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 13.25 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.
Figure13.25.Inverse Elements by Matt Farmer and Stephen Steward.
Carefully read the definition.
Definition13.26.
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 13.27 reproduce the definition by filling in the blanks.
Checkpoint13.27.Inverses.
Definition
Let S be a set and let * : S \(\times\) S \(\to\) S be a binary operation on S. We read a * b as ’a star b’.
Let e be
select
the identity with respect to * in S
some element in S
some odd element in S
some even element in S
some green element in S
.
An element b in S is an inverse of a in S with respect to * if
select
(a * b) * c = a * (b * c)
a * b = b * a
a * e = a and e * a = a
a * b = e and b * a = e
.
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 \(\bullet\text{.}\)
Theorem13.28.
Let \(S\) be a set and \(\bullet:S\times S \to S\) be a binary operation on \(S\text{.}\) If \(\bullet\) is associative and \(t\in S\) is the inverse of \(s\in S\) with respect to \(\bullet\text{,}\) then \(t\) is the only inverse of \(s\) 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.
Definition13.29.
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.
Theorem13.30.
Let \(e\) be the identity with respect to an associative binary operation \(\bullet:S\times S\to S\) on a set \(S\text{.}\) Then \(e^{-1\bullet}=e\text{.}\)
Proof.
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 \(t\text{.}\)
Theorem13.31.
Let \(S\) be a set and \(\bullet:S\times S \to S\) be a binary operation on \(S\text{.}\) If \(s\) has an inverse \(s^{-1\bullet}\) then
Example13.32.Inverses with respect to known binary operations.
We continue to consider the binary operations from Example 13.3:
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 \(s\text{.}\)
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 \(\N\text{.}\)
As there is no identity with respect to subtraction of integers, there cannot be any inverses.
Problem13.33.Additive inverse of \(3\).
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.
Example13.34.Inverses with respect to \(\star:T\times T\to T\).
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. Recall from Example 13.20 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.
Example13.35.Inverses with respect to \(\star:T\times T\to T\) revisited.
With the above comment in mind, we revisit Example 13.34. 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 \(\Tz^{-1\star} = \Tx\text{.}\)
Example13.36.Inverses with respect to \(\oplus:\Z_5\times\Z_5\to\Z_5\).
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 13.22). We explicitly give the inverse of each element in \(\Z_5=\{0,1,2,3,4\}\text{.}\)
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.
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 \(1\text{.}\)
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 13.37 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.
Checkpoint13.37.Find Inverses.
Fill in the operation table for the binary operation \(\oplus\) on the set \(\mathbb{Z}_{6}\) defined by \(a \oplus b = (a + b)\bmod 6\) :
\(\oplus\)
1
2
3
4
5
1
2
3
5
1
2
4
1
3
4
5
4
4
3
5
5
1
Complete the following:
In \(\mathbb{Z}_{6}\) with respect to \(\oplus\)
select
the identity element is 0
the identity element is 1
there is no identity element
.
Find the inverses of the elements of \(\mathbb{Z}_{6}\) with respect to \(\oplus\) . If an element does not have an inverse answer ’none’.
