Pascal's Triangle In Motion

By Sebastiaan

I recently created this Flash video for my (upcoming, nearly finished) site. It shows Pascal's Triangle in motion:

(If the space above is empty, you do not have QuickTime installed)

For more information on Pascal's Triangle, take a look at my previous post on this subject.

Odd Squares

By Sebastiaan

No, this is not a post about strange squares. It's all about odd squares though.

Warning! Non-mathematicians may find this post unreadable.

Let me list the first few:

1² = 1
3² = 9
5² = 25
7² = 49
9² = 81

At first glance, there does not seem to be a pattern. I got this puzzle from my school, where I study to be a mathematics teacher. It stated:

"The squares of all positive odd integers are eightfolds plus one."

First thing to do was translate this into proper mathematics language.

So, odd, would mean the integer can be written as "2k+1" with k any positive integer. Thus we get:

(2k+1)²

Also, an eightfold, is "8n" with n any positive integer. Thus we get:

(2k+1)² = 8n + 1

Now let's work out the brackets...

4k² + 4k + 1 = 8n + 1
4k² + 4k = 8n
for all positive integers k and n.

So now we have a relatively simple statement to proof. First thing that springs to mind is induction, but if you look just a little bit further... you'll find a much simpler and more elegant proof.

So, let's first start by dividing by four.

k² + k= 2n

What's left to proof is that the sum of an integer and its square is even. There are multiple approaches to solving this, I'll give you two.

Say, k is even. Then k² is most certainly even, for (2p)² is 4p² which is a multiple of four and thus even. When you add another even integer, it will always remain even.
Say, k is odd. Then k² is odd, for (2p+1)² is 4p² + 4p + 1, which are two even integers plus one, making it odd. Now add the same odd integer k, you will get an even integer, for two odds make an even.

So, for any positive integer k, we get an even integer. Thus every odd square is an eightfold plus one.
Q.E.D.

Another approach could be rewriting the left-hand of the equation, that would make k(k+1) = 2n. Now, an integer multiplied by the next integer, would always be 'odd' times 'even' or 'even' times 'odd'. This always makes even.
Q.E.D.

Symmetry

By Sebastiaan

Symmetry is a wondrous subject, which one not only finds in mathematics, but in other subjects as well, and of course, in nature.

First, let's start by defining symmetry. There are different kinds of symmetry, but the one that springs immediately to mind, is reflection symmetry. Look at the following kite:

It has a clear axis of symmetry right through the centre. But we could also look at a star like this one:

This one also has the symmetry axis, but as you can see, it can also be rotated - this is called rotational symmetry.

But there are other forms of symmetry too. What about palindromes? They are found in words, numbers or sequences of words and numbers... Here are a few examples:

eye
never odd or even
1475741
etcetera..

I absolutely love playing with words and numbers. That's why I love symmetry. You can find symmetry in many places, if you look hard enough. Gorgeous symmetry can be found inside music. In the lyrics, but most definitely in the instruments too. But that's for some future post.

Pythagorean Theorem

By Sebastiaan

It's been long since I wrote about math, so here's a rather simple subject which almost everyone knows of, but taken to a bit higher level. I'm talking about the Pythagorean Theorem.

Let me first bring back the long ago acquired knowledge of the theorem. When you have a right-angled triangle, or simply a right triangle, the theorem states that the sum of the squares of the legs belonging to the right angle are equal to the square of the side opposite the right angle.

Right, that's a pretty large piece of information, but what I'm actually saying is, with reference to the picture above, that:

AB² + AC² = BC²

or

b² + c² = a²

The classic example is AB = 3 and AC = 4. The squares of these sum up to 25 (9 + 16 = 25), and the square root of 25 is 5. So BC = 5.

Now, a more interesting question comes up when one asks oneself, "why?". For most of us this is pure magic, but it is most definately proveable. Not only is it proveable, it has been proven in many, many different ways. I will actually try to take some of these proofs and show them to you, in understandable language.


This first 'proof', is actually just to make the matter a bit more comprehensible. It's good to get a bit more knowledge about the theorem.

With reference to the picture above:

If AC = 3, then the area of the square with side AC, is 3 x 3 = 9.
If AB = 4, then the area of the square with side AB, is 4 x 4 = 16.
If AC = 5, then the area of the square with side BC, is 5 x 5 = 25, which is indeed 9 + 16. If you work this out neatly, you'll find by measuring that this is indeed true.

But this doesn't actually prove the theorem. If you replace '3' with 'b' and '4' with 'c' and '5' with 'a', then you get b² + c² = a².

A nicer prove, in my opinion, is the following:

With reference to the above picture:

(1) The area of each of the triangles are given by 1/2 times a times b, so 1/2*(a + b), so all four triangles sum up to 4*(1/2*(a + b)) which is 2*a*b or simply 2ab.
(2) The area of the square with sides equal to c is given by c*c, or c².
(3) The area of the square with sides equal to a + b is (a + b)*(a + b) or (a + b)².
(4) The area of the square is also given by combining (1) and (2), which makes c² + 2ab.
Now we can combine (3) and (4), and you get, magically (or perhaps not so magically?), the Pythagorean Theorem:

(a + b)² = c² + 2ab
a² + b² + 2ab = c² + 2ab
a² + b² = c²

I will probably post more proofs, hopefuly more elegant ones, in the future.

Math Puzzle: Solution

By Sebastiaan

This is the solution to this post.

The answer is 22 hours.

You can look at it in different ways, for example: n₀ = 1. Then n_i+1 = n_i * 3 + 1. So then we just enter the numbers, and work our way towards the answer.

One could also get the following formula: n_i+1 = 4*n_i - 3*n - 3 with n₀ = 1.

Lets take a closer look a the second formula. If you take a big n, say, 20, what does it look like? If n is that big, the second term (3*n) is insignificant, not even speaking of - 3. But what is 4*n_i? It's four times the previous n, which is four times the previous n, etc... The first term is 4*1 (so, 4¹), the second 4*4 (so, 4²), the third is 4*16, you can guess where we're going: 4ⁿ. This is a quick way to see how fast this progresses, within a day everyone knows what the secret is. But it does not provide you with the exact answer (22), since it's only an estimate.

If you calculate exactly, or use something like this calculating programme, you can easily find the answer. First number would be 1, the common ratio 3, and n is 22. Try progressing n up to 22, and see how fast it goes up.

Math Puzzle

By Sebastiaan

I came across a puzzle a few days ago. It is part of the (Dutch) National Science Quiz, my brother was reading the question out loud. I hadn't given it much thought, until now.

It is as follows: one person knows some sort of secret, he/she then tells three others. This takes an hour. Now these three others will each pass the secret on to again three others, which takes another hour. This process continues this way, until everyone in the world knows about the secret. Only the persons who were told about the secret last, can tell three more persons. How long will it take for the secret to be known by everyone in the world?

In the quiz, you can choose between a day, weekend or a week.

I won't give the answer to this problem, not just yet. I will wait until the 22^nd of December, for then the quiz doesn't take entries anymore. I will post the answer (at least, what I assume is the answer) before the 27^th, the day the official quiz answers are given.

So, let's take a look at the progress in this problem.

1 person knows about the secret. After 1 hour, 4 persons know about the secret. After 2 hours, 13 persons know about the secret.

So. in maths language:

n₀ = 1
n₁ = 4
n₂ = 13

Can we find a formula describing this process?

Please do not comment any answers.

First Encounter With Math: Pascal's Triangle

By Sebastiaan

Thinking back, the first time I fell in love with mathematics must have been reading the magnificent book "De Telduivel", the dutch translation of "The Number Devil" by H.M. Enzensberger.

The part I found most fascinating, was what I later learned is called "Pascal's Triangle" named after Blaise Pascal, a famous western mathematician, physicist and religious philosopher.

First let me explain a bit about the triangle. "1" is our top number, which comes in the uppermost 'corner' of the triangle. Then we go downwards and left, where we put another "1" and we go downwards right, where we put a third "1". Let me sketch the situation for you:

1

1   1

Now, we go down-left, and down-right from both ones. The number we put in the space, is the sum of both numbers above it, and in the leftmost and rightmost spaces we just put a "1". Continuing this process infinitely, we get Pascal's Triangle:

1

1   1

1   2   1

1   3   3   1

1   4   6   4   1

1   5  10  10  5   1

1   6  15  20  15  6   1

1   7  21  35  35  21  7   1

You get the general idea.

Now, the most interesting thing about the triangle, is that it has a wide variety of uses. (I understand that almost everyone who reads this will lose interest in a few moments, but I still would like to share this)

One practical use is found in the calculation of combinations. Say, we are looking for the number of combination of n things taken k at a time. Mathematicians call this n choose k. Let's take a practical example, which we find in the game of poker.

Take a look at my example: we have n = 52, for there are 52 cards in a deck of cards, and let's say k = 5 because we have a poker hand with 5 cards. We now have 52 choose 5. The official formula for combinations is n! divided by k! times (n-k)!, so let's put that into perspective...

...which means there are 2,598,960 combinations of poker hands.

Could we not have found this answer in a differnet way? The answer is yes. Take a look at the following -much simpler- example: 3 choose 1. After a few calculations we find the answer is the simple "3". Take a look at the 3rd row of Pascal's Triangle (don't count the top "1" as a row, this can be seen as the starting number, or row 0), and then the first number that isn't one (for that would be 3 choose 0). This is also 3!

This works for all combinations. 7 choose 3 would be 35, according to the triangle. According to the formula this is 7! (so 7 times 6 times 5 times 4 ..... times 1) divided by 3!4!:

(7 x 6 x 5 x 4 x 3 x 2 x 1) / (3 x 2 x 1 x 4 x 3 x 2 x 1)

...which is indeed 35.

As a kid, I used to write down the triangle as far as I could, which mostly meant 'till I ran out of paper... I especially used to do this during the more boring classes in primary school. It's ages ago, though I remember it as if it were yesterday.

I will probably post more about the triangle later on, since I so dearly love it. There are some great patterns in it...