My sources* inform me that Plutarch (c.40CE-120CE) wrote that the Pythagoreans** disliked (”abominated” in one translation) the number 17 for barricading 16 from 18. Why single out 17: doesn’t any number form a barricade between two others?
The Pythagorean love of number mysticism had them linking all sorts of numbers with all sorts of gods and goddesses (especially Egyptian ones), omens, portents, maleness, femaleness, and flatulence-causing beans. But much of the mysticism was rooted in the wonder they felt at being able to prove eternally beautiful things about whole numbers. 16 and 18 share a gorgeous distinction: they are the only two rectangular numbers whose perimeters are equal to the area they enclose. I remind you that a rectangular number is any number that is not a prime and so is expressible as a×b, where a and b are whole numbers greater than 1: you can think of a rows of b points (or conversely, naturally, and such a diagram itself is a nice demonstration of the commutative nature of the natural numbers). If a=b then you have a square number, like 16.
Can you prove this theorem?
Too easy? Try another puzzle on this blog.
Too hard? Just ask below!
*The Mathematical Experience by PJ Davis & R Hersh (1995), The Penguin Dictionary of Curious and Interesting Numbers by D Wells (1987).
** I like this quote from the Wikipedia article: “The Pythagoreans were called mathematikoi, which means “those that study all.” “
Filed under: mathematics | Tagged: 16, 17, 18, math, mathematics, maths, neoplatonism, number, platonism, plutarch, prime, problem solving, puzzle, pythagoras, rectangle, square
Nice task about those numbers 16 and 18, I’ll try to prove it later!
Good luck, then, edgarsr. What I love about that theorem is how it reveals that in the whole infinite universe of numbers, the only two with this property are 16 and 18 – small, familiar numbers.
OK, I guess, I proved this!
So, if the length and width of a rectangle is a and b, then the area of the rectangle is a*b, and the perimeter of the rectangle is 2(a+b). It is said these two numbers are the same:
a*b=2(a+b)
Or: a=2b/(b-2)
Both a and b are positive integers, so the expression 2b/(b-2) is a positive integer. It is easy to see that if b>2 (if b=1 or b=2, then a is not a positive integer), then the function f(b)=2b/(b-2) is a diminishing function, and:
lim 2b/(b-2) = 2
(b->infinity)
So the only possible integers being able to be the result of this function are the ones within the interval (2;f(3)=6], or:
f(3)=6
f(4)=4
f(6)=3
So, there are actually only two different rectangles possible – one with a=b=4, and the other one with a=3, and b=6 (or vice versa). Q.E.D.
Hi edgarsr. I like your proof! (I would also like to be pedantic for a moment and insert a couple of lines near the beginning that show that b=2 leads to a contradiction, so that dividing by b-2 is OK.) It takes the now-classic idea of replacing a function on the integers with a function on the reals. This would seem to make things harder, but in fact makes it easier (much like moving from the reals to the complex numbers can often help).
Can you – or anyone else – think of a more geometric proof?
[...] P.S. This is my 17th post in this blog, so let’s remember that thought about the 17 being an evil number! [...]