 3^2
3 x 8 + 1 = 25 = 5^2
6 x 8 + 1 = 49 = 7^2
10 x 8 + 1 = 81 = 9^2
15 x 8 + 1 = 121 = 11^2
21 x 8 + 1 = 169 = 13^2
28 x 8 + 1 = 225 = 15^2

How do we know this goes on forever? Well, let's try to prove it. We know that the triangular numbers follow the formula n(n+1)/2. You can click here to see the post where that was proved.

Also, notice the numbers being squared. They are 3, 5, 7, 9, 11, and so on, which are all odd numbers. They fit into the formula:

2n + 1.

So, we can set up the following equation, considering n is what term we are using in the sequence. So, for the second triangular number, n = 2.

8(n(n + 1)/2) + 1 = (2n + 1)^2

Now, we will multiply through on both sides, so there are no parentheses.

8(n(n + 1)/2) + 1 = (2n + 1)(2n + 1)
8(n^2 + n)/2 + 1 = 4n^2 + 2n + 2n + 1
4n^2 + 4n + 1 = 4n^2 + 4n + 1

And already, we see that the two things are equal.

I found this pattern especially cool because of the switch between sequences, which Fibonacci numbers do all of the time. Probably next spring, I will be talking about the fact that triangulars and squares are much closer related than you'd think. Just look at their names and you should be able to see what I mean.

Triangular Day: Tripled

I don't know if you noticed, but today is a triangular day (and a week away from our hundredth post). It is the sixth, and six is a triangular number. We haven't really tried to prove stuff in a little while, so I wanted to take the time and do that today.

First off, let me tell you what we are going to do. We will prove that any triangular number tripled plus the next consecutive triangular number equals a triangular number.

That was a lot of information there. Let's just do the pattern so you can see better.

1) 3(1) + 3 = 6
2) 3(3) + 6 = 15
3) 3(6) + 10 = 28
4) 3(10) + 15 = 45
5) 3(15) + 21 = 66
6) 3(21) + 28 = 91
7) 3(28) + 36 = 120

That is pretty cool. But proving it is even better. And just like lots before this one, we can prove it with algebra.

Did you notice any pattern in the numbers coming out? Look at the positions of the numbers in the sequence rather than the number itself.

1 ––> 3
2 ––> 5
3 ––> 7
4 ––> 9

It consistently works out to the following:

3T(n) + T(n+1) = T(2n + 1)

(Remember that T(n) means the nth triangular number).

Using the triangular number formula n(n+1)/2, how do we write these three terms?

3T(n) = 3n(n+1)/2
T(n+1) = (n+1)(n+2)/2
T(2n+1) = (2n+1)(2n+2)/2

So, we can write it as:

3n(n+1)/2 + (n+1)(n+2)/2 = (2n+1)(2n+2)/2

Since I don't love working with fractions, let's multiply both sides by two and get rid of any denominators.

3n(n+1) + (n+1)(n+2) = (2n+1)(2n+2)

Now, let's simplify all of these factored terms.

3n^2 + 3n + n^2 + 2 = 4n^2 + 3n + 2
4n^2 + 3n + 2 = 4n^2 + 3n + 2

Both sides simplify to the same thing, thereby proving the pattern to be correct.

We can also prove this geometrically. There were tiles in first grade that we would play with that had different shapes, like squares, triangles, trapezoids, hexagons, etc.

We always would try to create a really big triangle. A shortcut we used was instead of making the second row with three triangle pieces (two facing up and one facing down), we would just use a trapezoid. If we were to take a three-dotted triangle and turn it into a trapezoid, we would just do this:

•  •  •  •  •
  •  •  •  •

You can see the triangles better like this:


•  •    •    •  •
  •    •  •    •


This relates perfectly to this pattern. We are adding to a triangle (the second number) to get a bigger triangle (the total). And what exactly are we adding? Three triangles, which makes a trapezoid.

Let's use some logic here. The triangle we are adding onto has a length of n+1, as it is T(n+1). So, the trapezoid should have a smaller base of n+2.

The full triangle at the end should have a side length of 2n+1, meaning that is going to be the bigger base of the trapezoid.

To fully prove it, we think about the actual construction of the trapezoid. Remember how I said that two triangles would point up and one would point down?

For the smaller side, the two triangles that point towards it each add one dot to it. The one that points the other way adds n dots to it. You can see that better if you look at the diagram above.

1 + 1 + n = n + 2

For the bigger side, we have only one triangle pointing towards it with a side of one. The others add n dots to it.

1 + n + n = 2n + 1

And there is the proof. I found this fascinating because of the pattern itself, as well as the fact that you can prove it many ways, which is common with triangular numbers.

Learn to Add Over Fifteen Numbers in Under Fifteen Seconds

Since I haven't gone over any mental math tricks in an extremely long time, I wanted to get one in this week.

Before this post, I have done posts involving multiplication and squaring tricks. But today, I am going to talk about some addition.

There are ways of doing mental addition reasonably quickly, but some specific series of numbers (like the Fibonacci series, which I taught a way to add last December) can be added much quicker. So, we are going to need some more fastidiousness when we get our numbers to add.

The series we will be adding is the powers of two. Have your volunteer choose a number and list out that many powers of two. Let's say they say nine.

1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 =

Dont retrogress yet. All you have to do is double the final number in the sequence and then subtract one.

256 x 2 - 1

512 - 1

511

And there is your answer. If you have a little more prowess, you could try going up to sixteen numbers.

1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 + 2048 + 4096 + 8192 + 16384 + 32768 =

If you do the math, you will find that the answer is 65535.

This method is actually fairly easy to prove. Remember back when we did infinite series? Let's write that one we are subtracting as an infinite series.

2^0 + 2^1 + 2^2 + ... + 2^n-2 + 2^n-1 = 2^n - (2^-1 + 2^-2 + 2^-3 + ...)

Now, add that series to both sides. If we write it as an infinite series, we get:

2^n = 2^n-1 + 2^n-2 + ... + 2^1 + 2^0 + 2^-1 + 2^-2 + ...

The left side is already simplified at 2^n. How do we simplify the right side? I may have mentioned in a earlier post (I'm not sure which) how to simplify this type of series. Let me explain it again.

1. Find what number you multiply each number in the sequence by to get the next number in the sequence. Call that s.
2. Call the first and biggest number in the sequence x.
3. Use the formula x/(1-s) to get the total.

In this sequence, we are multiplying by 1/2. 1/2(2^n-1) can be written as 2^n-2.

The biggest number is the 2^n-1. So, we plug it into the formula to get:

(2^n-1)/(1 - 1//2)
(2^n-1)/(1/2)
2(2^n-1)

Using the law of exponents, we can simplify that to get:

2(2^n-1)
2^n-1+1
2^n

And there is the proof. Though the proof is kind of cool, I found it cool that you could add this many numbers up so quickly. Good luck fooling people with this one!