Binary Operations

These are operations involving two operands. Addition, subtraction, multiplication and division are all binary operations. Formally, for a set $S$, the binary operation $*$ assigns every ordered pair $(a,b)\in S\times S$ to a unique element $a*b\in S$.

Evaluating the results of a binary operation

We can find the result of a binary operation by just substituting according to what is given by the definition of the operation.

Example

Given the binary operation defined by $a*b=ab+1$, find the value of:

• $3*2$
• $-1*4$
• $5x*2$
• $b*a$

Solution In the first case, the $3$ is substituted as $a$ and the $2$ is substituted as $b$: $$ \begin{equation}\begin{aligned} 3*2&=(3)(2)+1\\ &=6+1\\ &=7\\ \end{aligned}\end{equation} $$

For the second, our $a$ is $-1$, and $b$ is $4$: $$ \begin{equation}\begin{aligned} -1*4&=(-1)(4)+1\\ &=-4+1\\ &=-3\\ \end{aligned}\end{equation} $$

Thirdly, we have an $a$ of $5x$ and a $b$ of $2$: $$ \begin{equation}\begin{aligned} 5x*2&=(5x)(2)+1\\ &=10x+1\\ \end{aligned}\end{equation} $$

Finally, the value of $a$ is now $b$ and that of $b$ is $a$: $$ \begin{equation}\begin{aligned} b*a&=(b)(a)+1\\ &=ba+1\\ &=ab+1\\ \end{aligned}\end{equation} $$

Closure

A number set is closed under an operation if the result of any two elements of the set ALWAYS produces another element in the same number set. A set $S$ is closed under an operation $*$ if $a*b\in S\forall a,b\in S$. The universal quantifier, $\forall$, is read as “for every” or “for all”.

Example: Real numbers under multiplication

Any real number multiplied by another real number will always produce another real number thus we say that the set of real numbers is closed under multiplication.

Example: Integers under division

We can easily pick any two integers e.g. $1$ and $2$ and show that the result of the division will NOT produce an integer: $$\frac12=0.5$$

Hence the set of integers is NOT closed under division.

Identity of an operation

The identity, $e$ of a binary operation, $\oplus$ is that single element which when operated on by every element $a$, produces that other element: $$ \begin{equation}\begin{aligned} a\oplus e&=a \text{ and }\\ e\oplus a&=a\\ \end{aligned}\end{equation} $$

We already know of the additive identity, $0$: $$ \begin{equation}\begin{aligned} a+0&=a \text{ and }\\ 0+a&=a\\ \end{aligned}\end{equation} $$ $0$ is the identity for addition because any number added to $0$ produces the number itself. We also know the multiplicative identity, $1$: $$ \begin{equation}\begin{aligned} a\times 1&=a \text{ and }\\ 1\times a&=a\\ \end{aligned}\end{equation} $$ $1$ is the identity for multiplication because any number times $1$ is the number itself.

Example

Given the operation $a\oplus b=a-3b+6$, determine if there is an identity element.

Solution Using the requirement for an identity: $$ \begin{equation}\begin{aligned} a\oplus e &= a\\ \therefore a-3e+6&=a\\ \end{aligned}\end{equation} $$

Finding for $e$: $$ \begin{equation}\begin{aligned} a-3e+6&=a\\ -3e&=a-6-a\\ -3e&=-6\\ \therefore e&=2\\ \end{aligned}\end{equation} $$

The universal right identity thus exists but we also have the check the other way: $$ \begin{equation}\begin{aligned} e\oplus a&=a\\ e-3a+6&=a\\ e&=a+3a-6\\ e&=4a-6\\ \end{aligned}\end{equation} $$

Because the identity of the operation is dependent on the value of the operand, we do not have a universal identity element for the operation $\oplus$.

The inverse of an element

This is the element, $a^{-1}$ which when operated on by another element ($a$) produces the identity for that operation: $$ \begin{equation}\begin{aligned} a\oplus a^{-1}=e\\ \end{aligned}\end{equation} $$

Example: Addition

Because $0$ is the additive identity, it is easy to determine the inverse of any number, $a$ with respect to addition: $$ \begin{equation}\begin{aligned} a+a^{-1}&=e\\ a+a^{-1}&=0\\ a^{-1}&=0-a\\ \therefore a^{-1}&=-a\\ \end{aligned}\end{equation} $$

Thus for any number, $a$, the additive inverse of that number is simply the negative version of that number (a.k.a. the product of the number and $-1$).

Example: Multiplication

$1$ has already been established as the multiplicative identity thus the inverse ($a^{-1}$) can be found for any number, $a$: $$ \begin{equation}\begin{aligned} a\times a^{-1}&=e\\ a\times a^{-1}&=1\\ a^{-1}&=1\div a\\ \therefore a^{-1}&=\frac{1}{a}\\ \end{aligned}\end{equation} $$

For any number, $a$, the multiplicative inverse of that number is its reciprocal, written as $\frac{1}{a}$.

It must be noted that we cannot find the inverse of an element without a universal identity element. Thus if asked to find the inverse of an element, we need to verify the existence of the identity element first.