3 Operations

Highlights of this Chapter: We begin axiomatizing the real numbers by axiomatizing their operations of addition and multiplication, leading to the field axioms. We give careful definitions of various notations from arithmetic, and do several example calculations (including a proof that $2 + 2 = 4$ and $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ ) to exhibit that all arithmetical facts are consequences of the field axioms.

The first step to axiomatizing numbers is to give a precise description of addition, subtraction, multiplication and division. These operations naturally group into two pairs (addition/subtraction as well as multiplication/division) of operation/inverse, so first we will formalize the notion of an invertible operation. Furthermore, the two operations are related to one another by the distributive law. Two invertible operations bonded together by the distributive law form a mathematical structure we call a field, which is what we axiomatize in this chapter.

Definition 3.1 (Binary Operation) A binary operation $⋆$ on a set $S$ is a rule that takes any two elements of $S$ and combines them to make a new element of $S$ .

Formally, this is a function $⋆ : S \times S \to S$ . Whereas we often write functions $f : S \times S \to S$ as $f (a, b)$ for a binary operation we traditionally write the function name in the middle so $a ⋆ b$ instead of $⋆ (a, b)$ .

Example 3.1 Addition is a binary operation on the natural numbers, integers, rationals, and real numbers. Subtraction is a binary operation on the integers, but not on the natural numbers, as $4 - 7 = - 3$ gives an element not in the original set.

Definition 3.2 (Commutativity and Associativity) An operation $⋆$ is commutative if the order the elements are combined does not affect the outcome: for all elements $a, b \in S$ $a ⋆ b = b ⋆ a$

An operation is associative if combinations of 3 or more terms can be re-grouped at will (not changing the order), without affecting the outcome: for all $a, b, c \in S$ $(a ⋆ b) ⋆ c = a ⋆ (b ⋆ c)$

Example 3.2 (Commutativity and Associativity) The operation of addition is commutative and associative, but the operation of subtraction is neither. The operation of matrix multiplication is associative, but is not commutative in general.

An operation which is commutative but not associative is given by the children’s game rock paper scissors: if $S = {r, p, s}$ we may define the operation $⋆$ to select the winning element of any pair. Thus, because paper beats rock, we have $r ⋆ p = p$ . Explain why this is commutative, and find an example proving it is not associative.

Definition 3.3 (Identity Element) Let $S$ be a set with binary operation $⋆$ . Then an element $e \in S$ is an identity for the operation if it does not change any elements under combination. Formally, for all $s \in S$

$e ⋆ s = s ⋆ e = s$

Given a binary operation $⋆$ on a set $S$ with identity $e \in S$ , an element $x \in S$ is invertible if it can be combined with something to produce the identity. That is, if there exists a $y \in S$ with $x ⋆ y = y ⋆ x = e$

This element $y$ is called the inverse of $x$ . An operation $⋆$ is called invertible if every element of $S$ has an inverse.

Example 3.3 (Identity Element) Zero is the identity of the operation of addition, 1 is the identity of multiplication (in any familiar number system you’d like to take as an example). The identity matrix $(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})$ is the identity of $2 \times 2$ matrix multiplication.

Not all operations have an identity. Can you see why there is no identity operation for exponentiation $x^{y}$ on the positive integers?

Example 3.4 (Inverse) The operation of addition is invertible, and its inverse is subtraction. The operation of multiplication is not invertible, because the number $0$ does not have an inverse (you can’t divide by zero! We’ll prove this soon)

Definition 3.4 (Group) A group is a set $G$ with an associative, invertible binary operation $e$ .

The concept of a group is ubiquitous in mathematics, as it formalizes the idea of a nice binary operation. But for analysis, we need more than this: numbers come with two binary operations (addition and multiplication) and so we need to describe how they interact.

Definition 3.5 (The Distributive Law) Let $S$ be a set with two commutative binary operations $+, \cdot$ . Then $\cdot$ distributes over $+$ if for all $a, b, c \in S$ we have $a \cdot (b + c) = (a \cdot b) + (a \cdot c)$

Definition 3.6 (Field) A Field is a set $F$ with two binary operations denoted $+$ (addition) and $\cdot$ (multiplication) satisfying the following axioms.

(Commutativity) If $a, b \in F$ then $a + b = b + a$ and $a \cdot b = b \cdot a$ .
(Associativity) If $a, b, c \in F$ then $(a + b) + c = a + (b + c)$ and $(a \cdot b) \cdot c = a \cdot (b \cdot c)$
(Identities) There are special elements denoted $0, 1 \in F$ where for all $a \in F$ , $a + 0 = a$ and $1 \cdot a = a$ .
(Inverses) For every $a \in F$ there is an element $- a$ such that $a + (- a) = 0$ . If $a \neq 0$ , then there is also an element $a^{- 1}$ such that $a \cdot a^{- 1} = 1$ .
(Distributivity) If $a, b, c \in F$ then $a \cdot (b + c) = (a \cdot b) + (a \cdot c)$

3.0.1 Shorthand Notations

There is a lot of notation that we use to simplify writing out basic arithmetic expressions in fields. I will attempt a list of these shorthands here. First, some relating to the operations themselves.

We often write the operation of multiplication simply as juxtaposition, without any intervening symbol. That is, we write $a b$ instead of $a \cdot b$ .
We make the convention that multiplication precedes addition, so we evaluate the expression $a b + c$ as $(a \cdot b) + c$ not $a \cdot (b + c)$ . This allows us to drop a lot of parentheses, making things easier to read.

Next, some notation for certain elements: - We define the symbol $2$ to be the result of the operation $1 + 1$ . - We define the symbol $3$ to be the result of the operation $2 + 1$ - We define the symbol $4$ to be the result of the operation $3 + 1$ - We define the symbol $5$ to be the result of the operation $4 + 1$ - Etc

This lets us easily write down repeated addition, as we will see below $3 x = x + x + x$ for any $x$ . It’s also useful to have some notation for repeated multiplication, which we denote with powers

The notation $x^{2}$ will mean the product $x \cdot x$ .
The notation $x^{3}$ will mean the product $x \cdot x \cdot x$ .
Etc…
The notation $x^{0}$ will denote the multiplicative identity $1$ .

We also introduce another notation for multiplicative inverse, to make formulas more readable:

If $c \neq 0$ we write $\frac{a}{c}$ for $a c^{- 1}$ .

3.0.2 Computations in Fields

Example 3.5 $2 x = x + x$

To prove this for an arbitrary $x \in F$ , we recall the definition $2 = 1 + 1$ and use the distributive property: $\begin{aligned} 2 x & = (1 + 1) x \\ = 1 x + 1 x \\ = x + x \end{aligned}$ Finally the last equality follows as $1$ is the multiplicative identity, so

Example 3.6 $0 x = 0$

To prove this for an arbitrary $x \in F$ , recall that $0$ is the additive identity so for any field element $c$ , we have $0 + c = c$ . Thus, when $c = 0$ we have $0 + 0 = 0$ . We can use this together with the distributive property to get $\begin{aligned} 0 x & = (0 + 0) x \\ = 0 x + 0 x \end{aligned}$

Now, we can take the additive inverse of $0 x$ and add it to both sides:

$0 x + (- 0 x) = 0 x + 0 x + (- 0 x)$ This gives the additive identity $0$ by definition on the left side, and cancels one of the factors of $0 x$ on the right, yielding

$0 = 0 x + 0$

Finally we use again that $0$ is the additive identity to see $0 x + 0 = 0 x$ , which gives us what we want: $0 x = 0$

Example 3.7 (The Zero-Product Property) Let $a, b$ be elements of a field and assume that $a b = 0$ . Then either $a = 0$ or $b = 0$ .

We assume that both $a$ and $b$ are nonzero, and see that we reach a contradiction. Since they’re nonzero, they have multiplicative inverses $a^{- 1}$ and $b^{- 1}$ , so we may multiply both sides of $a b = 0$ by these to get

$b^{- 1} a^{- 1} a b = b^{- 1} a^{- 1} 0$

On the left this simplifies to $b^{- 1} 1 b = b^{- 1} b = 1$ by definition, and on the right this becomes $0 (b^{- 1} a^{- 1}) = 0$ by the previous example. Thus, we’ve proven $0 = 1$ ! So this could not have been the case, and either $a$ or $b$ must have not been invertible to start with - they must have been zero.

Example 3.8 $- x = (- 1) x$

The definition of the symbol $- x$ is the element of $F$ which, when added to $x$ , gives $0$ . Thus, to prove that $- x = - 1 x$ we want to prove that if you add $(- 1) x$ to $x$ , you get $0$ . Since $1$ is the additive identity, we know $1 x = x$ so we may write

$x + (- 1 x) = 1 x + (- 1 x)$

Using the fact that multiplication is commutative and the distributive law, we may factor out the $x$ : $1 x + (- 1) x = (1 + (- 1)) x$

Now, by definition $1 + (- 1)$ is the additive identity $0$ , so this is just equal to $0 x$ . But by Example 3.6$ we know $0 x = 0$ ! Thus

$x + (- 1 x) = 0$ And so $- 1 x$ is the additive inverse of $x$ as claimed. Thus we may write $- x = (- 1) x$

Example 3.9 $(- 1) (- 1) = 1$

This is an immediate corollary of the above: we know that $(- 1) x$ is the additive inverse of $x$ , and so $(- 1) (- 1)$ is the additive inverse of $- 1$ . But this is just $1$ itself, by definition!

Exercise 3.1 For any $x \in F$ we have $- (- x) = x$

Exercise 3.2 Prove, using only the field axioms and the definitions of the symbols $0, 1, 2, 3, 4$ that the following is true: $2 + 2 = 4$

Example 3.10 $2 \cdot 2 = 4$

This is a corollary of ?exr-2-plus-2 above, as using the distributive law we see $2 \cdot 2 = 2 \cdot (1 + 1) = 2 \cdot 1 + 2 \cdot 1 = 2 + 2$ And we already know $2 + 2 = 4$

All of the standard arithmetic “rules” learned in grade school are consequences of the field axioms, and so you are welcome to use all of them in this course, without comment. However, to feel justified in doing this, its good to prove a couple of them yourself, to convince yourself that you could in fact trace and any all such manipulations back to the rigorous axioms we laid down.

Exercise 3.3 (The difference of squares) Prove that for any $a, b \in F$ $(a + b) (a - b) = a^{2} - b^{2}$

In your proof you may use the field axioms, the notational shorthands, and any of the example properties proved above in the notes. Anything else you need, you should prove from this.

Exercise 3.4 Prove, using the field axioms and our notational shorthands, for any $a, b$ and $c \neq 0$ $\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}$

Exercise 3.5 Prove that fraction addition works by finding a common denominator: for any $a, c$ and nonzero $b, d$ $\frac{a}{b} + \frac{c}{d} = \frac{a d + b d}{b d}$

In your proof you may use the field axioms, the notational shorthands, and any of the example properties proved above in the notes. Anything else you need, you should prove from this.