### 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)

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.

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.

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.

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.

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