## Multiplicative subsets of the Natural Numbers.

“Vert” should be another tog, I’ll come up with a definition later I suppose.

OK, here’s some terminology; a multiplicative subset of the natural numbers is one where if i and j are both in it then so is i.j. An additive subset is one where if i and j are in it then so is i+j. Obviously all additive and multiplicative subsets of the naturals have infinite cardinality. They are all elements of the power set of the natural numbers.

Apparently, there are uncountably many multiplicative subsets of the natural numbers. I’m not entirely convinced. There are at least countably many, since {the numbers greater than n} form a countably infinite family of multiplicative subsets. There are only countably many additive subsets of the naturals, however. Additive implies multiplicative since multiplication is just iterated addition. A multiplicative subset is a member of the power set of N. But there are uncountably many M-sets, so there must be a bijection between **P(N)** and M-sets. But there are infinitely many subsets of **N** that are not multiplicative…In fact for any M-set there is an infinite family of subsets which are not multiplicative. (i.e. for some set M, any set of the form M\{x} where x is in M and x=i.j some i,j in M) Are there really only countably many non-multiplicative subsets of **N**? I know infinity is thoroughly weird, but I’m having trouble with this.

The easiest explanation is that my Symbolic Logic notes are faulty. I’ll email Mr. Dude about that later…

In other news that question I was pondering a while ago about the set of all circles with a given arc as a chord has been solved! Well, we think that the answer is that the only parts of the plane not in the set are points on a line which is an extension of the chord.

Speaking of maths problems I’ve pondered before I realised that there are several other ways to define the sierpinski gasket, some of which were mentioned in a talk Ian Stewart gave at a maths society event last term… I’ll write some more about that later, right now it’s Simpsons O’clock!

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