$1,000,000 problem :: Riemann :: Two ways to skin an imaginary cat  
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I am going to introduce you to some new sorts of numbers. We shall think of them as different families of numbers. Some of these new families won't have unique prime building blocks. We are very familiar with the family consisting of the ordinary numbers …-3,-2,-1,0,1,2,3… But there are other families of numbers which mathematicians actually stumbled on in their endeavours to solve Fermat's Last Theorem. Consider for example all numbers of the form A+B2 where A and B are whole numbers. (When I write B2 this is the number B times 2.) So this family includes a number like 1+32 which is approximately 5.24264… But the special character of these numbers is more transparent if we stick to writing them as 1+32.

What makes a collection of numbers like this particularly nice is that I can add or multiply them together and I won't get any numbers which are outside the family. For example
(1+2)+(3+42)=4+52
(1+2)x(3+42)=(3+2x(42)) +(4+3)2=11+72.
In this last line I've slipped in a bit of algebra which requires you to know that (A+B)x(C+D)=AxC+BxD+AxD+BxC. This is as complicated as any of the algebra we'll need is going to get. Don't despair if the sight of such equations makes your eyes glaze over. It justifies that we really won't get any numbers outside of our family of numbers like A+B2.

So these numbers work like ordinary numbers in that we can add and multiply them together. They also have a concept of prime building blocks like ordinary numbers and in this family built from 2 the building blocks of each number are unique. But if we change the square root of 2 to something different, suddenly we find examples where numbers can be written in two different ways as products of prime building blocks.

To find such a family, let's turn the screw one more time. Consider the family of numbers of the form A+B-6 where A and B are whole numbers. Notice that these are now imaginary numbers

Using Gauss’s map of the imaginary numbers, we now have a picture where we can "see" our new numbers like 3+-6. This number is the point reached by moving 3 steps east-west followed by 6 (which is approximately 2.44949) moves in the north-south direction. We are using here that
-6=6 x v-1, or more generally that (AxB)=(A) x (B). So the imaginary bit -6 is broken up into a normal number 6 and this new imaginary number i=-1 which we interpret as heading one unit north in our expanded worldview of numbers.

How do we do arithmetic with these numbers? Addition is straightforward. Each number A+B-6 we can think of as a set of directions in this imaginary world: A steps east-west, followed by (B times 6) moves in the north-south direction. If I gave you two sets of directions A+B-6 and C+D-6 you do one followed by another and end up at a third point which you could have got to by being given the one direction (A+C)+(B+D)-6. This number is then the sum of the two original numbers. So for example here is a picture of (3+-6)+(1+2-6)=(4+3-6):

Multiplication is a little trickier. We can take the algebraic route offered by the way we saw the numbers A+B2 being multiplied together:

(A+B-6)x(C+D-6)
=(AC-6BD)+(AD+BC) -6.

For those interested in the algebra, note we are using that -6 times -6 is -6 since this was the definition of this imaginary number.

Notice though that this family of numbers of the form A+B-6 doesn’t get any bigger when we add or multiply them together.

The important thing for us now about these new sort of numbers A+B-6 where A and B are whole numbers is that we have two essentially different ways to write numbers as products of building blocks. An easy example is
(-6)x(-6)=-6= -2x3.
But maybe we can crack these numbers down further. For example in ordinary numbers consider the case of 14x10=140=4x35. In this case we hadn't hit the prime building blocks yet. One more step and both sides of the equation gave us 2,2,5 and 7. However unlike the case of 140, -6, 2 and 3 are all numbers which can't be split any further if we restrict ourselves to using numbers of the form (A+B-6) where A and B are whole numbers.

This is not so obvious given that we have allowed ourselves more possibilities for how to split numbers. After all, with more room to manoeuvre in this two dimensional picture, perhaps we can split these numbers further. Actually a picture will help here. These new numbers we think of as a collection of points in our two dimensional picture, where each number is got by going a certain number of units east-west then north-south. Lets mark all the points which are close to the 0 point in our picture:

Take one of the three points -2, 3 and -6 in this picture, for example -6. Let's show that it is an indivisible building block. Is there any way you could find two numbers closer to 0 which multiplied together got you to -6? A quick check inside the circle through -6 soon reveals that it is impossible to get to -6 by multiplying two other numbers closer to zero in this picture.

Hence -6 must be an indivisible building block in this set of numbers. The same analysis will reveal -2 and 3 are also indivisible. This makes them building blocks in this set of numbers. (Mathematicians won't call these prime building blocks because they only like to give them the name prime if there is unique way to decompose numbers. Mathematicians refer to these indivisible building blocks as irreducible numbers.)

Here is a picture of some the building blocks in this family of numbers A+B-6. Like the ordinary primes not a particularly orderly picture. In fact you might be forgiven for thinking this picture was produced by some fractal formula.

If you were brave enough to negotiate the proof that normal numbers do have unique sets of prime building blocks then the test of your understanding is to follow the same argument for our new set of numbers which doesn't have unique factorisation. Where does the argument break down? What do we mean by a smallest rogue here? Well, one nearest to the starting point 0 might suffice. But look back at how we made our rogue smaller in that proof. We chose p(1) to be the smallest prime appearing in either expression for the smallest rogue number N. Then we waved our wand and considered (q(1)-p(1))xq(2)x…xq(S), a number which was smaller than N. But in our two dimensional picture this magic doesn't work. Remember we had
(-6)x(-6)=-6= -2x3.
The number 2 is the nearest point to 0. But if we walk two paces east or west of the point at -6 we always move further away from 0. So in this two-dimensional world, we can't seem to shift down to cases we've already covered.

The fact that certain types of numbers like the family A+B-6 don't have a unique way to dismantle them into building blocks turned out to be one of the major hurdles in proving Fermat's Last Theorem. If it had been true it would have been Cauchy, Lamé or Kummer in the 19th century and not Wiles who picked up the coveted scalp of proving Fermat's tantilizing conjecture. Indeed some believe that this might be behind Fermat's belief that he had a proof of his Last Theorem, that he had mistakenly believed that all these numbers had unique factorisation.

Mathematicians have a way of giving families of numbers a score according to how badly they fail to have unique factorization into building blocks. The higher the score, the worse they behave. For example, according to this measure the family A+B-6 scores a 2.

Those well-behaved families which have unique factorisation like our ordinary numbers get the coveted score of one. But were there any such families beyond the ordinary numbers? Gauss made the remarkable discovery that these families have something to do about the special equations that Euler had discovered which produced so many primes. Euler was amazed to find that the equation X2+X+41 produced a list of 40 primes when fed the numbers from 0 to 39. The same happened if he took instead of 41, the equation X2+X+Q where Q=2,3,5,11 or 17. Gauss realised that the same Q could be used to build families of numbers with unique factorization into building blocks. He showed that like the ordinary numbers, the family of numbers A+B(1-4Q) had unique building blocks. (There is a slight twist here. The numbers A and B should be not just whole numbers but can also both be whole numbers plus 1/2, e.g. A=1+1/2 and B=3+1/2. Why? Because it turns out that multiplying and adding this larger family still doesn’t get you outside this family.) Gauss conjectured that in addition to these six families, the only other families of numbers A+B-D to score a one for uniquely factorising were D=1,2 and 3. Remarkably it took over 160 years to prove that Gauss was right. Between 1966 and 1967 the Cambridge mathematician Alan Baker and the American Harold Stark proved that Gauss had indeed found the only families of numbers A+B-D that scored a one for uniquely factorising.

Answering a problem that goes back to Euler or Gauss is generally considered as a true mark of a mathematician's metal. If you can knock off one of Hilbert's 23 problems too before the age of 40 then you're a dead cert for a Field's medal as Baker discovered. Baker had found a way to generalize Gelfond and Schneider's solution to the seventh problem on Hilbert's list. Hilbert had mistakenly believed that the seventh would outlast both the Riemann Hypothesis and Fermat's Last Theorem.

 
2.2.2.4 Any other new numbers? >
 
  Advanced sections
  2.2.2.1 The geometry of arithmetics...
  2.2.2.2 Why is the concept of the square root of -1...
  2.2.2.3 Two ways to skin an imaginary cat
  2.2.2.4 Any other new numbers?
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