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.

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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.

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Definition2.1.3

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

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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 \(a*b=b*a\text{.}\)

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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 \(x\in S\text{.}\)

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

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Theorem2.1.9

A binary structure \(\langle S, *\rangle\) has *at most one* identity element. That is, identity elements in binary structures, when they exist, are unique.

##### Proof

Assume that \(e\) and \(f\) are identity elements of \(S\text{.}\) Then since \(e\) is an identity element, \(e*f=f\) and since \(f\) is an identity element, \(e*f=e\text{.}\) Thus, \(e=f\text{.}\)

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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 \(a*b=b*a=e\text{.}\)

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Theorem2.1.13

Let \(\langle S, *\rangle\) be a binary structure with an identity element, where \(*\) is associative. Let \(a\in S\text{.}\) If \(a\) has an inverse, then its inverse is unique.

##### Proof

Let \(e\) be the identity element of \(S\text{.}\) Suppose \(a\) has inverses \(b\) and \(c\text{.}\) Then \(a*b=e\) so, multiplying both sides of the equation by \(c\) on the left, we have \(c*(a*b)=c*e=c\text{.}\) But since \(*\) is associative, we have \(c*(a*b)=(c*a)*b=e*b=b\text{.}\) But \(b=c\text{.}\) Thus, \(a\)'s inverse is unique.

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

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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 \(t_1,t_2\in T\text{.}\)

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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.

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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 \(AB\in T\text{.}\)