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Integers and Modern Applications for the Uninitiated

Sebastian Pauli

Section 13.4 Inverses

Two elements in a set are inverses when applying the binary operation to them gives the identity element for that operation.
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In the video in Figure 13.4.1 we explain when an element has an inverse for a binary operation. The video also gives examples. After the video, we give the formal definition of inverse elements and more examples. Then we give methods for deciding whether an element has an inverse for a binary operation.
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Figure 13.4.1. Inverse Elements by Matt Farmer and Stephen Steward.
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Carefully read the definition.
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Definition 13.4.2.

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 \(S\) for \(\bullet\text{,}\) and let \(s \in S\text{.}\) An element \(t\in S\) is an inverse of \(s\) for the operation \(\bullet\) if \(s\bullet t=e\) and \(t\bullet s=e\text{.}\)
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For the binary operation \(\bullet:S\times S\to S\text{,}\) inverses can exist only if \(S\) has an identity element for \(\bullet\text{.}\) This follows from the definition.
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When the inverse is unique, which is almost always the case in this course, we introduce a special notation for it.
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Definition 13.4.4.

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 for \(\bullet\text{,}\) we denote the inverse of \(s\) by \(\gexp{s}{-1}{\bullet}\text{.}\)
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The notation for inverses is similar to the notation for inverse functions. We place \({}^{-1}\) with the binary-operation symbol to show which operation the inverse belongs to.
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Proof.

Since \(e\) is the identity we have \(e\bullet e=e\text{.}\) So \(e\) satisfies all properties of the inverse of \(e\text{.}\)
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Also, 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 \(t\text{.}\)
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Example 13.4.7. Inverses with respect to known binary operations.

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 \(s\text{.}\)
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  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 \(\N\text{.}\)
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  3. As there is no identity with respect to subtraction of integers, there cannot be any inverses.
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Problem 13.4.8. Additive inverse of \(3\).

Find the inverse of \(3\) with respect to the addition of integers \(+:\Z\times\Z\to\Z\text{.}\)
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Solution.
We have \(3+(-3)=0\) and \((-3)+3=0\text{,}\) so \((-3)\) is the inverse of 3 with respect to addition of integers.
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Example 13.4.9. 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 13.1.4. Recall from Example 13.3.5 that \(\Ty\) is the identity element of \(T\) for \(\star\text{.}\) 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 \(\Tz^{-1\star}=\Tx\text{.}\) So every element in \(T\) has a unique inverse for \(\star\text{.}\)
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When an operation on a set is given by an operation table, first find the identity element (if there is one). Then find each place where the identity appears in the table. Trace that place to the header column on the left and the header row at the top to find elements that are inverses of each other.
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Example 13.4.10. Inverses with respect to \(\star:T\times T\to T\) revisited.

With the above comment in mind, we revisit Example 13.4.9. 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 matching element in the header column is \(\Ty\) and in the header row is \(\Ty\text{.}\)
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\(\star\) \(\Tx\) \(\color{red}\Ty\) \(\Tz\)
\(\Tx\) \(\Tz\) \(\Tx\) \(\Ty\)
\(\color{red}\Ty\) \(\Tx\) \(\color{gray}\Ty\) \(\Tz\)
\(\Tz\) \(\Ty\) \(\Tz\) \(\Tx\)
So, we see that \(\Ty \star \Ty = \Ty\) and conclude that \(\Ty^{-1\star} = \Ty\text{.}\)
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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 matching element in the header column is \(\Tz\) and in the header row is \(\Tx\text{.}\) Also, we find that for the second shaded \(\Ty\text{,}\) the matching element in the header column is \(\Tx\) and in the header row is \(\Tz\text{.}\)
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\(\star\) \(\color{red}\Tx\) \(\Ty\) \(\Tz\)
\(\Tx\) \(\Tz\) \(\Tx\) \(\Ty\)
\(\Ty\) \(\Tx\) \(\Ty\) \(\Tz\)
\(\color{red}\Tz\) \(\color{gray}\Ty\) \(\Tz\) \(\Tx\)
\(\star\) \(\Tx\) \(\Ty\) \(\color{red}\Tz\)
\(\color{red}\Tx\) \(\Tz\) \(\Tx\) \(\color{gray}\Ty\)
\(\Ty\) \(\Tx\) \(\Ty\) \(\Tz\)
\(\Tz\) \(\Ty\) \(\Tz\) \(\Tx\)
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 \(\Tz^{-1\star} = \Tx\text{.}\)
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Example 13.4.11. 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 for \(\oplus\) is 0 (compare Example 13.3.7). 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.
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  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 \(1\text{.}\)
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  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 \(2\text{.}\)
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