Definition 13.1.
A binary operation \(\bullet\) on a set \(S\) is a function \(\bullet:S\times S\to S\text{.}\) For the image of \((a,b)\in S\times S\) under the function \(\bullet\) we write \(a\bullet b\) (read ‘\(a\) dot \(b\)’).
Section 13.1 Definition of a Binary Operation
A binary operation can be considered as a function whose input is two elements of the same set \(S\) and whose output also is an element of \(S\text{.}\) Two elements \(a\) and \(b\) of \(S\) can be written as a pair \((a,b)\text{.}\) As \((a,b)\) is an element of the Cartesian product \(S\times S\) we specify a binary operation as a function from \(S\times S\) to \(S\text{.}\)
We use symbols to represent binary operations instead of function names, just as we do with addition and multiplication of integers. Addition uses the symbol \(+\) and multiplication uses the symbol \(\cdot\text{.}\) We will use symbols such as \(\star\) and \(\bullet\) to represent arbitrary (non-specific) binary operations, and we will also define new binary operations using the symbols \(\oplus\) and \(\otimes\text{.}\)
We give an overview over the remainder of the section in the video in Figure 13.2.
We give examples for binary operations that we have encountered before.
Example 13.3. Known binary operations.
- The addition of integers \(+:\Z\times\Z\to\Z\) is a binary operation on \(\Z\text{.}\) We denote the image of \((a, b) \in \Z \times \Z\) by \(a+b\text{.}\)
- The multiplication of natural numbers \(\cdot:\N\times\N\to\N\) is a binary operation on \(\N\text{.}\) We denote the image of \((a, b) \in \N \times \N\) by \(a\cdot b\text{.}\)
- The subtraction of integers \(-:\Z\times\Z\to\Z\) is a binary operation on \(\Z\text{.}\) We denote the image of \((a, b) \in \Z \times \Z\) by \(a- b\text{.}\)
As is the case for other functions, there are several ways of specifying a binary operation. If the set is small, we sometimes specify the binary operation by a table.
Example 13.4. A binary operation given by a table.
Let \(T:=\{\Tx,\Ty,\Tz\}\text{.}\) The binary operation \(\star:T\times T \to T\) is given by the operation table:
| \(\star\) | \(\Tx\) | \(\Ty\) | \(\Tz\) |
|---|---|---|---|
| \(\Tx\) | \(\Tz\) | \(\Tx\) | \(\Ty\) |
| \(\Ty\) | \(\Tx\) | \(\Ty\) | \(\Tz\) |
| \(\Tz\) | \(\Ty\) | \(\Tz\) | \(\Tx\) |
From the table, we can obtain \(a\star b\) (read “\(a\) star \(b\)”) for each \(a,b \in T\text{:}\)
To determine the value of \(\Ty\star\Tz\) we go to the \(\Ty\) row which is
| \(\star\) | \(\Tx\) | \(\Ty\) | \(\Tz\) |
|---|---|---|---|
| \(\cdots\) | |||
| \(\Ty\) | \(\Tx\) | \(\Ty\) | \(\Tz\) |
| \(\cdots\) | |||
In the \(\Tz\) column of this row we now find the value of \(\Ty\star\Tz\text{,}\) namely \(\Tz\text{.}\)
When we go through all possible combinations we obtain:
\begin{gather*}
\Tx \star \Tx=\Tz\\
\Tx\star \Ty=\Tx\\
\Tx\star \Tz=\Ty\\
\Ty \star \Tx=\Tx\\
\Ty\star \Ty=\Ty\\
\Ty\star \Tz=\Tz\\
\Tz \star \Tx=\Ty\\
\Tz\star \Ty=\Tz\\
\Tz\star \Tz=\Tx
\end{gather*}
As before we use parenthesis to indicate order of operations. We first evaluate the expression in the parenthesis.
Checkpoint 13.5. Binary operation.
Let the binary operation \(\Box\) (box) on the set \(F=\lbrace\) n, o, p, q, r, s, t, u, v, w\(\rbrace\) be defined by:
| \(\Box\) | n | o | p | q | r | s | t | u | v | w |
|---|---|---|---|---|---|---|---|---|---|---|
| n | n | n | n | n | n | n | n | n | n | n |
| o | o | q | u | r | w | v | t | p | s | n |
| p | p | v | r | q | n | p | v | r | q | n |
| q | q | r | v | p | n | q | r | v | p | n |
| r | r | p | q | v | n | r | p | q | v | n |
| s | s | p | t | v | w | r | u | q | o | n |
| t | t | r | o | p | w | q | s | v | u | n |
| u | u | v | s | q | w | p | o | r | t | n |
| v | v | q | p | r | n | v | q | p | r | n |
| w | w | n | w | n | w | n | w | n | w | n |
We read o \(\Box\) u as o box u.
Find the following.
o \(\Box\) u =
u \(\Box\) o =
r \(\Box\) o =
o \(\Box\) q =
(r \(\Box\) o) \(\Box\) q =
r \(\Box\) (o \(\Box\) q) =
Sometimes it can be useful to generate the operation table from a binary operation given by an algebraic rule.
Example 13.6. The binary operation \(\oplus:\Z_5\times\Z_5\to\Z_5\).
The operation table for the binary operation \(\oplus:\Z_5\times\Z_5\to\Z_5\) given by \(a\oplus b=(a+b)\fmod 5\) is:
| \(\oplus\) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| 0 | 0 | 1 | 2 | 3 | 4 |
| 1 | 1 | 2 | 3 | 4 | 0 |
| 2 | 2 | 3 | 4 | 0 | 1 |
| 3 | 3 | 4 | 0 | 1 | 2 |
| 4 | 4 | 0 | 1 | 2 | 3 |
We read \(a\oplus b\) as “\(a\) mod plus \(b\text{.}\)”
In Checkpoint 13.7 complete the operation table for a binary operation.
Checkpoint 13.7. Operation table.
Fill in the operation table for the binary operation \(\otimes\) on the set \(\mathbb{Z}_{6}\) defined by \(a \otimes b = (a \cdot b)\bmod 6\) :
The left column represents the \(a\) values and the top row represents the \(b\) values.
| \(\otimes\) | 1 | 2 | 3 | 4 | 5 | |
|---|---|---|---|---|---|---|
| 1 | 5 | |||||
| 2 | 4 | 2 | 4 | |||
| 3 | 3 | |||||
| 4 | 2 | |||||
| 5 | 4 |
