Section13.4Inverses

When a binary operation is performed on two elements in a set and the result is the identity element of the set, with respect to the binary operation, the elements are said to be inverses of each other.

In the video in Figure 13.4.1 we say when an element has an inverse with respect to a binary operations and give examples. Following the video we present the formal definition of inverse elements, give examples, and discuss methods for determining whether an element has an inverse with respect to a binary operation with.

Definition13.4.2.

Let $$S$$ be a set and $$\bullet:S\times S \to S$$ be a binary operation on $$S\text{.}$$ Suppose that $$e$$ is the identity element of $$S$$ with respect to $$\bullet\text{,}$$ and let $$s \in S\text{.}$$ An element $$t\in S$$ is an inverse of $$s$$ with respect to the operation $$\bullet$$ if $$s\bullet t=e$$ and $$t\bullet s=e\text{.}$$

In Checkpoint 13.4.3 reproduce the definition by filling in the blanks.

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 $$a \otimes b = \left(a\cdot b\right) \bmod 15\text{.}$$

The binary operation $$\otimes$$ is

• select

• commutative

• not commutative

.

Counterexample: The statement is false, because for $$a=$$ $$\in \mathbb{Z}_{15}$$ we have $$a \otimes 9 \ne 9 \otimes a\text{.}$$

(2) Let the binary operation $$\ominus:\mathbb{Z}_{14}\times \mathbb{Z}_{14}\to \mathbb{Z}_{14}$$ be given by $$a \ominus b = \left(a-b\right) \bmod 14\text{.}$$

The binary operation $$\ominus$$ is

• select

• commutative

• not commutative

.

Counterexample: The statement is false, because for $$a=$$ $$\in \mathbb{Z}_{14}$$ we have $$a \ominus 12 \ne 12 \ominus a\text{.}$$

(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 $$\ominus$$ is

• select

• commutative

• not commutative

.

Counterexample: The statement is false, because for $$a=$$ $$\in \mathbb{Z}_{10}$$ we have $$a \ominus 3 \ne 3 \ominus a\text{.}$$

Hint.

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 $$b\in \mathbb{Z}_{10}\text{.}$$

$$\text{commutative}$$

$$\text{}$$

$$\text{not commutative}$$

$$-2.34679\times 10^{9}$$

$$\text{not commutative}$$

$$-2.34679\times 10^{9}$$

It follows directly from the definition that inverses with respect to a binary operation $$\bullet:S\times S\to S$$ can only exist if the set $$S$$ contains an identity element with respect to $$\bullet\text{.}$$

When the inverse is unique, which is almost always the case in this course, we introduce a special notation for it.

Definition13.4.5.

Let $$S$$ be a set and $$\bullet:S\times S \to S$$ be a binary operation on $$S\text{.}$$ If $$s\in S$$ has exactly one inverse with respect to $$\bullet\text{,}$$ we denote the inverse of $$s$$ by $$\gexp{s}{-1}{\bullet}\text{.}$$

The notation for inverses uses notation similar to what we used for function inverses. The symbol used for the binary operation is shown with the $${}^{-1}$$ to remind you with respect to which binary operation it is the inverse.

Since $$e$$ is the identity we have $$e\bullet e=e\text{.}$$ So $$e$$ satisfies all properties of the inverse of $$e\text{.}$$

Furthermore, from the condition $$s\bullet t=e$$ and $$t\bullet s=e$$ in the definition, we know that if $$t$$ is the inverse of $$s\text{,}$$ then $$s$$ is the inverse $$t\text{.}$$

We continue to consider the binary operations from Example 13.1.3:

1. We consider the addition of integers $$+:\Z\times\Z\to\Z\text{.}$$ Recall that the identity element of $$\Z$$ with respect to addition is 0. Let $$s\in\Z\text{,}$$ and note that its negative $$-s$$ is also in $$\Z\text{.}$$ Since $$s + (-s)=0$$ and $$(-s) + s=0\text{,}$$ we may conclude that $$-s$$ is an inverse of $$s$$ in $$\Z$$ with respect to $$+\text{.}$$ In fact, it is the only such inverse, and we call $$-s$$ the additive inverse of $$s\text{.}$$

2. We consider the multiplication of natural numbers $$\cdot:\N\times\N\to\N\text{.}$$ Recall that the identity element of $$\N$$ with respect to multiplication is 1. For $$2 \in \N\text{,}$$ we are looking for an element $$t$$ such that $$2\cdot t=1$$ and $$t\cdot 2=1\text{.}$$ The only choice would be $$t = \frac{1}{2}\text{;}$$ however, $$\frac{1}{2}$$ is not a natural number. So, 2 does not have a multiplicative inverse in the set of natural numbers. In fact, for each natural number $$n>1\text{,}$$ we have that $$\frac{1}{n} \notin \N\text{,}$$ implying that that each natural number $$n>1$$ does not have a multiplicative inverse in $$\N\text{.}$$

3. As there is no identity with respect to subtraction of integers, there cannot be any inverses.

Find the inverse of $$3$$ with respect to the addition of integers $$+:\Z\times\Z\to\Z\text{.}$$

Solution.

We have $$3+(-3)=0$$ and $$(-3)+3=0\text{,}$$ so $$(-3)$$ is the inverse of 3 with respect to addition of integers.

Let $$T=\{\Tx,\Ty,\Tz\}\text{,}$$ and let the binary operation $$\star:T\times T\to T$$ be given by the table in Example 13.1.4. Recall from Example 13.3.6 that $$\Ty$$ is the identity element of $$T$$ with respect to $$\star\text{.}$$ As is always the case, the inverse of the identity element is itself, so the unique inverse of $$\Ty$$ is $$\Ty^{-1\star}=\Ty\text{.}$$ Also, since $$\Tx\star \Tz=\Ty$$ and $$\Tz \star \Tx=\Ty\text{,}$$ $$\Tx$$ and $$\Tz$$ are inverses of each other. Since there are no other elements in $$T$$ that satisfy the requirements to be an inverse of either $$\Tx$$ or of $$\Tz\text{,}$$ we may communicate the uniqueness by writing $$\Tx^{-1\star}=\Tz$$ and $$\Tz^{-1\star}=\Tx\text{.}$$ Thus every element in $$T$$ has a unique inverse with respect to $$\star\text{.}$$

When we have an operation on a set given by an operation table, we can determine which elements are inverses of each other by first determining the identity element (if there is one). Then, we locate the identity element within the table and trace back to the header column on the left side of the table and the header row on the top of the table to find elements that are inverses of each other.

With the above comment in mind, we revisit Example 13.4.10. Recall that the identity element is $$\Ty\text{.}$$ First, we trace back to the header column on the left side of the table and the header row on the top of the table from the following shaded $$\Ty$$ within the table. We find that the corresponding element in the header column is $$\Ty$$ and in the header row is $$\Ty\text{.}$$

 $$\star$$ $$\Tx$$ $$\color{red}\Ty$$ $$\Tz$$ $$\Tx$$ $$\Tz$$ $$\Tx$$ $$\Ty$$ $$\color{red}\Ty$$ $$\Tx$$ $$\color{gray}\Ty$$ $$\Tz$$ $$\Tz$$ $$\Ty$$ $$\Tz$$ $$\Tx$$

So, we see that $$\Ty \star \Ty = \Ty$$ and conclude that $$\Ty^{-1\star} = \Ty\text{.}$$

Now, we trace back to the header column on the left side of the table and the header row on the top of the table from each of the following two shaded $$\Ty$$'s within the table. We find that for the first shaded $$\Ty\text{,}$$ the corresponding element in the header column is $$\Tz$$ and in the header row is $$\Tx\text{.}$$ Furthermore, we find that for the second shaded $$\Ty\text{,}$$ the corresponding element in the header column is $$\Tx$$ and in the header row is $$\Tz\text{.}$$

 $$\star$$ $$\color{red}\Tx$$ $$\Ty$$ $$\Tz$$ $$\Tx$$ $$\Tz$$ $$\Tx$$ $$\Ty$$ $$\Ty$$ $$\Tx$$ $$\Ty$$ $$\Tz$$ $$\color{red}\Tz$$ $$\color{gray}\Ty$$ $$\Tz$$ $$\Tx$$
 $$\star$$ $$\Tx$$ $$\Ty$$ $$\color{red}\Tz$$ $$\color{red}\Tx$$ $$\Tz$$ $$\Tx$$ $$\color{gray}\Ty$$ $$\Ty$$ $$\Tx$$ $$\Ty$$ $$\Tz$$ $$\Tz$$ $$\Ty$$ $$\Tz$$ $$\Tx$$

From the first highlighted table, we see that $$\Tz \star \Tx = \Ty\text{,}$$ and from the second highlighted table, we see that $$\Tx \star \Tz = \Ty\text{.}$$ Since $$\Tz \star \Tx = \Ty$$ and $$\Tx \star \Tz = \Ty\text{,}$$ we simultaneously conclude that $$\Tx^{-1\star} = \Tz$$ and that $$\Tz^{-1\star} = \Tx\text{.}$$

Consider the binary operation $$\oplus:\Z_5\times\Z_5\to\Z_5$$ given by $$a\oplus b=(a+b)\fmod 5\text{.}$$ The identity element with respect to $$\oplus$$ is 0 (compare Example 13.3.8). We explicitly give the inverse of each element in $$\Z_5=\{0,1,2,3,4\}\text{.}$$

1. As $$0\oplus 0=(0+0)\fmod 5=0\fmod 5=0$$ the inverse of $$0$$ with respect to $$\oplus$$ is $$0\text{.}$$ This illustrates our earlier observation that the inverse of the identity element is the identity element.

2. As $$1\oplus 4=(1+4)\fmod 5=5\fmod 5=0$$ and $$4\oplus 1=(4+1)\fmod 5=5\fmod 5=0$$ the inverse of $$1$$ with respect to $$\oplus$$ is $$4\text{.}$$ This also shows that the inverse of $$4$$ with respect to $$\oplus$$ is $$1\text{.}$$

3. As $$2\oplus 3=(2+3)\fmod 5=5\fmod 5=0$$ and $$3\oplus 2=(3+2)\fmod 5=5\fmod 5=0$$ the inverse of $$2$$ with respect to $$\oplus$$ is $$3\text{.}$$ This also shows that the inverse of $$3$$ with respect to $$\oplus$$ is $$2\text{.}$$

In Checkpoint 13.4.13 complete the operation table for the given binary operation and then use the table to find the identity element and the inverses with respect to that binary operation.

A set S with a binary operation * on S is a commutative group if

$$\bullet$$ there is

• select

• a complement

• an element

• an identity

• an inverse

• a set

• an operation

with respect to * in S and

$$\bullet$$ for each a in S there is

• select

• a complement

• an element

• an identity

• an inverse

• a set

• an operation

with respect to * in S and

$$\bullet$$ the operation * is

• select

• associative and commutative

• associative and transitive

• commutative and symmetric

.

$$\text{an identity}$$
$$\text{an inverse}$$
$$\text{associative and commutative}$$