Elementary number theory is largely about the * ring
of integers*, denoted by the symbol . The integers are an example of an algebraic
structure called an * integral domain*. This
means that satisfies the following axioms:

(a) has operations + (addition) and (multiplication). It is * closed*
under these operations, in that if , then
and .

(b) Addition is * associative*: If , then

(c) There is an * additive identity* : For all ,

(d) Every element has an * additive inverse*: If
, there is an element such that

(e) Addition is * commutative*: If , then

(f) Multiplication is * associative*: If , then

(g) There is an * multiplicative identity* : For all ,

(h) Multiplication is * commutative*: If , then

(i) The * Distributive Laws* hold: If , then

(j) There are * no zero divisors*: If and , then either
or .

* Remarks.*

(a) As usual, I'll often abbreviate to .

(b) The last axiom is equivalent to the *
Cancellation Property*: If ,
, and , then .

Here's the proof:

Since there are no zero divisors, either or . Since by assumption, I must have , so .

Notice that I didn't *divide* both sides of the equation by a
--- I cancelled a from both sides. This shows that division and
cancellation aren't "the same thing".

* Example.* If , prove
that .

Adding to both sides, I get

By associativity for addition,

Then using the fact that and are additive inverses,

Finally, 0 is the additive identity, so

* Example.* If , prove
that .

In words, the equation says that the additive inverse of n (namely
) is equal to . What
*is* the additive inverse of n? It is the number which gives 0
when added to n.

Therefore, I should add and see if I get 0:

By the discussion above, this proves that .

* Example.* Give an example of a set of objects
with a "multiplication" which is not commutative.

If you have had linear algebra, you know that matrix multiplication is not commutative in general. For instance, considering real matrices,

The integers are * ordered* --- there is a notion
of greater than (or less than). Specifically, for , is defined to mean that is a * positive integer*: an
element of the set .

Of course, is defined to mean . and have the obvious meanings.

There are several order axioms:

(k) The positive integers are closed under addition and multiplication.

(l) (* Trichotomy*) If , either , , or .

* Example.* Prove that if and , then .

, so is a positive integer. means is a positive integer, so by closure is a positive integer.

By a property of integers (which you should try proving from the axioms), . Thus, is a positive integer. So is a positive integer, which means that .

* Well-Ordering Axiom.* Every nonempty subset of
the positive integers has a smallest element.

Your long experience with the integers makes this principle sound
obvious. In fact, it is one of the deeper axioms for . Some consequences include the *
Division Algorithm* and the principle of *
mathematical induction*.

* Example.* Prove that is not a rational number.

The proof will use the Well-Ordering Property.

I'll give a proof by contradiction. Suppose that is a rational number. In that case, I can write , where a and b are positive integers.

Now

(To complete the proof, I'm going to use some divisibility properties of the integers that I haven't proven yet. They're easy to understand and pretty plausible, so this shouldn't be a problem.)

The last equation shows that 2 divides . This is only possible if 2 divides a, so , for some positive integer c. Plugging this into , I get

Since 2 divides , it follows that 2 divides . As before, this is only possible if 2 divides b, so for some positive integer d. Plugging this into , I get

This equation has the same form as the equation , so it's clear that I can continue this procedure indefinitely to get e such that , f such that , and so on.

However, since , it follows that ; since , I have , so . Thus, the numbers a, c, e,
... comprise a set of positive integers *with no smallest
element*, since a given number in the list is always smaller than
the one before it. This contradicts Well-Ordering.

Therefore, my assumption that is a rational number is wrong, and hence is not rational.

Copyright 2017 by Bruce Ikenaga