Hippasus & The Discovery of Irrational Numbers

In November I gave a speculative talk to the New England Section of the Mathematical Association of America on the discovery of irrational numbers by the Pythagorean Hippasus through an examination of the mystic pentagram, the sacred symbol of the Pythagoreans. I have expanded it a bit, adding a method of recursively computing the golden ratio, phi. If you want to see how these topics are related, please check my paper on Scribd: https://www.scribd.com/doc/294006886/Irrational-Numbers-the-Mystic-Pentagram-and-Eigenvectors.

E32. Chimes

The following problem is given in Jacobs' Geometry: If it takes a clock 3 seconds to ring 3:00 (3 chimes), how long does it take the same clock to ring 6:00 (6 chimes)? The answer is not 6 seconds. The answer depends on making some assumptions, which I think are reasonable ones.

Embarrasing mistake

I just realized that my last post was far off the mark. There is a super-obvious example that shows that f(n) >= n/2, and so in fact f(n) = n/2. The example is the subset {n/2 + 1, n/2 + 2, ... , n} which contains n/2 elements and clearly has the non-divisible property. The problem has very little to do with prime numbers. The previously-published result was sharp.

A simple (?) number theory conjecture

Call a set S of positive integers non-divisible if, whenever a and b belong to S, it is not true that (a|b or b|a). For example, the set of prime numbers is non-divisible. Let S(n) be the set of the first n positive integers, and let f(n) be the cardinality of the largest non-divisible subset of S(n). Then clearly f(n) >= pi(n), where pi(n) denotes the number of prime numbers <= n. Furthermore, I know a pretty proof (published, but not by me) that shows f(n) <= n/2 (n even). That is, in any subset of n/2 + 1 positive integers all <= n, at least one integer divides another integer in the set. This proof involves the pigeonhole principle. So, pi(n) <= f(n) <= n/2. (n even) My conjecture is that, in fact, f(n) = pi(n) for all n > 1. Does anyone have a counterexample or a proof? It seems like it should be very simple.

E31. A packing polynomial

Funny how one comes across math problems. I was reading my Reed College alumni magazine and came across an article about Maddie Grant, class of '15, whose undergraduate thesis is apparently a substantial generalization of a 1923 result by Fueter and Polya. Feuter and Polya evidently discovered a polynomial function that maps the non-negative integers one-to-one and onto the pairs of non-negative integers. That is, they found a simple formula to express the inverse Cantor's "diagonal" mapping of pairs of non-negative integers to non-negative integers. Their function ‐ a so-called packing polynomial ‐ is given by

f(x,y) = (1/2)[(x + y)² + x + 3y],
where x and y are non-negative integers.

The proof that this function is one-to-one and onto the non-negative integers is actually pretty elementary if you look at it the right way. I will print a hint as a comment in a few days.

A11. Splitting a triangle and a tetrahedron

János Kurdics published this interesting problem in The Math Connection Linkedin group:
Given a triangle, find the shortest line segment that divides the triangle into two regions of equal area. A solution that does not involve calculus is preferred.
The solution that I know is very nice, and gives quite a workout in elementary trigonometry.

János says that he is working on three-dimensional analog (plane that divides a tetrahedron into two regions of equal volume and gives smallest possible cross-sectional area with the tetrahedron) but has only solved it for a regular tetrahedron.

A10. Triangle with sides in arithmetic sequence.

Find a number n such that there is a triangle with sides n, n + 1, and n + 2 in which the largest angle is twice the smallest angle. How many such numbers n are there? Note that n does not have to be an integer.

This problem came up in a high school textbook during a tutoring session, except there were several hints given that made the problem much easier. See if you can do it without hints.

E 30. A Cryptarithm

Introduction: Here is a problem that I think would be good for any child who knows how to use the standard algorithm to do multi-digit multiplication. It is a cryptarithm from The Moscow Book of Puzzles, an amusing book by Boris Kordemsky, available in a very inexpensive Dover edition. For those of you who don't know (as I didn't) a cryptarithm is like a cryptogram, except the unknowns are digits, not letters. One you may have seen is (S  E  N  D) + (M  O  R  E) = (M  O  N  E  Y), where each letter stands for a digit, with different letters standing for different digits, and the expressions in parentheses standing for multi-digit numbers.

Problem: In the following multiplication, each * stands for a prime number digit (2, 3, 5, 7). Replace each * by a prime digit so that the multiplication is correct.

      * * *
     x  * *
   * * * *
 * * * *
* * * * *
Strangely enough, you have all the information you need to find the unique solution.

Multiplication table problem

Sometimes mathematics problems that are interesting to the professional mathematician can arise out of the simplest questions. We all learned a multiplication table in school, which lists all 100 products of integers from 1 * 1 to 10 * 10. I've been told that in England, students are required to memorize up to 12 * 12. In the multiplication table for 1 to 10, there are 100 products, but it is only necessary to learn 55, because of the commutative property. In other words, we only need to learn the products a * b, where a ≤ b, and the number of such products is 1 + 2 + ... + 10 = T(10) = (11)(10)/2, where T(n) is the n-th triangular number, (n + 1) * n / 2. Therefore, the number of different products is at most 55. Actually it is less. For example, 6 appears twice in the triangular table, as 1 * 6 and as 2 * 3. The number of different products in a 10 by 10 multiplication table is 42. (Shades of Douglas Adams!)

A mathematician would naturally be interested in knowing how many different products there are in an n by n multiplication table, in other words, the number of different products of the form ab, where a and b are positive integers less than or equal to n. Call this sequence a(n). Then a(n) ≤ T(n), so the lim sup of a(n)/n^2 as n goes to infinity is less than or equal to 1/2. In fact, Erdös gave a very nice proof that the limit is 0, and he and others obtained more accurate asymptotic formulas for a(n).

See the Online Encyclopedia of Integer Sequences, where a(n) is given as sequence A027424. Links provide much information about the sequence. Also, the question of how many different numbers appear in a multiplication table could be given to students at almost any level.

E29. Divisible by Sixteen

Here's an easy one:

You have sixteen balls, numbered consecutively 1 through 16, and 4 boxes. You are asked to put 4 of the balls into each box, so that the product of the numbers in each box is divisible by 16. Give a solution, or prove that this cannot be done.