Elementary math, beautiful math: Introduction and a property of Fibonacci numbers
Introduction to the series
Math has a terrible reputation. Given how math is taught, especially in elementary schools, this isn’t surprising. Math is thought of as boring, hard, pointless and so on. And, finding out how much 8 + 4 is, or what the square root of 49 is, is, in fact, boring.
Many people don’t know what mathematicians do all day. They do a lot, but two things they do is solve problems and prove theorems. And when they describe these theorems they use words like “beautiful”, “elegant” and “charming”. Unfortunately, many of these theorems are inaccessible to all but a few people, because they hide behind formidable math. However, there is an abundance of problems that are accessible to anyone who understands some basic math (the kind you learn in elementary school).
In this series, I hope to expose you to such problems, with a lot of guidance for the math phobic.
The Fibonacci numbers
The Fibonacci numbers are those in the Fibonacci series, one of the most famous series in mathematics. Entire books have been written about them (for a fairly elementary guide, see The Fabulous Fibonacci Numbers by Alfred Posamentier).
Very briefly, the series starts with two 1’s and then each successive number is the sum of the previous two. The Fibonacci series starts:
1, 1, 2, 3, 5, 8, 13 ….
Today’s problem
I got today’s problem from James Tanton (who posts a lot of neat problems on Twitter, he is also here on Medium where he is James Tanton).
Take the first N Fibonacci numbers and try to divide them into two piles so that the sum of each pile is the same. For which N is this possible?
For example, when N = 2 the two numbers are the first two Fibonacci numbers (1 and 1) and dividing them into two piles automatically makes them equal because 1 = 1. When N = 3 we have (1, 1, 2) and it is again easy to divide them into two equal-sum piles: (1, 1) and (2). Continuing:
N = 4: there is no solution. (1, 1, 2, 3) cannot be put into two equal piles. Try all the combinations.
N = 5: Divide (1, 1, 2, 3, 5) into (1, 2, 3) and (1, 5).
N = 6: Divide (1, 1, 2, 3, 5, 8) into (1, 1, 8) and (2, 3, 5)
N = 7: There is no solution
and so on.
Do you see a pattern? Try with N = 8 and 9 and 10. Play around. Use a calculator if you want. See if you can find out for which N this is possible. Don’t look at the next section until you have played a bit. Math is all about this kind of playing. Get a little frustrated but don’t try yourself crazy.
The pattern
There is a solution when N = 2, 3, 5, 6, 8, 9 …… That is, whenever N/3 does not have a remainder of 1. There is no solution when N = 4, 7, 10….. e.g. 4/3 has a remainder of 1.
Now that you either worked out or saw the solution, can you prove that it is right? Because mathematicians love noticing patterns, but they aren’t satisfied unless they have proof that the pattern continues. Play around some more.
Think about it. Go away, come back, mull it over. Proofs don’t show up instantly in your head.
The proof
There are many types of mathematical proof. One type of proof is mathematical induction. For such a proof, you start off by showing that something is true for one example, then you prove that if it is true for one example (say, N = n) then it is true for N = n + 1. I will use a variety of this method.
First, we prove it is true for a specific example. We have already seen that it is true when N =2 and N = 3. and that it is false when N = 4. Now, Given that it is true for N = n, we will show that it is also true for N = n+3.
First, divide the first a Fibonacci numbers as you did already. Then put Fn+1 and Fn+2 into one pile and Fn+3 into the other pile and the two piles will still be equal because Fn+3=Fn+1+Fn+2. Example: Start with n = 2 and two piles (1) and (1). Now put 2 and 3 in one pile and 5 in the other. Since 2+3 = 5, the two new piles (1, 2, 3) and (1, 5) are equal.
We also need to show that it *not* possible when n/3 has a remainder of 1. Since we showed that it is impossible when n = 4, there is no way to add the next three numbers in ways that make the two piles equal, because of the same property as above.
What now?
Well, lots of things! Play some more! What about more than 2 piles? What if you start somewhere other than the first Fibonacci number? What about if you allowed subtraction? Play!
