Binary Operations Introduction

The operations acted upon variables in Abstract Algebra is generalized.

For example, $a\ast b$ where $\ast$ is a generalized operator on two objects.

Definition 1.3 - Binary Operation

A binary operation $\ast$ on a set $S$ is a function mapping $S\times S$ into $S$.

For every $(a,b)\in S\times S$, we denote the element $\ast ((a,b))$ of $S$ as $a\ast b$.

Intuitively, the usual addition operator $+$ on $\mathbb{R}$ gives a binary operation on $\mathbb{R}$.

The term binary refers to that we are mapping pairs of elements from $S$ into $S$.

Example - Min Operation

On $Z^{+}$ we define $\ast$ by $a\ast b$ as giving the smaller of $a$ and $b$ or the common value if $a=b$.
Intuitively then, $\ast =min(a,b)$.

Definition 1.4 - Closure

Let $\ast$ be an operation on $S$ and let $H\subseteq S$.
The subset $H$ is closed under $\ast$ if for all $a,b\in H$ we have $a\ast b\in H$.

The operation on $H$ given by restricting $\ast$ to $H$ is called the induced operation of $\ast$ on $H$.

Example - Distributive Property for Operation

Prove the following statement or give a counter example:

If $\ast$ and ${\ast }^{\prime }$ are two binary operations on $S$, then

a * (b *' c) = (a * b) *' (a * c)

This is shown false by giving the following counterexample.

Set $\ast =+$ and ${\ast }^{\prime }=-$ on the integers $\mathbb{Z}$.

Then we have that $a+(b-c)$.
However, we have that $a+(b-c)\ne (a+b)-(a+c)$
Since $(a+b)-(a+c)=b-c$.

Definition 1.11 Commutative Operation

An operation $\ast$ on a set $S$ is commutative iff $a\ast b=b\ast a$ for all $a,b\in S$.

Example - Min Commutative

$min(a,b)$ is the same as $min(b,a)$, so $min$ is commutative

Definition 1.12 - Associative Operation

An operation on a set $S$ is associative iff $(a\ast b)\ast c=a\ast (b\ast c)$ for all $a,b,c\in S$

Function Composition

The composition of two functions is defined as $(f\circ g)(x)=f(g(x))$

The composition is a binary operation on the set $F$ of all real-valued functions $f$ in $\mathbb{R}$
because for every function $f,g\in F$ we have a mapping from $F\times F$ into $F$ by means of the composition of functions $(f\circ g)(x)$.

Theorem 1.13 - Associativity of Function Composition

Let $S$ be a set and $f,g,h$ be functions mapping $S$ into $S$.

Then, $f\circ (g\circ h)=(f\circ g)\circ h$.

Proof idea: No matter which you simplify function composition of three functions (fg then h, or gh then f), you get the same value for all x in S.

Definition 1.14 - Identity Element

Let $S$ be a set with binary operation $\ast$.

If $e\in S$ has the property that for all $a\in S$, $a\ast e=e\ast a=a$ then $e$ is called the identity element.

For example, $a+0=0+a=a$ for all $a\in \mathbb{Z}$.

Theorem 1.14 - Uniqueness of Identity Element

A set $S$ with binary operation $\ast$ has at most one identity element.

Proof: Let $S$ be a set with binary operation $S$. Suppose that $e$ and $e^{\prime }$ are both identity elements.
Then it follows that $e\ast e^{\prime }=e^{\prime }$ and also that $e\ast e^{\prime }=e$.
As a result, we have that $e^{\prime }=e$, and there is only one identity element.

Definition 1.17 - Inverse

If $\ast$ is an operation on the set $S$ and $S$ has an identity $e$,

then for any $x\in S$, the inverse of x is an element $x^{\prime }$ where

$x\ast x^{\prime }=x^{\prime }\ast x=e$.

Example - Equality of Inverse Function and Inverse Property for Group

Inverse function: $I_A=\{(x,x)|x\in A\}$

Using composition:
If $f$ is a bijection, then $g=f^{-1}$ if and only if $g\circ f=I_A$ or $f\circ g=I_B$.

Why do these definitions match up with the inverse property above?

Cayley Tables

Tables provide a visual way to assess binary operations using the following rule for operations:

(ith entry on the left) * (jth entry on top) = (entry in ith row and jth column)

Commutativity via Table

A binary operation is commutative if and only if the entries in the table are symmetric with respect to the diagonal from the upper left to bottom right.

Conditions for Binary Operations

1. Binary operations must be defined everywhere for all elements on $S$
   ($a\ast b=\frac{a}{b}$ is not defined on $\mathbb{Q}$ where $b=0$)

2. Binary operations must yield only one value
   (Define operation on $\mathbb{Z}$ where $a\ast b=c$ and $c\ge a$ and $c\ge b$, but there are infinitely many possibilities)

3. Closure under $\ast$ (every operation must produce an element also in $S$)
   ($a\ast b=\frac{a}{b}$ on ${\mathbb{Z}}^{+}$, but $\frac{1}{2}\notin {\mathbb{Z}}^{+}$)

Example

Determine whether $\ast$ defined by $a\ast b=2a-b$ is a binary operation on ${\mathbb{R}}^{+}$.

Let $a=1,b=4$. Then $a\ast b=2(1)-4=-2$ is not in ${\mathbb{R}}^{+}$.
Thus, this operation is not well-defined by condition 3 of closure.