# Summary

We started today by getting to know the policies and expectations in the course. All of this is available already on the syllabus, but if you have any questions don't be shy about moc.liamg|ztluhcs.c.werdna#ydnA gniliame. We also spent some time introducing ourselves briefly; this will be continued as you post your own profiles for Homework 0. Afterwards, we started talking about the basics in number theory, starting with the axioms. We finished by introducing the notion of divisibility for the integers.

# The Axioms of Number Theory

When trying to build a mathematical discipline from the ground up, one needs to describe the fundamental objects and operations in the discipline and then define the basic properties these objects will obey. These properties are called axioms, and they are the "ground rules" the objects and operations must satisfy. With axioms in place, one can then start proving theorems by manipulating the axioms.

In number theory, the basic objects of interest are integers. You might know these objects as whole numbers. In this class we'll denote the set of all integers as $\mathbb{Z}$:

(1)The basic operations we have on the integers are addition and multiplication; notice that we don't include subtraction, since it's just a special case of addition. More important, we've avoided division since division doesn't behave very well on the integers: the quotient of two integers is frequently not another integer. We also have basic tools for comparing integers, namely equality and inequality. The operations and comparisons will follow the usual notation that you're used to, so I won't bother reintroducing them.

With the fundamental objects, operations and comparisons in place, we can start writing down the basic rules they all satisfy. Here's the list that we were able to come up with in class:

- for any $a \in \mathbb{Z}$$a = a$ (reflexivity of equality);
- for any $a,b,c \in \mathbb{Z}$, if $a=b$ and $b=c$ then $a=c$ (transitivity of equality);
- for any $a,b \in \mathbb{Z}$, $a=b$ implies $b=a$ (symmetry of equality — I think we forgot this one in class);
- for any $a,b \in \mathbb{Z}$, the sum $a+b$ is an integer (closure under addition);
- for any $a,b \in \mathbb{Z}$, the product $ab$ is an integer (closure under multiplication);
- for any $a,b,c \in \mathbb{Z}$, $(a+b)+c = a+(b+c)$ (associativity of addition);
- for any $a,b,c \in \mathbb{Z}$, $(ab)c = a(bc)$ (associativity of multiplication);
- for any $a,b,c \in \mathbb{Z}$, $(a+b)c = ac + bc$ (distributivity);
- for any $a,b \in \mathbb{Z}$, $a + b = b+a$ (commutativity of addition);
- for any $a,b \in \mathbb{Z}$, $ab = ba$ (commutativity of multiplication);
- for any $a \in \mathbb{Z}$, $a+0 = a$ (additive identity);
- for any $a \in \mathbb{Z}$, $a\cdot 1 = a$ (multiplicative identity);
- for any $a \in \mathbb{Z}$ there exists $-a \in \mathbb{Z}$ so that $a + (-a) = 0$ (additive identity);
- for any $a,b,c \in \mathbb{Z}$ with $a \neq 0$, then $ab = ac$ implies $b =c$ (cancellation of multiplication);
- for any $a,b,c \in \mathbb{Z}$, $b=c$ implies $ab = ac$ (substitution for multiplication);
- for any $a,b,c \in \mathbb{Z}$, $b=c$ if and only if $a+b = a+c$ (substitution for addition; cancellation of addition);
- for any $a \in \mathbb{Z}$, exactly one of the following is true
^{(1)}$a<0$^{(2)}$a=0$^{(3)}$a>0$ (Trichotomy law); - for any $a,b,c \in \mathbb{Z}$, if $b<c$ then $a+b<a+c$;
- for any $a,b,c \in \mathbb{Z}$, if $b<c$ and $a>0$, then $ab<ac$;
- for any $a,b,c \in \mathbb{Z}$, if $b<c$ and $a<0$, then $ab>ac$; and
- if $S$ is a nonempty set of positive integers, then $S$ has a least element (Well ordering principle).

The last axiom — the well ordering principle — probably sticks out as the ugly duckling of the bunch, and it certainly isn't one which most people think of when rattling off basic properties of the integers. It is, however, essential to what we'll be doing in class, as it is logically equivalent to mathematical induction — a tool that we'll be using with some frequency in this course.

## Playing around with the axioms

Our list of axioms is almost certainly a little redundant, meaning that we could probably prove some of the axioms we've listed in terms of the other axioms. In this sense, it doesn't pass the usual mathematical aesthetic. We did manage, however, to excise one of the axioms from the book and prove it as a consequence of the axioms we wrote down.

Specifically, we proved

Theorem: For any $a \in \mathbb{Z}$, $a\cdot 0 = 0$

Proof: We started by noting that

(2)we were able to conclude this by using the additive identity property together with substitution for multiplication. Using the distributive property we then have

(3)adding the additive inverse of $a\cdot 0$ to both sides of the equation using the additive substitution property then gives

(4)which becomes $a\cdot 0 = 0$ after applying the additive identity property.$\square$

As you can see, proofs which are born straight from axioms require a lot of care. In practice, we won't be so explicit when we use the axioms above in class; chances are we won't mention them, or that we'll apply several of these axioms at once without comment. This shouldn't be too bothersome, since you are more accustomed to manipulating these axioms then perhaps you realize.

# Putting the Elementary in Number Theory

With the basic ground rules set, we had a chance to talk about the most important property of integers in this whole class: **divisibility**. It is the study of this property which makes the number theory we'll study "elementary." One can think of divisibility as the attempt to carry division into the realm of the integers, made appropriately cautious to reflect the fact that the integers don't always behave so well under division.

An integer

dis said to divide an integeraif there exists an integerqso thata = dq. Ifddividesawe write $d \mid a$, and ifddoes not divideathen we write $d \nmid a$.

## An example

This definition should agree with your own intuitive notion of divisibility in the integers, so hopefully it isn't too surprising. To see an example in action, notice that $2 \mid 18$ since we can find an integer *q* to solve the equation

in this case, the integer *q* is simply *9*.$\square$

## A non-example

Let's try to prove that $2 \nmid 5$. For this, we need to show that we cannot find an integer *q* satisfying the equation

For this notice that $5 > 2\cdot 2$; for any integer *q* satisfying $2 \geq q$ we have $2\cdot 2 \geq 2\cdot q$ (a slight modification on one of our axioms), and hence we have

Likewise we know that $5 < 2 \cdot 3$, and for all integers *q* satisfying $q \geq 3$ we get

Since all integers fall into one of the categories we have described, we conlude that $5 \neq 2\cdot q$ for any integer *q*, and so $2 \nmid 5$.$\square$