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Section 1.1 Binary Operations and Fields
“I never say of an operation that it is without danger.”
―August Bier
Definition 1.1 . Binary Operation.
A binary operation on a set \(X\) is a function from \(X × X\) into \(X\text{.}\)
Definition 1.2 . Binary Operation Properties.
A binary operation on a set \(S\) is commutative if \(a*b=b*a\) for all \(a,b\in S\text{.}\)
A binary operation on a set \(S\) is associative if \((a*b)*c=a*(b*c)\) for all \(a,b,c\in S\text{.}\)
An element \(e\) is called an identity if \(e * x=x=x * e\) for all \(x\in S\text{.}\)
An element \(y\) is called an inverse of \(x\) if \(y * x=e=x*y\text{.}\)
Definition 1.3 . Field.
A field is a set \(F\) with two operations
addition: \(F\times F\rightarrow F:(x,y)\mapsto x+y\)
multiplication \(F\times F\rightarrow F: (x,y)\mapsto xy\) (or \(x\cdot y\) )
such that
\(\displaystyle (a+b)+c=a+(b+c)\,\forall\,a,b,c\in F\)
\(\displaystyle a+b=b+a\,\forall\, a,b,\in F\)
\(\exists 0\in F\) st \(a+0=a=0+a\,\forall\, a\in F\) (exercise: 0 is unique)
\(\forall\,a\in F\exists -a\in F\) such that \(a+(-a)=0=(-a)+a\)
\(\displaystyle (ab)c=a(bc)\,\forall\, a,b,c\in F\)
\(\displaystyle ab=ba \,\forall\, a,b,\in F\)
\(\exists 1\in F\) such that \(\forall a\in F\, a1=1=1a\) (\(1\neq 0\text{,}\) exercise to show \(1\) unique)
\(\forall\,a\in F\setminus\{0\},\exists a^{-1}\) such that \(aa^{-1}=1=a^{-1}a\) (exercise show \(a^{-1}\) unique)
\(\displaystyle a(b+c)=ab+ac\,\forall\, a,b,c\in F\)
\(\displaystyle (a+b)c=ac+bc\,\forall\, a,b,c\in F\)
Definition 1.4 . Subtraction, Division.
Let \(x\) and \(y\) be elements of a field \(F\text{.}\) We define binary operations \(-\) and \(÷\) on \(F\) such that \(x-y=x+(-y)\) and \(x÷y=x\cdot\frac 1y\text{.}\)
Theorem 1.6 . Unique Identity.
Any binary operation has at most one identity.
Theorem 1.7 . Multiplication by \(0\) .
\(x · 0 = 0\) for all \(x\) in \(F\text{.}\)
Proof.
\(x·(0 + 0) = x·0\) by axiom 4. On the other hand \(x·(0 + 0) = x·0 + x·0\) by axiom 11. Thus \(x·0 = x·0 + x·0\text{.}\) Now
\begin{equation*}
x·0 + [–(x·0)] = (x·0 + x·0) + [–(x·0)]
\end{equation*}
Again using axiom 5 and axiom 2, we have \(0 = x·0 + 0\) and so by axiom 4 we have \(0 = x · 0\text{.}\)
Proposition 1.8 . Field Arithmetic.
The field axioms imply the following statements, for any \(x, y,z \in F\text{.}\)
\(\displaystyle -0=0\)
\(\displaystyle -(-x)=x\)
\(\displaystyle −(x + y) = −x − y\)
\(xy = 0\) if and only if \(x = 0\) or \(y = 0\text{.}\)
if \(xy = xz\) and \(x ≠ 0\text{,}\) then \(y = z\text{.}\)
\(\displaystyle −(xy) = x(−y) = (−x)y\)
\(\displaystyle (−1)x = −x\)
\(\displaystyle (-x)(-y) = xy.\)
