Roxbury Math Epsilon Club Lecture1

Lecture 1

Roxbury Math Epsilon Club

Welcome to Roxbury Epsilon Math Club. My Name: Igor Zomb My Email: [email protected] My Hobby: Math My Objective: Make math “not scary,” instead make it fun… Your Job: Pay attention, ask lots of questions. A copy of this lecture can also be found on: http://igorzomb.blogspot.com/ Why am I doing this? I grew up in a different country (can you guess which one? My name should be a pretty good giveaway). For many years in school, I was fortunate enough to have a number of very good teachers… They taught me to understand math at its depth. In many cases, you don’t really need to memorize anything, but you do need to understand the fundamental concepts really well. Math is like a high-rise building: without the foundation, you can’t build anything. Here, we start with some very basic concepts (what is a concept ?). We do this by studying some unusual problems and hope that they will inspire us to go further. Understanding is the key…

Lecture 1

Roxbury Math Epsilon Club

3-by-3 Magic Square Problem: Arrange the numbers 1 through 9 on a 3-by-3 grid so that: 1) Each number occurs only once 2) Sum of entries in each row, each column and across the diagonals be the same • • •

Very, very old problem. Lo Shu square dates back to 650 BC (i.e. 2008 + 650 = ? ) years ago 3 by 3 is not the only kind of Magic Square --- simplest one is 1-by-1, but there are also 4-by-4, 5-by-5, etc. Today, we only discuss 3-by-3

Basic Question: need to place numbers 1-9 in correct boxes. How to do this?

Let’s start with what we know. All the numbers 1-9 will have to be inside somewhere. Let’s figure out the sum of all the numbers inside: We will work on a general solution to this type of problem a bit later today, but for now, let’s just find the answer. One simple way to do it in this case is to regroup numbers as follows:

Lecture 1

Roxbury Math Epsilon Club

If all the rows have the same sum and there are 3 rows, then the sum of each row must be: Ok… If all the rows and all the columns and both diagonals add up to the same number… and each row adds up to 15, then all the rows and columns and diagonals always add up to 15. Very important to get this far!!! Now we know that if any combination of numbers we put in does not add up to 15, then this combination will not work. Let’s write down again our main conditions (fancy math term is constraints) • • •

We can only use numbers: All the numbers have to be used up and no number can repeat All rows, columns and diagonals must add to 15

Here are some valid combinations:

And here are some invalid combinations:

Let’s first try to figure out what to put in the middle square. This is the most important position, because it will be used in 4 different groups: middle row, middle column and both diagonals --- lots of pressure on this number to stand up to this test. In other words this number will have to be used with all the numbers. Let’s see if this will give us an important clue, what this number should be. Let’s see what combinations we can use with the smallest of all the numbers (i.e. 1) as well as the largest of all the numbers (i.e. 9):

Lecture 1

Roxbury Math Epsilon Club

What is the only number which appears in both groups: the “magic number” is 5. We could have guessed that from symmetry, but above argument actually proves it. So far, our magic square looks like this:

5

This does not look like much, but actually, we’re very close to solving this… Let’s see where we can put the largest number “9”. There are only really 2 choices: corner or not-corner. Let’s try a corner and see what happens: 9 5 If 9 is in the upper-left corner, then the number is the lower right corner has to be: of course, it is 1, since the diagonal needs to add up to 15. 9 5 1 We’re on a roll… Let’s see how we can fill in the top row and left column. Remember, that everything has to add up to 15 and no number can be used 2 times. 9 4 2

3 5

3 1

Whoops… There does not seem to be a way to make this work… Does this mean that we will not find our solution??? Actually, not. This just means that one of our assumptions was not correct. Well, 5 has to be in the center (we proved it for sure). On the other hand, we did assume that 9 can go in the corner --- this is what needs to be changed. If 9 can’t go in the corner, it must be going into not-corner. Let’s see what happens:

Lecture 1 9

Roxbury Math Epsilon Club 5

1

Let’s figure out numbers which will “surround” 9. Remember, that the combination has to add up to 15 and can’t include 5 or 1. There is only one valid choice:

2 9 4

5

1

… and now we can fill the numbers on diagonals as well --- we’re almost there!!! 2 9 4

5

6 1 8

… and the final touch, the missing two numbers in the top and bottom rows: Presenting the complete Roxbury Epsilon Magic Square: 2 9 4

7 5 3

6 1 8

… something else to think about… is this the only way to complete this? How many actual combinations are there?

Lecture 1

Roxbury Math Epsilon Club

Mathematical Vocabulary Word-Of-The-Day: Transitive Relation. This sounds quite scary… but it means something very simple. Let’s see an example: If and , then . That’s all that this means. Any mathematical expression which has this “transition property” is called transitive. Can you think of another example of something transitive? How about something non-transitive? Note: Relation Operations in math are different from the normal Operations (like +, -,etc.). Relation Operations is simply: true or false. For example is a true relation, while is a false relation. Another Math Vocabulary Word --- something everyone has to learn at one point or another: Theorem. A theorem is a rule which we can prove. So, it is not true simply because we say it is true… It is true, because we can show why it is true. Another kind of statement in math is called: Axiom. Axioms we do not prove. We assume them to be always true.

Lecture 1

Roxbury Math Epsilon Club

Let’s discuss another interesting Math Problem. What is the sum of the first 100 numbers? For that matter, what is the sum of the first “N” numbers: Can we figure out a formula so that we don’t need to do many-many-many-…many (100 times) additions? Let’s start with a smaller, simpler example. Perhaps, we can see a pattern and extend it to more numbers: If we start with a middle number 3 and include one number to the left and one number to the right, we get: If we include one more group of numbers again going one left and one right, we get: Do you see a pattern? If we just replace every pair of numbers with the “middle number,” we get the same result as before, but it is much easier to calculate. So, the trick is to find the “middle-number” and just multiply it by the “number-ofnumbers.” Let’s do another simple example: In this case, the middle number was 4 and the number of numbers was 7. It could also happen that the middle number may not be an integer. For example: In this case, the middle number was 2.5 and the number of numbers was 4. By the way, when would the middle number be “exact” and when would it be “with-ahalf”? Let’s go back to our original problem and see if we can apply this idea somehow. What is the number of numbers here: Clearly, we have one hundred numbers, so the number of numbers is 100. What is the middle number? We can figure out that the middle number is exactly in the middle between 50 and 51… which makes it 50.5. So (drum roll….):

Lecture 1

Roxbury Math Epsilon Club

By the way, we really should be calling our “middle number” a mean, since that’s exactly what it is. One thing we can also note is that our mean is “always” exactly in-between the very first number (which is always 1) and the very last number (which we’re calling N). So, we can always find the mean using this formula:

mean 

1 N N 1  2 2

For example, when

mean 

100  1 100 1    50  0.5  50.5 2 2 2

Sidebar… Why am I allowed to compute this fraction in such a way? Answer: Distributive Law… Along with Commutative Law and Associative Law, these are the most important laws in Math. We will discuss them more in the future (not today) Now, we’re finally ready to generalize our idea to any number of numbers. In other words, we can find a general formula for:

1  2  3  ...  N  mean  N 

N 1 N 2

Let’s apply this to something we could not possibly do by hand:

1, 000, 000  1  1, 000, 000 2  500, 000.5  1, 000, 000  500, 000,500, 000

1  2  3  ...  1, 000, 000 

What is this number: five hundred billion five hundred thousand!!! By the way, we’ve just proved our very first Theorem. Not a big deal, but still quite a nice accomplishment!!!

Lecture 1

Roxbury Math Epsilon Club

Let’s see yet another way how to solve this same problem using triangles. Let’s build a special “marble” triangle in the following way. I will put one marble on line one, two marbles on line two, three on line three and four on line four. Here is what my triangle looks like: * * * * * * * * * * How many marbles do I have here… Well, if the entire square was filled with marbles, then we would have: of them. But here, we only have about half of square… actually we have a bit more than half of square. Let’s see what our picture would look like if we had “exactly” half of square of marbles: * * * *

* * *

x *

x

The only difference between the two pictures is that I removed one half of the diagonal. Wait a minute… how many squares are there on a diagonal. Well, in our 4x4 square, there are exactly 4 little squares on a diagonal. So, let’s review one more time all the facts here: 1) Our original square represents the number of marbles equals: 2) This number can be broken into two parts: a. Half of the marbles in the big square b. Half of the marbles on the diagonal 3) …but we know how many marbles in the whole square and we also know how many marbles are on the diagonal Conclusion:

1 2  3  4 

4 4 4   8  2  10 2 2

We’re in luck!!! We’re getting the same result as before. It would be really bad, if we didn’t...

Lecture 1

Roxbury Math Epsilon Club

This approach also works when we’re adding the odd number of numbers, like:

In that case, half of square will include half of marble in the very middle, but adding half of a diagonal will put the middle marble back together and we will end up with the proper integer result. Using the formula, we get:

1 2  3  4  5 

5 5 5   12.5  2.5  15 2 2

Here is a picture: * * * * *

* * * *

* *

x *

x

Lecture 1

Roxbury Math Epsilon Club

Homework… I would like everybody to think of really clever solutions to these very formidable math problems… 1) a. Hint: try to find:

This is really the same as , but a little different. Once you figure out how to compute the smaller example, use the same method on the original “big” problem.

2) a. Hint: how is “related” to . Again, use the same idea to solve the “big” problem 3) Consider 4-by-4 magic square. Instead of 9 numbers in it, it will have different numbers. Can you find what each row, column and diagonal must add up to? Do the same for 5-by-5 magic square. If you’re really brave, try to figure out a formula for N-by-N magic square We will discuss all the solutions next time…