One possible answer to this famous question of Kronecker is "solutions to equations." For example, the negative integers, which were so confusing to Renaissance mathematicians that they were sometimes called "imaginary numbers," could be defined as solutions to equations like
We would say the solution is x = -6, but if you weren't comfortable with the notion of "-6", you could just say it is a new symbol defined by that equation. You just carry it along as if it were a number, and if you ever run into the combination "(-6)+6", you can replace it by "0". We could compute (-6)+7=(-6)+6+1=0+1=1, for example, meaning (-6)+7=1. This shows how to add negative integers, etc. Every property of (-6) follows from the equation above that it satisfies. So a way of introducing new, interesting numbers is by equations. The idea is not to solve the equations, but to define the numbers! New types of equations give rise to new numbers.
The equation
defines a new number, which we usually call 5/7. All we really need to know about this number, though, is that it satisfies the above relation. The arithmetic of fractions is entirely determined when you say that they are quantities defined by relations of the form
for integers p and q, meaning, of course, the fraction we usually call p/q. Suppose we had two fractions x and y, about which we know only
Then what would we mean by xy? by x+y? Well, since you can multiply equals by equals and the results are still equal, we find
That shows that xy is another new number of the same type (i.e., a fraction) and defines it in the same way as the others, by the relation that it satisfies. We could multiply the defining relations for x and y by 8 and 7 respectively and find
Then, adding,
and this defines x+y in the same way that other fractions are defined, by a relation, or equation, that it satisfies. In particular, when you add two fractions, you get a fraction. This is, of course, just the usual way that one adds fractions, but slightly disguised, so that it takes a little thought to see that it is the same.
Other numbers we have thought about are also defined by equations. For example the square root of 2, the first irrational number to be discovered is defined as (one of the) solutions of
We can just introduce it as a new symbol, and if it ever occurs squared, we can use the above relation and replace its square by the integer 2. That's all the usual symbol is anyway!
We know of lots of irrational numbers now, all the square roots of integers that aren't perfect squares, for example. Of course that is only a countable infinity, and we know the irrationals are not countable. One can go further and introduce new and more complicated relations, which define new numbers, for example
We could say we define a new number c which satisfies that relation. All the numbers you can get in this way, i.e. numbers which satisfy polynomial relations (where the polynomials have integer coefficients) form a nice set of numbers, called the algebraic numbers. Like the rationals, if you add two algebraic numbers you get another algebraic number, and if you multiply two algebraic numbers you get another algebraic number. You can even divide (just like the rationals) and of course there are many more algebraic numbers than rationals, because the rationals satisfy a polynomial equation of degree 1, but the algebraic numbers just satisfy polynomial equations of some degree, and there are many more such polynomials -- or are there??? As you might have suspected, Cantor proved that the algebraic numbers are countable! That means the non-algebraic numbers are uncountable. The usual name for non-algebraic numbers is transcendental numbers. About the only familiar example of a transcendental number is "pi", and it was only proved to be transcendental a little over a hundred years ago. And yet, by the above arguments, almost all real numbers are transcendental. These are precisely the numbers we cannot write down exactly, and cannot even describe by equations! It's a very peculiar result about the number line -- most of it is, in a way, "beyond" us.
Now that the number line is thoroughly understood, -- i.e., since the invention of calculus, and the more or less satisfactory resolution of the puzzles that it raised -- number theory has changed. It was essential to understand the real numbers -- the number line -- but that part of the subject is essentially complete, and forms the basis for subjects like calculus. Meanwhile, there are many questions involving the integers and the rationals which are still unsolved. Nowadays "number theory" is more likely to be about such questions. A polynomial equation always has solutions, especially if you use the equation to define the numbers -- then it has solutions by definition! But it may not have integer solutions. If the solutions are required to be integers, then the equation is called a Diophantine equation, in honor of Diophantus, a late Hellenistic mathematician, active at Alexandria.
A famous example of a Diophantine equation is
where the word "Diophantine" means we seek a solution in integers (x,y,z). As we already noted once, there are integer solutions, called "Pythagorean triples". They are geometrically interesting, because they are the sides of a right triangle with commensurate sides (no irrational sides). There are trivial solutions as well, in which x or y is 0, but those are not considered Pythagorean triples, and of course with one side of length zero, they don't make a triangle either. A Pythagorean triple must be a non-trivial solution, corresponding to a real right triangle: the smallest one is (3,4,5), since 9+16=25. Two more are (5,12,13) and (8,15,17). Another one is (6,8,10), but you notice right away that this one is just (3,4,5) scaled up by a factor of 2, which is really the same triangle again, just measured in different units, or what is called in geometry a similar triangle, so it isn't really new. To avoid getting essentially the same triangles over and over, it is customary to single out the primitive Pythagorean triples, which are Pythagorean triples that have no common factor. In the example above, (6,8,10) is not primitive, because the factor 2 is common to all sides. If we divide through by all common factors, we get a primitive Pythagorean triple.
The familiar notion of even and odd can be extremely useful in Diophantine equations! We don't usually get to use this idea, because in most equations the solutions will be non-integers, so even and odd wouldn't make any sense, but in Diophantine equations everything is an integer, so the special properties of integers, like being either even or odd, come into play. Let us see how this works in the Pythagorean equation above. You might have noticed that all the primitive triples we wrote down had exactly two odd numbers and one even one. Was that just an accident? Let us see, first recalling how evens and odds behave when you add and multiply:
| + | even | odd |
| even | even | odd |
| odd | odd | even |
| * | even | odd |
| even | even | even |
| odd | even | odd |
This means, for example that even+even=even, odd+odd=even, odd*even=even, odd*odd=odd, etc.
If we look at the Pythagorean equation and ask for a primitive triple, we realize the (x,y,z) can't all be even, because that would mean they share a common factor of 2. Thus at least one of them is odd. Now even*even=even, and odd*odd=odd, so if x is odd, then x^2 is also odd, and if x is even, x^2 is also even, etc. But that means it is impossible for only one of them to be odd. If only x were odd, we would have odd+even=even, which is NOT TRUE, and if only z were odd, we would have even+even=odd, also NOT TRUE. Also all three of them odd is impossible, because it would require odd+odd=odd, which is NOT TRUE. Therefore exactly 2 odd numbers in a Pythagorean triple is the only possibility, just what the examples had already suggested.
In noticing this, we haven't solved the problem of finding more primitive Pythagorean triples, or even shown that there ARE any more, but we have found a property that they MUST HAVE if they exist. With not much effort, we have gotten a kind of partial result, a quick insight. That is frequently how "even" and "odd" work.
The notions of even and odd are so easy and so useful, it is natural to ask if there are generalizations of this idea that could work the same way, and give more information about Diophantine equations. Since "even" just means "divisible by 2," we could invent a notion of "threven," meaning "divisible by 3." The other numbers would be "throdd." Is there a similar behavior of these numbers? Multiplication turns out to work the same way:
| * | threven | throdd |
| threven | threven | threven |
| throdd | threven | throdd |
as you can quickly see: when you multiply two integers, the product will have 3 as a factor if either of the integers did, otherwise it will not. But addition does not work! It would go like this:
| + | threven | throdd |
| threven | threven | throdd |
| throdd | throdd | threven??/throdd?? |
It fails in the last box of the addition table. When you add throdd + throdd, you can't tell for sure what you will get. Either it will be like 4+1=5 (result is throdd), or like 4+2=6 (result is threven). Thus there really isn't a good addition law. Since the result could (apparently) be anything, it will never tell us anything. The whole idea seems to fail.
If you look in more detail, with a finer magnifying glass, so to speak, you see something that explains what is going on. In 4+1=5, each of 4 and 1 is one more than a multiple of 3. (Note zero is a multiple of 3!) When you add them you get 5, which is two more than a multiple of 3. But in 4+2=6, we have 4 (one more than a multiple of 3) and 2 (two more than a multiple of 3) adding to something that is three more than a multiple of 3, i.e., something which is again a multiple of 3 (namely 6). We have to realize that there are two different kinds of "throdd" numbers, those like {1,4,7,10,13,16,19,22,...} which are 1 more than a multiple of 3, and others like {2,5,8,11,14,17,20,23,...} which are 2 more than a multiple of 3. It matters which set the throdd number is in. So let us define three sets instead of two,
and now work out the addition and multiplication tables: that is your problem. I will lay it out as tables for you fill in:
| + | threven | throdd_1 | throdd_2 |
| threven | |||
| throdd_1 | |||
| throdd_2 |
| * | threven | throdd_1 | throdd_2 |
| threven | |||
| throdd_1 | |||
| throdd_2 |