Proving theorems and special cases (Part 6): The Goldbach conjecture

In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:

Is it ever possible to prove a statement or theorem by proving a special case of the statement or theorem?

Usually, the answer is no. In this series of posts, we’ve already seen that a conjecture could be true for the first 40 cases or even the first $10^{316}$ cases yet ultimately prove false for all cases.

For the next few posts, I thought I’d share a few of the most famous unsolved problems in mathematics… and just how much computational work has been done to check for a counterexample.

1. The Goldbach conjecture (see here and here for more information) claims that every even integer greater than 4 can be written as the sum of two prime numbers. For example,

4 = 2 + 2,

6 = 3 + 3,

8 = 3 + 5,

10 = 3 + 7,

12 = 5 + 7,

14 = 3 + 11, etc.

This has been verified for all even numbers less than $4 \times 10^{18} = 4,000,000,000,000,000,000$ . A proof for all even numbers, however, has not been found yet.

Here are some results related to the Goldbach conjecture that are known:

1. Any integer greater than 4 is the sum of at most six primes.

2. Every sufficiently large even number can be written as the sum or two primes or the sum of a prime and the product of two primes.

3. Every sufficiently large even number can be written as the sum of two primes and at most 8 powers of 2.

4. Every sufficiently large odd number can be written as the sum of three primes.

Non-geometric infinite series (Part 12)

I conclude this series of posts with thoughts about infinite series which use reciprocals of positive integers. I offer this post for the enrichment of talented Precalculus students who have exhibited mastery of geometric series.

Geometric. As we’ve discussed at length, the series

$1 + \displaystyle \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots$

converges and is in fact equal to $2$ .

Harmonic. Including the reciprocals of all positive integers is called a harmonic series:

$1 + \displaystyle \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \dots$

As shown in the link to the MathWorld website, this series actually diverges, even though the terms get smaller and smaller.

So we’ve made an observation: if too many reciprocals are included, the series diverges. But if we take enough of them away, then we can still end up with a series that is infinite but converges.

Squares. Let’s now consider the reciprocals of perfect squares:

$1 + \displaystyle \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \dots$

Clearly, we’ve taken away a lot of the terms of the harmonic series? Have we taken enough away so that the series converges? It turns out that the answer is yes. And the answer is precisely what you’d think it should be (not): $\pi^2/6$ . This is just another way that the circumference of a circle has an odd way of appearing in the most unexpected of places.

The proof that this series equals $\pi^2/6$ requires the clever use of Parseval’s theorem from Fourier analysis.

Fourth Powers. Let’s now turn to the reciprocals of fourth powers:

$1 + \displaystyle \frac{1}{16} + \frac{1}{81} + \frac{1}{256} + \frac{1}{625} + \dots$

By the Direct Comparison Test and the series for reciprocals of squares, this series converges. Using Parseval’s theorem, it can be shown that the answer is $\pi^4/90$ .

Cubes. Now let’s investigate the reciprocals of cubes:

$1 + \displaystyle \frac{1}{8} + \frac{1}{27} + \frac{1}{64} + \frac{1}{125} + \dots$

Again by the Direct Comparison Test and the series for reciprocals for squares, this series must converge. This sum is called Apéry’s constant. However, and amazingly, no one knows what the answer is. Of course, a computer can be programmed to evaluate this series to as many decimal places as desired. According to Wikipedia, this sum was evaluated to over 100 billion decimal places in 2010. However, to the best of my knowledge, no one has figured out if there’s a simple way of writing the answer, like $\pi^2/6$ or $\pi^4/90$ .

So if you figure out a simple way to evaluate Apéry’s constant, feel free to call me collect.

The previous four series are example of Riemann’s zeta function, which is of central importance in number theory and is the focus of the celebrated Riemann Hypothesis, for which a solution is worth a cool $1 million.

Primes. Now let’s consider the reciprocals of primes:

$\displaystyle \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{11} + \dots$

As noted above, the harmonic series diverges, but if we remove enough terms from the harmonic series, then it’s possible to make an infinite series that converges. So the central question is, did we remove enough fractions (by taking away all of the composite denominators) so that the series converges?

Surprisingly, the answer is no: the sum of the reciprocals of the primes actually diverges. The proof actually requires a graduate-level class in analytic number theory.

By no means would I expect high school students to master all of the above facts. As noted above, the subject of this post is mostly for the enrichment of high school students who have mastered infinite geometric series.

That said, students who know the following facts from Precalculus will be well-served when they reach calculus and other university-level mathematics courses.

They should either know the formula for an infinite geometric series or else be able to quickly derive it.
They should know that not every infinite geometric series is geometric.
They should know that not every infinite series converges.
They should be familiar with the meaning of the terms converge and diverge.

Unsolved problems: the Collatz conjecture

Students at all levels — elementary, middle, secondary, and college — tend to think that either (1) all the problems in mathematics have already been solved, or else (2) some unsolved problems remain but only an expert can understand even the statement of the problem.

There are plenty of famous unsolved problems in mathematics. And the Collatz conjecture is an easily stated unsolved problem that can be understood by most fourth and fifth graders.

Here’s the statement of the problem.

Start with any positive integer.
If the integer is even, divide it by $2$ . If it’s odd, multiply it by $3$ and then add $1$ .
Repeat until (and if) you reach $1$ .

That’s it. From Wikipedia:

For instance, starting with 6, one gets the sequence 6, 3, 10, 5, 16, 8, 4, 2, 1.

Starting with 11, for example, takes longer to reach 1: 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1.

The sequence for 27 takes 111 steps, climbing to 9232 before descending to 1: 27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1

The longest progression for any initial starting number less than 100 million is 63,728,127, which has 949 steps. For starting numbers less than 1 billion it is 670,617,279, with 986 steps, and for numbers less than 10 billion it is 9,780,657,630, with 1132 steps.

Here’s the question: Does this sequence eventually reach $1$ no matter the starting value? Or is there a number out there that you could use as a starting value that has a sequence that never reaches $1$ ?

Like I said, this is an easily stated problem that most fourth graders could understand. And no one knows the answer. Every number that’s been tried by computer has produced a sequence that eventually reaches $1$ . But that doesn’t mean that there isn’t a bigger number out there that doesn’t reach $1$ .

I’ll refer to the above Wikipedia page (and references therein) for further reading about the Collatz conjecture. Pedagogically, I suggest that casually mentioning this unsolved problem in class might inspire students to play with mathematics on their own, rather than think that all of mathematics has already been solved by somebody.

Source: http://www.xkcd.com/710/