# Section2.1Binary operations and structures¶ permalink

So far we have been discussing sets. These are extremely simple objects, essentially mathematical “bags of stuff.” Without any added structure, their usefulness is very limited. Imagine, for instance, living with friends in a two-story house without rooms, stairs, closets, or hallways. You have no privacy, cannot access the second floor, etc. A set with no added structure will not help us, say, solve a linear equation. What will help us with such things are objects such as groups, rings, fields, and vector spaces. These are sets equipped with binary operations which allow us to combine set elements in various ways. We first rigorously define a binary operation.

##### Definition2.1.1

A binary operation on a set $S$ is a function from $S\times S$ to $S\text{.}$ Given a binary operation $*$ on $S\text{,}$ for each $(a,b)\in S\times S$ we denote $*((a,b))$ in $S$ more simply by $a*b\text{.}$ (Intuitively, a binary operation $*$ on $S$ assigns to each pair of elements $a,b \in S$ a unique element $a*b$ of $S\text{.}$)

A set $S$ equipped with a binary operation $*$ is called a binary (algebraic) structure, and is denoted by $\langle S,*\rangle\text{,}$ or just by $S\text{,}$ if $*$ is understood from context.

##### Remark2.1.2

1. For $*$ to be a binary operation on $S\text{,}$ $a*b$ must be well-defined and in $\mathbf{S}$ for each $a,b\in S\text{.}$ For instance, we cannot define a binary operation on $\R$ by

\begin{equation*} a*b=\text{ the greatest number less than $a+b$} \end{equation*}

since there is no such “greatest number.” Nor can we define a binary operation on $\Z$ by $a*b=ab/2\text{,}$ since for, say, $a=b=1 \in \Z\text{,}$ $ab/2=1/2 \not\in \Z\text{.}$

2. Not every binary operation is denoted by $*\text{.}$ In fact, many already have common notations: for instance, $+$ on $\Z$ or $\circ$ on the set of functions from $\R$ to $\R\text{.}$ We will assume these common notations represent the “usual” binary operations we know them to mean, unless otherwise noted.

3. Do not mix up the $*$ that indicates a binary operation and the superscript $^*$ that indicates that we are only considering the nonzero elements of a given set (e.g., $\R^*$). You should be able to tell which type of $*$ we are using from context and placement. Also, make sure you correctly place these symbols!

##### Definition2.1.3

A binary operation $*$ on a set $S$ is associative if $(a*b)*c=a*(b*c)$ for all $a,b,c\in S\text{.}$

##### Remark2.1.4

When a binary operation is associative, we can omit parentheses when operating on set elements. For example, $+$ is associative on $\Z\text{,}$ so we can unambiguously write the (equal) expressions $1+(2+3)$ and $(1+2)+3$ as $1+2+3\text{.}$

##### Definition2.1.5

A binary operation $*$ on set $S$ is commutative if

\begin{equation*} a*b=b*a \end{equation*}

for all $a,b\in S\text{.}$ We say that specific elements $a$ and $b$ of $S$ commute if $a*b=b*a\text{.}$

##### Definition2.1.6

Let $\langle S,*\rangle$ be a binary structure. An element $e$ in $S$ is an identity element of $\langle S,*\rangle$ if $x*e=e*x=x$ for all $x\in S\text{.}$

##### Note2.1.7

Sometimes an identity element of $\langle S,*\rangle$ is referred to as an identity element of $S$ under $*$, or, when $*$ is clear from context, simply as an identity element of $S$.

##### Warning2.1.8

Not every binary structure contains an identity element! (Ex: $\langle \Z,-\rangle\text{.}$)

A natural question to ask is if a binary structure can have more than one identity element? The answer is no!

##### Definition2.1.10

Let $\langle S, *\rangle$ be a binary structure with identity element $e\text{.}$ Then for $a\in S\text{,}$ $b$ is an (two-sided) inverse of $a$ in $\langle S,*\rangle$ if $a*b=b*a=e\text{.}$

##### Note2.1.11

We can also refer to such an element $b$ as an inverse for $a$ in $S$ under $*$, or, when $*$ is clear from context, simply as an inverse of $a$.

##### Warning2.1.12

• Not every element in a binary structure with an identity element has an inverse!

• If a binary structure does not have an identity element, it doesn't even make sense to say an element in the structure does or does not have an inverse!

##### Proof

We end by discussing the idea of closure under a binary operation.

##### Definition2.1.14

Let $\langle S,*\rangle$ be a binary structure and let $T \subseteq S\text{.}$ Then $T$ is said to be closed under $*$ if $t_1 * t_2 \in T$ whenever $t_1,t_2\in T\text{.}$

##### Example2.1.15

Consider the binary structure $\langle \M_2(\R), +\rangle\text{,}$ where $+$ denotes matrix addition. Let $T=\{A\in M_2(\R): A \mbox{ is invertible}\}.$ We claim that $T$ is not closed under $+\text{.}$ Indeed, if we denote the identity matrix in $\M_2(\R)$ by $I_2\text{,}$ then we observe that $I_2, -I_2\in T\text{,}$ but $I_2+(-I_2)\not\in T\text{,}$ since the zero matrix isn't invertible.

##### Example2.1.16

Consider the binary structure $\langle \M_2(\R), \cdot\rangle\text{,}$ where $\cdot$ denotes matrix multiplication. Again, let $T$ be the set of all invertible matrices in $\M_2(\R)\text{.}$ We claim that $T$ is closed under $\cdot\,\text{.}$ Indeed, let $A,B\in T\text{.}$ Then $A$ and $B$ are invertible, so their determinants are nonzero. Thus, $\det(AB)=(\det A)(\det B)\neq 0,$ so $AB$ is invertible, and hence $AB\in T\text{.}$