Binary operation is the starting point of group theory. Before we can define a group, we must first know how two elements of a set are combined and whether the result remains inside the same set. This is the first structural idea in abstract algebra, because the set alone does not determine the algebraic system. The same set may behave very differently under addition, multiplication, subtraction, composition, or a specially defined rule. In this lecture, students will learn the meaning of a binary operation, the closure requirement, and the method for checking whether a given rule is a binary operation on a given set. :::definition[Binary Operation] Let $S$ be a non-empty set. A rule $*$ is called a $binary operation$ on $S$ if $*$ assigns to every ordered pair $(a,b)\in S\times S$ a unique element $a*b\in S$. Equivalently, $*$ is a function $$\begin{equation*} *:S\times S\to S. \end{equation*}$$ Thus a binary operation on $S$ must satisfy the following conditions: (i) $a*b$ is defined for every $a,b\in S$. (ii) $a*b\in S$ for every $a,b\in S$. ::: The second condition in the definition is called closure. A set $S$ is said to be closed under a rule $*$ if the result of combining any two elements of $S$ again belongs to $S$. This condition is not a decorative condition; it decides whether the operation is truly an operation on the stated set. Students often look only at the formula and forget the set. For example, ordinary subtraction is a binary operation on $\mathbb{Z}$, but it is not a binary operation on $\mathbb{N}$ when $\mathbb{N}=\{1,2,3,\ldots\}$. A binary operation must keep every ordered-pair result inside the same set. Enter a finite set of integers and choose an operation. The preview builds the operation table and marks each output as inside or outside the set. One outside entry is enough to prove that the rule is not a binary operation on the chosen set. :::scientific-preview[Binary Operation Closure Grid] ::: :::example Let $S=\mathbb{Z}$ and define $*$ on $S$ by $$\begin{equation*} a*b=a+b \quad \forall\, a,b\in\mathbb{Z}. \end{equation*}$$ Then $*$ is a binary operation on $\mathbb{Z}$. [1] The rule is defined for every ordered pair $(a,b)\in\mathbb{Z}\times\mathbb{Z}$. [2] Since the sum of two integers is an integer, we get $$\begin{equation*} a+b\in\mathbb{Z}\quad \forall\, a,b\in\mathbb{Z}. \end{equation*}$$ Hence, addition is a binary operation on $\mathbb{Z}$. ::: :::example Let $S=\mathbb{N}$ and define $*$ on $S$ by $$\begin{equation*} a*b=a-b \quad \forall\, a,b\in\mathbb{N}. \end{equation*}$$ Then $*$ is not a binary operation on $\mathbb{N}$. Take $a=2$ and $b=5$. Then $$\begin{align*} a*b&=2-5\\ &=-3. \end{align*}$$ But $-3\notin\mathbb{N}$. Therefore the closure condition is not satisfied. Hence, subtraction is not a binary operation on $\mathbb{N}$. ::: :::theorem Let $S$ be a non-empty set and let $*$ be a rule defined on ordered pairs of elements of $S$. If for every $a,b\in S$, the element $a*b$ exists uniquely and belongs to $S$, then $*$ is a binary operation on $S$. ::: :::proof Given that $S$ is a non-empty set and $*$ is a rule defined on ordered pairs of elements of $S$. To prove that $*$ is a binary operation on $S$. Let $a,b\in S$. By the given condition, $a*b$ exists uniquely and $$\begin{equation*} a*b\in S. \end{equation*}$$ Therefore the rule $*$ assigns to each ordered pair $(a,b)\in S\times S$ a unique element of $S$. Thus $*$ is a function from $S\times S$ into $S$. Hence, $*$ is a binary operation on $S$. $\square$ ::: To check whether a rule is a binary operation on a set, one should not begin with identity elements or inverse elements. Those ideas belong to the definition of a group and will be used only after a binary operation is confirmed. The first question is: whenever two elements are selected from the set, does the rule produce exactly one element of the same set? In practice, this means that the set and the rule must be read together. The formula $a+b$ behaves differently on $\mathbb{Z}$, on $\mathbb{Q}$, and on a finite set such as $\{0,1,2\}$. :::solved-problem Let $S=\mathbb{Q}-\{1\}$ and define $*$ on $S$ by $$\begin{equation*} a*b=a+b-ab. \end{equation*}$$ Determine whether $*$ is a binary operation on $S$. ::: :::solution Let $S=\mathbb{Q}-\{1\}$ and let $a,b\in S$. To determine whether $*$ is a binary operation on $S$. Since $a,b\in\mathbb{Q}$, we get $$\begin{equation*} a*b=a+b-ab\in\mathbb{Q}. \end{equation*}$$ It remains to prove that $a*b\neq 1$. If possible let $$\begin{equation*} a*b=1. \end{equation*}$$ Then $$\begin{align*} a*b=1 &\implies a+b-ab=1\\ &\implies ab-a-b+1=0\\ &\implies (a-1)(b-1)=0. \end{align*}$$ Since $a\in S$ and $b\in S$, we get $a\neq 1$ and $b\neq 1$. Therefore $$\begin{equation*} (a-1)(b-1)\neq 0. \end{equation*}$$ A contradiction. Hence, $a*b\neq 1$. Therefore $a*b\in\mathbb{Q}-\{1\}$ for every $a,b\in S$. Hence, $*$ is a binary operation on $S$. $\square$ ::: :::remark A common mistake is to say that every formula gives a binary operation. A formula gives a binary operation only after the set is fixed and closure is verified. The rule $a/b$ is meaningful for many ordered pairs of real numbers, but it is not defined when $b=0$. Therefore division is not a binary operation on $\mathbb{R}$. It becomes a binary operation on $\mathbb{R}-\{0\}$ because the quotient of two nonzero real numbers is again nonzero. ::: Use the calculator below to test closure for a finite set of integers. Enter a finite set separated by commas, choose an operation, and press the button. The calculator checks all ordered pairs from the set. A single failed ordered pair is enough to show that the rule is not a binary operation on the given set. This activity is useful before studying groups because every group operation must first be a binary operation. :::calculator[Closure Test Calculator] ::: :::exercise [1] Determine whether addition is a binary operation on $\mathbb{N}$. [2] Determine whether subtraction is a binary operation on $\mathbb{Z}$. [3] Determine whether multiplication is a binary operation on $\mathbb{Q}$. [4] Determine whether division is a binary operation on $\mathbb{Q}$. [5] Let $S=\{0,1\}$. Define $a*b=a+b$. Determine whether $*$ is a binary operation on $S$. ::: :::answer [1] Yes. If $a,b\in\mathbb{N}$, then $a+b\in\mathbb{N}$. [2] Yes. If $a,b\in\mathbb{Z}$, then $a-b\in\mathbb{Z}$. [3] Yes. If $a,b\in\mathbb{Q}$, then $ab\in\mathbb{Q}$. [4] No. Division is not defined when the second element is $0$. [5] No. Since $1*1=1+1=2$ and $2\notin S$, closure fails. ::: :::faq Q: Is every rule a binary operation? A: No. A rule is a binary operation on $S$ only when it combines every ordered pair of elements of $S$ and gives a unique result in $S$. Q: Is closure the same as associativity? A: No. Closure says $a*b\in S$. Associativity says $(a*b)*c=a*(b*c)$. Q: Can the same rule be a binary operation on one set but not on another? A: Yes. Subtraction is a binary operation on $\mathbb{Z}$ but not on $\mathbb{N}$. Q: Why is binary operation needed before group? A: A group is a non-empty set together with a binary operation satisfying further axioms. ::: :::call-to-action[Group] subtitle: After learning about binary operations, continue to the formal definition of a group and its four structural conditions. button: Next Lesson | /abstract-algebra/introduction-to-groups/definition-and-examples-of-groups/group-definition button-ghost: Section Overview | /abstract-algebra/introduction-to-groups/definition-and-examples-of-groups/ :::