Professor Stewart's Hoard of Mathematical Treasures (26 page)

So, although you can’t add two fractions this way, the formula has its uses, and we can define the mediant
provided that the fractions are in lowest terms. One problem with them not being in lowest terms is that different versions of the same fraction can lead to different results. For example,
which is different.

Farey sequences are widely used in number theory, and also show up in non-linear dynamics - ‘chaos theory’.

Alice and Betty owned adjacent market stalls, and both were selling cheap plastic bracelets. Each had 30 bracelets. Alice had decided to price hers at 2 for £10, while Betty was thinking of charging 3 for £20. So together they would make £150 + £200 = £350, provided that they both sold all their bracelets.

Worried that the competition might destabilise the market, they decided to pool their resources, and reasoned that 2 for £10 and 3 for £20 combine to give 5 for £30. At that price, if they sold all 60 bracelets, then their total income would be £360, which was £10 better.

Just across the way, Christine and Daphne were also selling bracelets, and also had 30 each to sell. Christine was thinking of selling hers at 2 for £10, while Daphne was thinking of undercutting the competition severely by selling hers at 3 for £10. When they got wind of what Alice and Betty were doing, they too decided to pool their resources, and sell their 60 bracelets at 5 for £20.

Was this a good idea?

 

A rep-tile is more properly known as a replicating polygon, and it is a shape in the plane that can be dissected into a number of identical copies, each the same shape but smaller. The shapes are allowed to have their boundaries in common, but do not otherwise overlap. If the polygon has s sides and it dissects into c copies, it is called a c-rep s-gon. Several different 4-sided rep-tiles (4-gons) are known. Most are 4-rep, but there are k-rep 4-gons for every k.

Top: replicating 4-gons. If the parallelogram at the bottom has sides 1 and √k, then it is rep-k.

Every triangle (3-gon) is 4-rep. Some special triangles are 3-rep or 5-rep.

Replicating 3-gons. The first can be any shape. The second has sides 1 (vertical) and √3 (horizontal). The third has sides 1 (vertical) and 2 (horizontal).

Only one 5-sided rep-tile has yet been discovered: the sphinx. It requires four copies. There is a unique 5-rep 3-gon (triangle), and exactly three 4-rep 6-gons are known.

The only 4-rep 5-gon, the sphinx, and the three known 4-rep 6-gons.

There are several rep-tiles that stretch ‘polygon’ to the limit. And some go beyond that, having infinitely many sides - but, hey, let’s be broad-minded.

More exotic rep-tiles.

The first 4-rep 4-gon in the first picture is also rep-9. Can you dissect it into nine copies of itself? As far as I am aware, every known rep-4 tile is also rep-9, but this has not been proved in general.

 

Now, I’m going to set the utilities puzzle (Cabinet, page 199; and Hoard, page 117) for the third time, with a new twist. Metaphorically and literally. Three houses have to be connected to three utility companies - water, gas, electricity. Each house must be connected to all three utilities. Can you do this without the connections crossing? Assume there is no third direction to pass pipes over or under cables, and you are not allowed to pass the connections through a house or a utility company building. Note: connections. No quibble-cooks (see page 116) allowed!

Connect houses to utilities, on a torus and a Möbius band.

What’s the difference this time? I’m not asking you to work in the plane. Try it on a torus (metaphorical twist) and a Möbius
band (literal twist). A torus is a surface with a hole, like a doughnut. A Möbius band is formed by joining the ends of a strip of paper with a half-twist (Cabinet, page 111).

By the way: mathematicians think of a surface like the Möbius band as having zero thickness, so that the utilities, houses and lines connecting them lie in it, not on it. But a real sheet of paper actually has two distinct surfaces, very close together. You can either think of the surface as being transparent, or (better) imagine that the lines are drawn on paper with ink that soaks through, so that everything is visible on both surfaces of the paper.
²⁸

If you don’t use this convention, then some of the lines in my answer end up on the back of the strip and don’t join up with the houses or utilities. You are then trying to solve the analogous problem on a cylindrical band with a double-twist. Topologically, this is the same as an ordinary cylindrical band, and in particular it has two distinct sides. Now there is no solution. Why not? A cylinder can be flattened out topologically in the plane, to form an annulus - the region between two circles. So any solution of the puzzle on a cylindrical band also provides a solution in the plane. But no solution in the plane exists without cooking (Cabinet, page 199).

 

Anyone who plays around with numbers soon notices that the consecutive integers 8 and 9 are both perfect powers (higher than the first power, of course). In fact, 8 is 2-cubed and 9 is 3-squared.

Are there any other positive whole numbers with this property—consecutive or not? (Powers higher than the cube are permitted, and strictly speaking 0 is not positive: it is non-negative. So this rules out 1
^m
- 0
ⁿ
= 1.) In 1844, the Belgian mathematician Eugène Catalan conjectured that the answer is no - that is, the Catalan equation
has only the above solutions in positive integers x and y when a and b are integers ≥ 2. In a mathematical publication known as Crelle’s Journal,
²⁹
he wrote: ‘Deux nombres entiers consécutifs, autres que 8 et 9, ne peuvent être des puissances exactes; autrement dit: l’équation
x
^m
-
y
ⁿ
= 1, dans laquelle les inconnues sont entières et positives, n’admet qu’une seule solution.’

The problem has a long history. The composer Philippe de Vitry (1291-1361) stated that the only powers of 2 and 3 that differ by 1 are (1,2), (2,3), (3,4) and (8,9). Levi ben Gerson (1288-1344) provided a proof that de Vitry was right: 3
^m
± 1 always has an odd prime factor if m > 2, so it cannot be a power of 2. By 1738, Euler had completely solved the equation
x
²
-
y
³
= 1 in whole numbers, proving that the only positive solution is x = 3,
y
= 2. But Catalan’s conjecture allows higher powers than the cube, so these earlier results were not sufficient to prove it.

In 1976, Robert Tidjeman proved that Catalan’s equation has only finitely many solutions; indeed, any solution must have x, y < exp exp exp exp 730, where exp
x
= e
^x
. However, this upper limit on the size is almost inconceivably gigantic - and in particular far too large for a computer search to eliminate all other potential solutions. In 1999, M. Mignotte proved that in any hypothetical solution, a < 7.15×10
¹¹
and b < 7.78×10
¹⁶
, but the gap is still too big for a computer to fill. A solution
seemed hopeless. But, in 2002, the mathematical world was stunned when the Romanian-born German mathematician Preda Mihailescu proved that Catalan was right, with a clever proof based on cyclotomic numbers - complex nth roots of 1. So the conjecture has now been renamed Mihailescu’s theorem.