Identity and Inverse Elements of Binary Operations

Identity and Inverse Elements of Binary Operations

Recall from the Associativity and Commutativity of Binary Operations page that an operation $* : S \times S \to S$ is said to be associative if for all $a, b, c \in S$ we have that $a * (b * c) = (a * b) * c$ (nonassociative otherwise) and $*$ is said to be commutative if $a * b = b * a$ (noncommutative otherwise). We will now look at some more special components of certain binary operations.

 Definition: An element $e \in S$ is said to be the Identity Element of $S$ under the binary operation $*$ if for all $a \in S$ we have that $a * e = a$ and $e * a = a$.

For example, the identity element of the real numbers $\mathbb{R}$ under the operation of addition $+$ is $e = 0$ since for all $a \in \mathbb{R}$ we have that:

(1)
\begin{align} \quad a + 0 = a \quad \mathrm{and} \quad 0 + a = a \end{align}

Similarly, the identity element of $\mathbb{R}$ under the operation of multiplication $\cdot$ is $e = 1$ since for all $a \in \mathbb{R}$ we have that:

(2)
\begin{align} \quad a \cdot 1 = a \quad \mathrm{and} 1 \cdot a = a \end{align}

We should mntion an important point regarding the existence of an identity element on a set $S$ under a binary operation $*$. We have asserted in the definition of an identity element that $e$ is unique. We will prove this in the very simple theorem below.

 Theorem 1: Let $S$ be a set and $* : S \times S \to S$ be a binary operation on $S$. If $e$ is an identity element on $S$ under $*$ then $e$ is unique.
• Proof: Suppose that $e$ and $e'$ are both identity elements of $S$ under $*$. Then we have that for all $a \in \mathbb{R}$ that:
(3)
\begin{align} \quad e = e * e' = e' \end{align}
• Therefore $e = e'$, so $e$ is unique. $\blacksquare$
 Definition: Let $S$ be a set and $* : S \times S \to S$ be a binary operation on $S$. If an identity element $e$ exists and $a \in S$ then $b \in S$ is said to be the Inverse Element of $a$ if $a * b = e$ and $b * a = e$. In such instances, we write $b = a^{-1}$.

For example, standard addition on $\mathbb{R}$ has inverse elements for each $a \in \mathbb{R}$ which we denoted as $-a \in \mathbb{R}$, which are called additive inverses, since for all $a \in \mathbb{R}$ we have that:

(4)
\begin{align} \quad a + (-a) = 0 = e_{+} \quad \mathrm{and} (-a) + a = 0 = e_{+} \end{align}

Similarly, standard multiplication on $\mathbb{R}$ has inverse elements for each $a \in \mathbb{R}$ EXCEPT for $a = 0$ which we denote as $a^{-1} = \frac{1}{a} \in \mathbb{R}$, which are called multiplicative inverses, since for all $a \in \mathbb{R}$ we have that:

(5)
\begin{align} \quad a \cdot a^{-1} = a \cdot \left ( \frac{1}{a} \right ) = 1 = e_{\cdot} \quad \mathrm{and} \quad a^{-1} \cdot a = \left ( \frac{1}{a} \right ) \cdot a = 1 = e^{\cdot} \end{align}

Note that an additive inverse does not exist for $0 \in \mathbb{R}$ since $\frac{1}{0}$ is undefined.

For another more complicated example, recall the operation of matrix multiplication on the set of all $2 \times 2$ matrices with real coefficients, $M_{22}$. The identity element on $M_{22}$ under matrix multiplication is the $2 \times 2$ identity matrix. Recall that for all $A \in M_{22}$.

For the matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ to have an inverse $A^{-1} \in M_{22}$ we must have that $\det A \neq 0$, that is, $ad - bc \neq 0$. to which we define $A^{-1}$ to be:

(6)
\begin{align} \quad A^{-1} = \begin{bmatrix} \frac{d}{ad - bc} & -\frac{b}{ad - bc} \\ -\frac{c}{ad -bc} & \frac{a}{ad - bc} \end{bmatrix} \end{align}

Therefore not all matrices in $M_{22}$ have inverse elements.