2.5 Subtraction and Division

2.89   Definition (Subtraction.) In any field $F$, we define a binary operation $-$ (called subtraction) by

$$\mbox{ for all } x,y\in F \qquad x-y=x+(-y).$$

Unfortunately we are now using the same symbol $-$ for two different things, a binary operation on $F$, and a symbol denoting additive inverses.

2.90   Exercise (Distributive laws.) A Let $F$ be a field, and let $a,b,c\in F$. Prove that

1. [a)] $a\cdot(b-c)=(a\cdot b)-(a\cdot c).$
2. [b)] $-(a-b)=b-a.$

2.91   Definition (Division.) Let $F$ be a field. If $a\in F$ and $b\in F\backslash\{0\}$ we define

$$a/b=a\cdot b^{-1}.$$

We also write $${{a\over b}}$$ for $a/b$. If $a,b$ are both in $F\backslash\{0\}$, then $a\cdot b^{-1}\in F\backslash \{0\}$ so $/$ defines a binary operation on $F\backslash\{0\}$. Also, if $b\neq 0$, then $${{1\over b}=1/b=1\cdot b^{-1}=b^{-1}}$$.

2.92   Example. In the field $(\mbox{{\bf Z}}_5,\oplus_5,\odot_5)$, $3^{-1}=2$ and $4^{-1}=4$. Hence

$${1\over 3}=2 \mbox{ and } {2\over 3}=2\cdot 2=4.$$

Thus,

$${1\over 3}+{2\over 3}=2+4=1.$$

2.93   Exercise. A Let $F$ be a field, and let $a,b,c,d$ be elements of $F$ with $b\neq 0$ and $d\neq 0$. Prove all of the following propositions. In doing any part of this problem, you may assume that all of the earlier parts have been proved.

1. [a)] $${ {{a\cdot d}\over {b\cdot d}}={a\over b}}.$$
2. [b)] $${ {{d\cdot a}\over {d\cdot b}}={a\over b}}.$$
3. [c)] $${ -\left( {a\over b}\right)={{-a}\over b}}.$$
4. [d)] $${ {a\over b}+{c\over b}={{a+c}\over b}}.$$
5. [e)] $${ {a\over b}+{c\over d}={{a\cdot d+b\cdot c}\over {b\cdot d}}}.$$
6. [f)] $${ {a\over b}-{c\over d}={{a\cdot d-b\cdot c}\over {b\cdot d}}}.$$
7. [g)] $${ {a\over b}\cdot {c\over d}={{a\cdot c}\over {b\cdot d}}}.$$
8. [h)] $${ \left( {b\over d}\right)^{-1}={d\over b}}$$.

I will now start the practice of using steps in proofs that involve multiple uses of the associative and commutative laws. For example, I'll write statements such as

$$(b-a)+(d-c)=(b+d)-(a+c)$$

with no explanation, because I believe that you recognize that it is correct, and that you can prove it. I'll also write $ab$ for $a\cdot b$ when I believe that no confusion will result, and I'll use distributive laws like

$$(x+y)\cdot z=x\cdot z+y\cdot z$$

and

$$(x-y)\cdot z=x\cdot z-y\cdot z$$

even though we haven't proved them. I will write

$$(a+b)(c+d)=(a+b)c+(a+b)d=ac+bc+ad+bd$$

and assume that you know (because of our conventions about omitting parentheses; cf. Remark 2.50) that the right side of this means

$$\left(\left((a\cdot c)+(b\cdot c)\right)+(a\cdot d)\right)+(b\cdot d)$$

and you also know (by exercise 2.30) that the parentheses can be rearranged in other sensible orders without changing the value of the expression.

2.94   Exercise. Let $F$ be a field. Show that for all $a,b,x$ in $F$

1. [a)] $(a+b)^2=a^2+2ab+b^2.$
2. [b)] $(a-b)^2=a^2-2ab+b^2.$
3. [c)] $(a-b)\cdot (a+b)=(a^2-b^2).$
4. [d)] $(x-a)(x-b)=x^2-(a+b)x+a\cdot b.$
5. [e)] $(2a+b)^2=4a^2+4ab+b^2.$

2.95   Theorem. Let $F$ be a field. Then for all $x,y\in F$,

$$(x^2=y^2)\hspace{1ex}\Longleftrightarrow\hspace{1ex}(x=y \mbox{ or } x=-y).$$

Proof:

$$\begin{eqnarray*} x^2=y^2 &\mbox{$\Longleftrightarrow$}& x^2-y^2=0 \\ &\mbox{$\... ...eftrightarrow$}& x=y \mbox{ or } x=-y.\mbox{ $\mid\!\mid\!\mid$} \end{eqnarray*}$$

2.96   Theorem (Quadratic formula.) Let $F$ be a field such that $2\neq 0$ in $F$. Let $A\in F\backslash\{0\}$, and let $B$, $C$ be elements of $F$. Then the equation

$$Ax^2+Bx+C=0$$ (2.97)

has a solution $x$ in $F$ if and only if $B^2-4AC$ is a square in $F$ (i.e., if and only if there is some element $y\in F$ such that $y^2=B^2-4AC$). If $y$ is any element of $F$ satisfying

$$y^2=B^2-4AC,$$

then the complete set of solutions of (2.97) is

$$\left\{ {{-B+y}\over {2A}},{{-B-y}\over {2A}}\right\}.$$

(This corresponds to the familiar quadratic formula

$$x={{-B\pm\sqrt{B^2-4AC}}\over {2A}}.\mbox{)}$$

Proof: The proof uses the algebraic identity

$$(px+q)^2=p^2x^2+2pqx+q^2 \mbox{ for all } p,q,x\in F.$$

Since $A\neq 0$ and $2\neq 0$, we have $2^2A\neq 0$ and hence

$$\begin{eqnarray*} Ax^2+Bx+C=0 &\mbox{$\Longleftrightarrow$}& 2^2A(Ax^2+Bx+C)=0\\... ...+B^2=B^2-4AC\\ &\mbox{$\Longleftrightarrow$}&(2Ax+B)^2=B^2-4AC. \end{eqnarray*}$$

Hence if $B^2-4AC$ is not a square, then (2.97) has no solutions. If $B^2-4AC=y^2$ for some $y\in F$, then

$$\begin{eqnarray*} Ax^2+Bx+C=0&\mbox{$\Longleftrightarrow$}&(2Ax+B)^2=y^2\\ &\mb... ...2A}} \mbox{ or } x={{-y-B}\over {2A}}.\mbox{ $\mid\!\mid\!\mid$} \end{eqnarray*}$$

2.98   Entertainment. $\mbox{{\bf Z}}_7$ is a field. (You can take my word for it or check it for yourself.) Find all solutions to the quadratic equations below in $\mbox{{\bf Z}}_7$.

a)

$x^2-x+2=0.$

b)

$3x^2+5x+2=0.$

c)

$2x^2+x+5=0.$

2.99   Note. The definition of field that we use is roughly equivalent to the definition given by H. Weber in 1893 [48, p526]. Weber does not give the zero-one axiom but he remarks that $0$ is different from $1$ except in the uninteresting case where the field has only one element. He includes commutativity of addition as an axiom, and he also appears to take $a(-b)=-(ab)$ as an axiom. Individual fields, both finite fields and subfields of the real and complex numbers, had been studied before Weber's paper, but Weber's definition provided an abstraction that included both finite and infinite fields.

There are many other choices we could have made for the field axioms. In [29], Edward Huntington gives eight different sets of axioms that are equivalent to ours. (Two sets of propositions ${\cal A, B}$ are equivalent if every statement in ${\cal A}$ can be proved using statements in ${\cal B}$, and every statement in ${\cal B}$ can be proved from statements in ${\cal A}$.)