Arithmetic Operations in a Field

All arithmetic properties of real numbers can be deduced from the axioms stated earlier. Now, we explore the consequences of the first six axioms (I-VI) which holds for every every (even unordered) field $F$.

Definition 1.2.1. Given two elements $x$ and $y$ of a field $F$. we define their difference, \begin {align*} x-y = x +(-y). \end {align*}

In other words, to subtract an element $y$ means to add its additive inverse, $-y$.
If $y \neq 0$, we also define the quotient of $x$ by $y$, \begin {align*} \frac {x}{y} = x \cdot (y^{-1}) \end {align*}

Also denoted by $x/y$. In other words, to divide $x$ by $y$ means to multiply $x$ by the reciprocal of y

Hence we have defined two new operations: subtraction and division, that are just special cases of addition and multiplication, we can apply the axioms to these new operations, in the following corollaries.

In Symbols: \begin {align*} (\forall x,y \in \mathbb {R}^1) \;\;\; (x-y )\in \mathbb {R}^1, \;\; (x/y) \in \mathbb {R}^1 \;\; (\text {the latter if $y\neq 0$}). \end {align*}

Corollary 2. If $a,b,c$ are elements of field $F$, with $a=b$, then \begin {align*} a+c=b+c \;\; \text {and}\;\; ac=bc \end {align*}

(In other words, we may add one and the same element $c$ to both sides of the equation $a=b$; Similarly for multiplication.)

In Symbols: \begin {align*} (\forall a,b,c \in F)\;\;a=b \implies a+c =b+c\;\; \text {and}\;\; ac=bc \end {align*}

Proof. By the properties of equality, we have $a+c = a+c$. Now, as $a=b$, we may replace $a$ by $b$ on the right side. This yields $a+c =b+c$ as required. Similarly for $ac=bc$. □

Corollary 3. (Cancellation Law). If $a,b,c$ are elements in a field $F$, then \begin {align*} a+c = b+c \implies a=b. \end {align*}

If, further, $c\neq 0$, then, \begin {align*} ac=bc \implies a=b. \end {align*}

(In other words, we may cancel a summand and a nonzero factor on both sides of the equation.)

Proof. Let $a+c = b+c$. By corollary 2, we may add $(-c)$ on both sides of the equation to get \begin {align*} (a+c) +(-c) = (b+c) + (-c) \end {align*}

By Associativity (Axiom III), \begin {align*} a+[c +(-c)]=b+[c+(-c)]. \end {align*}

As $c+(-c) =0$ (by Axiom V), we get $a+0=b+0$, i.e., $a=b$ by (Axiom IV); Similarly for multiplication. □

Theorem 1. Given two elements, $a$ and $b$, of a field $F$, there always exists a unique element $x$ such that $a+x =b$; This element equals the difference $b-a$.
(Thus $a+x =b$ means that $x=b-a$.)
If, further, $a\neq 0$, there also is a unique element $y \in F$, with $ay=b$; This element equals the quotient $b/a$. (Thus $ay=b$, $a\neq 0$, means that $y =b/a$.)

In symbols: \begin {align*} (\forall a,b \in F)\;(\exists ! \, x,y \in F ) \;\;a+x =b \;\; ay =b \;\; (\text {the latter if $a\neq 0$}) \end {align*}

Proof. It is easily checked that the equation $a+x=b$ is satisfied by $x=b-a$. In fact, we have: \begin {align*} a+x &= a+(b-a)\\ &= (b-a)+a &(\text {II})\\ &= [b+(-a)]+a \\ &= b+[(-a)+a] &(\text {III})\\ &= b+ 0 &(\text {V})\\ &= b &(\text {IV}) \end {align*}

Thus, the equation $a+x=b$ has at least one solution for $x$. To prove that this solution is unique, suppose that we have still another solution, $x'$, say. Then, we obtain $a+x=b$ and $a+x'=b$, so that $a+x=a+x'$ or $x+a=x'+a$, Cancelling $a$ (by Corollary 3), we see that $x=x'$, so that the two solutions coincide.

For Multiplication, similarly, it can be checked that the equation $ay=b$ is satisfied by $y=b/a$, given $a\neq 0$. \begin {align*} ay &=a \cdot (b\cdot a^{-1}) \\ &= (b\cdot a^{-1}) \cdot a &(\text {II})\\ &= b\cdot (a^{-1}\cdot a) &(\text {III})\\ &= b\cdot 1 & (\text {V})\\ &= b &(\text {IV}) \end {align*}

Hence, we can again conclude that the equation, $ay=b$ has at least one solution for $y$. For another solution $y'$, $ay'=b$ implies $ay= ay'$, Canceling $a$ (by Corollary 3), we can indeed see that $y=y'$. Thus, both the existence and the uniqueness of the solutions have been proven. □

Theorem 1 shows that subtraction and addition are inverse operations to addition and multiplication. Now, we can move a summand or a factor from one side of the equation to another side.

Proof. We have by definition, \begin {align*} 0-x &=0+(-x)\\ &= -x &(\text {IV}) \end {align*}

Similarly, \begin {align*} 1/x &= 1\cdot x^{-1}\\ &= x^{-1} &(\text {IV}) \end {align*} □

(Hence we never have $0\cdot x=1$; This is why $0$ cannot have multiplicative inverse)

Proof. By distributivity (Axiom VI) and Axiom IV, we get \begin {align*} 0x+0x=(0+0)x =0x=0+0x \end {align*}

Thus $0x+0x=0+0x$. Cancelling $0x$ from both sides (Corollary 3), we get $0\cdot x=0$, and by commutativity (Axiom II), we also get $x\cdot 0=0$. □

Proof. Formula (i) means that $a(-b)$, and similarly $(-a)b$, equals the additive inverse of $ab$. Hence to prove (i), we have to show that $a(-b) +ab=0$. By distributivity (Axiom VI), Axiom V, and Corollary 5 we have \begin {align*} a(-b) +ab = a[(-b)+b] = a \cdot 0 = 0 \end {align*}

Also, similarly, \begin {align*} (-a)b +ab = b[(-a)+a] = b\cdot 0 =0 \end {align*}

For (ii), we need to show that $-(-a) =a$. Because, $-a$ is the additive inverse of $a$, we have $a+(-a)=0$. Also, by definition, $-(-a)$ is the additive inverse of $(-a)$, hence, $-a+(-(-a))=0$. By Corollary 2, we can add $a$ to both sides of the equation: \begin {align*} a+ (-a) +(-(-a))&=a+0\\ \implies 0+(-(-a))&=a+0 \quad \text {(V)}\\ \implies -(-a)&=a \quad \text {(IV)} \end {align*}

Formula (iii) can be derived from (i) and (ii), by first using (i) twice replacing $a$ with $(-a)$: \begin {align*} (-a)(-b) = -((-a) \cdot b) = -(-(a\cdot b)) \end {align*}

Because of closure (Axiom I), $a\cdot b \in F$, Hence Formula (ii), applies, giving $-(-(a\cdot b)) = a \cdot b$ □

1.2 Arithmetic Operations in a Field