1 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 \((a+b)^2=a^2+2ab+b^2\)) 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.
1.1 Binary Operations
Definition 1.1 (Binary Operation) A binary operation \(\star\) 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 \(\star\colon S\times S\to S\). Whereas we often write functions \(f\colon S\times S\to S\) as \(f(a,b)\) for a binary operation we traditionally write the function name in the middle so \(a\star b\) instead of \(\star(a,b)\).
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 1.2 (Commutativity & Associativity) An operation \(\star\) is commutative if the order the elements are combined does not affect the outcome: for all elements \(a,b\in S\) \[a\star b = b\star a\]
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 \(a,b,c\in S\) \[(a\star b)\star c = a\star(b\star c)\]
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 \(\star\) to select the winning element of any pair. Thus, because paper beats rock, we have \(r\star p = p\). Explain why this is commutative, and find an example proving it is not associative.
Definition 1.3 (Identities & Inverses) Let \(S\) be a set with binary operation \(\star\). Then an element \(e\in S\) is an identity for the operation if it does not change any elements under combination. Formally, for all \(s\in S\)
\[e\star s = s\star e = s\]
Given a binary operation \(\star\) 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 \[x\star y = y\star x = e\]
This element \(y\) is called the inverse of \(x\). An operation \(\star\) is called invertible if every element of \(S\) has an inverse.
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 \(\left(\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\right)\) 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?
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 1.4 (Group) A group is a set \(G\) with an associative, invertible binary operation \(e\).
1.2 Fields
We’ve defined what a nice binary operation is. Numbers have two of these!
Definition 1.5 (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 1.6 (Field) A Field is a set \(\FF\) with two binary operations denoted \(+\) (addition) and \(\cdot\) (multiplication) satisfying the following axioms.
- (Commutativity) If \(a,b\in \FF\) then \(a+b=b+a\) and \(a\cdot b = b\cdot a\).
- (Associativity) If \(a,b,c\in\FF\) then \((a+b)+c=a+(b+c)\) and \((a\cdot b)\cdot c = a\cdot(b\cdot c)\)
- (Identities) There are special elements denoted \(0,1\in \FF\) where for all \(a\in \FF\), \(a+0=a\) and \(1\cdot a =a\).
- (Inverses) For every \(a\in\FF\) 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 \(a,b,c\in\FF\) then \(a\cdot (b+c)=(a\cdot b) + (a\cdot c)\)
Example 1.1 The rational numbers \(\QQ\) form a field, but the integers \(\ZZ\) do not, as they do not contain multiplicative inverses.
1.2.1 Shorthand Notation
We will work with fields and their operations throughout the course, so it is useful to introduce some shorthand notation that is familiar to us from previous mathematics classes, and put it on rigorous foundations in terms of the field axioms.
- Since addition and multiplication are associative, we will drop parentheses when three or more terms are combined using the same operation. That is, we will write \(a+b+c\) for both \((a+b)+c\) and \(a+(b+c)\) when convenient.
- We will adopt the convention that multiplication takes precedence over addition; that is, we drop parentheses in \((a\cdot b)+c\) to allow ourselves to write \(a\cdot b + c\); but we require parentheses to write \(a\cdot(b+c)\).
- We will denote multiplication by simple juxtaposition when convenient, dropping the \(\cdot\) symbol. That is, we will write \(ab\) for \(a\cdot b\) and \(a(b+c)\) for \(a\cdot (b+c)\).
- We use a bar \(\tfrac{a}{b}\) to denote multiplication by the inverse: that is \(a(b^{-1})\).
We also have a special shorthand for numerals, familiar to all
- The numerals \(0\) and \(1\) denote the special elements of any field guaranteed to exist by the axioms.
- We write \(2\) as a shorthand for \(1+1\).
- We write \(3\) as a shorthand for \(2+1\).
- We write \(4\) as a shorthand for \(3+1\).
For large integers, we use the base 10 system unless otherwise specified. That is, we interpret \(364\) as \(3\cdot 10^2 + 6\cdot 10^1 + 4\cdot 10^0\).
Exercise 1.1 (\(2+2\) and \(2\cdot 2\)) Prove, using only the field axioms and the definitions of the symbols \(0,1,2,3,4\) that \(2+2=4\) and \(2\cdot 2 =4\).
Show that \(2\) is the only natural number where \(x+x=x\cdot x\).
1.3 Elementary Computations
Example 1.2 (Multiplication by Zero) \[0x=0\]
To prove this for an arbitrary \(x\in \FF\), 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 \[\begin{align*} 0x &= (0+0)x\\ &= 0x+0x \end{align*}\]
Now, we can take the additive inverse of \(0x\) and add it to both sides:
\[0x+(-0x)=0x+0x+(-0x)\] This gives the additive identity \(0\) by definition on the left side, and cancels one of the factors of \(0x\) on the right, yielding
\[0 = 0x+0\]
Finally we use again that \(0\) is the additive identity to see \(0x+0=0x\), which gives us what we want: \[0x=0\]
Example 1.3 (The Zero-Product Property) Let \(a,b\) be elements of a field and assume that \(ab=0\). 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 \(ab=0\) by these to get
\[b^{-1}a^{-1}ab=b^{-1}a^{-1}0\]
On the left this simplifies to \(b^{-1}1b=b^{-1}b=1\) 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 1.4 (Additive Inverses and Negatives) \[-x = (-1)x\]
The definition of the symbol \(-x\) is the element of \(\FF\) which, when added to \(x\), gives \(0\). Thus, to prove that \(-x=-1x\) we want to prove that if you add \((-1)x\) to \(x\), you get \(0\). Since \(1\) is the additive identity, we know \(1x=x\) so we may write
\[x+(-1x)=1x+(-1x)\]
Using the fact that multiplication is commutative and the distributive law, we may factor out the \(x\): \[1x+(-1)x=(1+(-1))x\]
Now, by definition \(1+(-1)\) is the additive identity \(0\), so this is just equal to \(0x\). But by Example 1.2 we know \(0x=0\)! Thus
\[x+(-1x)=0\] And so \(-1x\) is the additive inverse of \(x\) as claimed. Thus we may write \(-x=(-1)x\)
Example 1.5 (Negative times a Negative) \[(-1)(-1)=1\]
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 1.2 (Negative of a Negative) For any \(x\in\FF\) we have \[-(-x)=x\]
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. 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 1.3 (The difference of squares) Prove that for any \(a,b\in \FF\) \[(a+b)(a-b)=a^2-b^2\]
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 1.4 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 1.5 Prove that fraction addition works by finding a common denominator: for any \(a,c\) and nonzero \(b,d\) \[\frac{a}{b}+\frac{c}{d}=\frac{ad+bd}{bd}\]
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.
1.4 Problems
Exercise 1.6 Fix some number \(r\neq 1\) in a field, and prove by induction that
\[1+r+r^2+\cdots + r^n = \frac{1-r^{n+1}}{1-r}\]
FURTHER TOPICS
Other Fields (define binary field just with symbols 0,1: show it satisfies axioms)
Show R2 is not a field
Field of Fractions
1.4.1 Base Systems
Euclidean division with remainder
Existence of base 10 expansion for any natural number
Base 5, base 2, other bases.
Examples finding base 16 expressions.