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.