In Part I we have already discussed the commutativity of addition and multiplication of integers. Commutativity of addition meant that, for example, \(2+7=9\) and also \(7+2=9\text{.}\) Also recall that this property does not hold for subtraction, as is proved by the counterexample \(2-7=-5\) but \(7-2=5\text{.}\) In the following we introduce the commutative property for general binary operations.
In the video in Figure 13.38 we introduce the commutative property for general binary operations and give examples. Following the video we present the formal definition of commutativity, give examples, and discuss methods for determining whether an binary operation is commutative in detail.
Let \(S\) be a set and \(\bullet:S\times S \to S\) be a binary operation on \(S\text{.}\) Then, \(\bullet\) is commutative if \(a \bullet b=b \bullet a\) for all \(a\in S\) and \(b\in S\text{.}\)
We already know that addition and multiplication of integers are commutative. In the following example we also investigate whether subtraction of integers is commutative.
We have \(5-2=3\) and \(2-5=(-3)\text{.}\) As \(3\ne(-3)\) we have \(5-2\ne 2-5\text{.}\) This counterexample shows that the binary operation \(-:\Z\times\Z\to\Z\) is not commutative.
Example13.42.Commutativity of \(\star:T\times T\to T\).
Let \(T=\{\Tx,\Ty,\Tz\}\text{,}\) and let the binary operation \(\star:T\times T\to T\) be given by the table in Example 13.4. To prove that \(\star\) is commutative, we exhaust all possibilities. We verify that for all \(a\in T\) and \(b\in T\text{,}\)
\begin{equation*}
a\star b\text{ is equal to } b \star a
\end{equation*}
by separately computing \(a\star b\) in the left column and \(b\star a\) in the right column and noticing that the two computations in each row match.
In the case where one of the general elements is the identity element, there is a shortcut. We can handle several cases at the same time by setting one of the two general elements equal to the identity element and using a variable for the other general element. Recall that the identity element is \(\Ty\) for \(T\) with respect to \(\star\text{.}\) Then, for all \(a\in T\) we have:
Now, note that if the two general elements are the same, there is nothing to check. For all \(a \in T\text{,}\) we trivially have that \(a\star a = a\star a\text{.}\) So, the only remaining case to check is covered here:
We have shown that \(a\star b=b\star a\) for all \(a\in T\) and \(b\in T\text{.}\) Thus, the binary operation \(\star:T\times T \to T\) is commutative.
When we have an operation on a set given by an operation table, we can determine whether or not the operation is commutative by observing whether or not the operation table possesses a particular symmetry. We locate the diagonal of the table from the operation symbol in the top left corner of the table to the bottom right corner of the table. Then, we determine whether or not that diagonal acts as a mirror for the other entries in the table. If so, the operation is commutative.
Example13.43.Commutativity of \(\star:T\times T\to T\) revisited.
With the above comment in mind, we revisit Example 13.42. We shade the diagonal that must act as a mirror for the other entries in the table if the operation is commutative. Then, we individually verify the symmetry by pointing out the pairs of entries that need to match and noting that they do, in fact, match.
Example13.44.Commutativity of \(\oplus:\Z_5\times\Z_5\to\Z_5\).
Consider the binary operation \(\oplus:\Z_5\times\Z_5\to\Z_5\) defined by \(a\oplus b=(a+b)\fmod 5\text{.}\) We follow an approach that is similar to that from Example 13.13 to show that \(\oplus\) is commutative. Let \(a\in\Z_5\) and \(b\in\Z_5\text{.}\) By the definition of \(\oplus\) and the commutativity of addition of integers we have
The operation \(\diamond\) is commutative if for all \(a\) and \(b\) in \(A\) we have \(a\diamond b=b\diamond a\text{.}\) In particular we must have \(\Td\diamond\Tc=\Tc\diamond\Td\text{.}\) Since \(\Td\diamond\Tc=\Th\) we must also have \(\Tc\diamond\Td=\Th\text{.}\) Hence the element \(\Th\) in the box makes \(\diamond\) commutative.
Consider the binary operation subtraction \(-:\Z\times\Z\to\Z\text{.}\) Since \(3-2=1\) and \(2-3=-1\text{,}\) and \(1\ne -1\text{,}\) the binary operation \(-\) is not commutative.
Problem13.47.Is this binary operations commutative ?
Decide whether the binary operation \(\otimes:\Z_{11}^\otimes\times \Z_{11}^\otimes\to \Z_{11}^\otimes\) given by \(a\otimes b = (a\cdot b) \fmod 11\) is commutative.
When we suspect that a binary operation is not commutative, we look for a counterexample. If we find a counterexample we have proven that the binary operation is not commutative.
We know that subtraction of integers is not commutative. So we suspect that the binary operation \(\ominus\) that is based on subtraction is not commutative. We find a counterexample. Let \(a:=1\) and \(b:=0\text{.}\) Then
Decide whether the following binary operations are commutative. If the binary operation is not commutative, give a counterexample, otherwise leave the field empty.
(1) Let the binary operation \(\otimes:\mathbb{Z}_{15}\times \mathbb{Z}_{15}\to \mathbb{Z}_{15}\) be given by \(a \otimes b = \left(a\cdot b\right) \bmod 15\text{.}\)
(2) Let the binary operation \(\ominus:\mathbb{Z}_{14}\times \mathbb{Z}_{14}\to \mathbb{Z}_{14}\) be given by \(a \ominus b = \left(a-b\right) \bmod 14\text{.}\)
(3) Let the binary operation \(\ominus:\mathbb{Z}_{10}\times \mathbb{Z}_{10}\to \mathbb{Z}_{10}\) be given by \(a \ominus b = \left(a-b\right) \bmod 10\text{.}\)
The binary operation \(\otimes\) is commutative if \(a \otimes b = b \otimes a\) for all \(a\in \mathbb{Z}_{15}\) and for all \(b\in \mathbb{Z}_{15}\text{.}\)
The binary operation \(\ominus\) is commutative if \(a \ominus b = b \ominus a\) for all \(a\in \mathbb{Z}_{14}\) and for all \(b\in \mathbb{Z}_{14}\text{.}\)
The binary operation \(\ominus\) is commutative if \(a \ominus b = b \ominus a\) for all \(a\in \mathbb{Z}_{10}\) and for all \(b\in \mathbb{Z}_{10}\text{.}\)