If you are reading this page, then you are likely in that happy minority who know better. Therefore, you might find things here that you will appreciate.
The prettiest illustration (not proof) of this theorem that I have ever seen was in the Ontario Science Center in Toronto, Canada. The exhibit showed a large right triangle, made out of transparent plastic, about an inch thick. Each side of the triangle was also one side of a square. Unlike the triangle, these three squares were hollow and connected together. The square on the hypotenuse was filled with water colored blue and the whole figure rotated 180 degrees about the hypotenuse every ten seconds. You can guess the rest.
This demonstration is both elegant and straightforward. Just start with any prime number, M, then construct the number
N = M! + 1 = [M*(M-1)*(M-2)*(M-3)*...*2*1] + 1.
So, what can one say about N? Obviously, N is bigger than M. It is also obvious that N cannot have any factor, from 2 through M, because there will always be a remainder of one, by construction. But, is N prime?
Answer: Either N is, itself, a prime number or it has a prime factor, J, which must necessarily be bigger than M (see last paragraph). Either way, we have found a new prime number, N or J, that is bigger than M. Since we could repeat this process indefinitely, with each new prime that we find, there must be an infinity of primes.
This proof is an illustration of one form of mathematical induction.
Footnote:
Several years ago, at a party, another guest started telling me about his "proof" of Fermat's Last Theorem. He said that he had tried a lot of integers and, so far, the theorem seemed to be correct. I asked how long he intended to keep this up. I do not recall his answer verbatim but got the impression that, when his patience was exhausted, that would constitute some kind of proof. Naturally, my pedagogical instincts took over and I tried to explain that, in mathematics, this was not quite how it was done, that the theorem was supposed to be perfectly general so the proof must be likewise.
He didn't get it. Most people don't.
I should have shown him something like the following proof that numbers can sometimes be irrational.
Incidentally, this is an excellent counterexample to the proposition that you cannot prove a negative. This proof shows conclusively that the given hypothesis is impossible.
Thus, for example, the number 8 has three cube roots, all solutions of x3 = 8. How many can you find?
Finally, if a polynomial is written with the exponents in numerical order and the coefficients are all real numbers, then Descartes' Rule of Signs states that there are at most M positive, real roots where M is the number of sign changes in the sequence of coefficients.
This is a proof that, even though there are an infinity of positive integers, these are still not enough to count the irrational numbers. In other words, the infinity of irrationals is "larger" than the infinity of positive integers. This, of course, means that there are "levels" of infinity. A truly mind-boggling concept!
The proof is actually quite simple (in hindsight) and may be paraphrased as follows:
First, assume that the fractional parts of all the irrational numbers are listed in a table, in no particular order. Also, to keep things simple, let these numbers be written in binary notation, as shown below.
Realize that, since these numbers are irrational, their fractional parts neither terminate nor repeat. [If they did either, then they would be rational.] Thus, there are an infinity of columns in the array above. Likewise, the rows could be numbered and, since there are an infinity of positive integers to count with, you might think that this would be enough. The big question, then, is whether or not this infinity of rows contains all of the irrational numbers or if, perhaps, one or more of the latter is missing.
Now for the clever part.
Construct an irrational number, IRR, by going through the list above and flipping all of the diagonal elements. In other words, change digit[0][0] from 1 to 0, digit[1][1] from 0 to 1, etc., and imagine going through the entire diagonal in this fashion. Is the resulting irrational number, IRR, anywhere in the list (which is infinite and, therefore, presumably complete)?
The answer is clearly No. IRR cannot be the first number listed because their first digits are different. It cannot be the second number listed because their second digits are different. In general, IRR cannot be the kth number in the list because the two will differ, at least, in digit k. Even though there are an infinity of irrational numbers in our list, nevertheless, IRR is missing! In plain English, there are "more" irrationals than there are positive integers.
Further commentary would be superfluous. You will just have to think about it, or go here for additional information.
This conjecture has never been proven or disproven and, if you can find a counterexample, then you will become famous! Here is a link to further information.
In society at large, statistics has a poor reputation. Again, part of this stems from a pervasive and disheartening innumeracy. The rest derives from the fact that, even with the best of intentions, probability and statistics are not easy to comprehend. Perhaps some of the items presented here will help to clarify this subject.
I've placed my own contribution last but please do not take this as a value judgment! Although modeling is a principal activity of researchers the world over, there are few, if any, general discussions comparable to "...the very game..." This (.pdf) tutorial is intended for novices but does require some mathematical understanding.
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