Definition 1.1. Binary Operation.
Assume \(S\) is a set. A binary operation on \(S\) is a function \(\ast \colon S \times S \to S\text{.}\)
Chapter 1 Binary Operations
Section 1.1 Basic Definitions and Examples
For \(a,b \in S\text{,}\) one writes \(\ast(a,b)\) for the result of the operation as a function. However, this notation becomes cumbersome, so we write \(a \ast b\) instead. When the operation is clear from context, we will frequently omit the operation and simply write \(ab\text{.}\)
We say our operation is written using multiplicative notation and we will frequently refer to the operation as multiplication. However, it is important to note that using multiplicative notation does not imply the operation is actually the familiar notion of multiplication of numbers.
Example 1.2. Arithmetic Binary Operations on Sets of Numbers.
-
The arithmetic operations of multiplication, \(\times\text{,}\) and addition, \(+\text{,}\) are binary operations on all your favorite number systems:\begin{align*} \N \amp= \{1,2,3,4,\ldots\},\\ \Z \amp= \{\ldots, -4,-3,-2,-1,0,1,2,3,4,\ldots\},\\ \Q \amp= \left\{\frac{n}{d} \;\middle\vert\; n,d \in \Z\right\},\, \text{and}\\ \R \amp= \left\{x \;\middle\vert\; x\ \text{is a real number}\right\} \end{align*}
-
The usual complex addition\begin{equation*} (a + ib) + (c + id) = (a + c) + i(b + d) \end{equation*}and complex multiplication\begin{equation*} (a + ib)(c + id) = (ac - bd) + i(ad + bc) \end{equation*}are both binary operations on \(\C = \left\{a + ib \;\middle\vert\; a,b \in \R\right\}\text{.}\)
Definition 1.3.
Assume \(X\) and \(Y\) are sets. Define the set of all set functions with domain \(X\) and codomain \(Y\) to be
\begin{equation*}
Y^X = \left\{f \colon X \to Y \;\middle\vert\; f\ \text{is a set function}\right\}\text{.}
\end{equation*}
Definition 1.4. The Set of Set Endomorphisms.
For any nonempty set \(X\text{,}\) we say \(f \in X^X\) is a set endomorphism of \(X\) and we write \(\End_{\Set}(X) = X^X\text{.}\)
Example 1.5. Composition of Endomorphisms.
For all \(f,g \in \End_{\Set}(X)\text{,}\) the function
\begin{align*}
f \circ g \colon X \amp\to X\\
x \amp \mapsto f\left(g\left(x\right)\right)
\end{align*}
is called the composition of \(f\) and \(g\). Composition defines a binary operation on \(\End_{\Set}(X)\text{.}\)
Warning 1.6.
Composition is not a binary operation in general. Suppose we drop the requirement that our functions are endomorphisms. We could, for example, compose the functions \(f \in Z^Y\) and \(g \in Y^X\) to create a new function \(f \circ g \in Z^X\text{.}\) However, these three functions each lie in different sets! That is to say, composition is a function
\begin{align*}
Z^Y \times Y^X \amp\to Z^X\\
(f,g) \amp\mapsto f \circ g
\end{align*}
involving three different sets, which does not conform to Definition 1.1.
Example 1.7. Addition of Vectors.
Fix \(n \in \N\text{.}\) Recall from linear algebra the sum of the vectors \(\mathbf{v} = (v_1, v_2, \ldots, v_n)\) and \(\mathbf{w} = (w_1, w_2, \ldots, w_n)\) is \(\mathbf{v} + \mathbf{w} = (v_1 + w_1, v_2 + w_2, \ldots, v_n + w_n)\text{.}\) This is a binary operation (called addition) on the set \(\R^n\) of \(n\)-dimensional vectors.
Definition 1.8. Matrix.
We denote by \(\operatorname{M}_{m \times n}(\R)\) the set of all \(m \times n\) matrices with entries from \(\R\text{.}\) When \(m = n\text{,}\) we simply write \(\operatorname{M}_n(\R)\) for the set of all \(n \times n\) matrices with entries from \(\R\text{.}\)
Example 1.9. Arithmetic Operations on Sets of Matrices.
Given \(M \in \operatorname{M}_{m \times n}(\R)\text{,}\) write \(\mathbf{v}_1, \ldots, \mathbf{v}_n \in \R^m\) for the columns of \(M\text{.}\) Then we can write
\begin{equation*}
M = \begin{bmatrix} \mathbf{v}_1 \amp \mathbf{v}_2 \amp \cdots \amp \mathbf{v}_n\end{bmatrix}\text{.}
\end{equation*}
Similarly, if we write \(\mathbf{w}_1, \ldots, \mathbf{w}_m \in \R^n\) for the rows of \(M\text{,}\) then we can write
\begin{equation*}
M = \begin{bmatrix}
\mathbf{w}_1\\
\mathbf{w}_2\\
\vdots\\
\mathbf{w}_m
\end{bmatrix}\text{.}
\end{equation*}
-
The usual addition of matrices\begin{equation*} \begin{bmatrix} \mathbf{v}_1 \amp \cdots \amp \mathbf{v}_n\end{bmatrix} + \begin{bmatrix} \mathbf{w}_1 \amp \cdots \amp \mathbf{w}_n\end{bmatrix} = \begin{bmatrix} \mathbf{v}_1 + \mathbf{w}_1 \amp \cdots \amp \mathbf{v}_n + \mathbf{w}_n\end{bmatrix} \end{equation*}is a binary operation on \(\operatorname{M}_{m \times n}(\R)\text{.}\)
-
Fix \(n \in \N\text{.}\) Recall the dot product (or scalar product) of vectors in \(\R^n\) is defined to be\begin{equation*} (v_1, \ldots, v_n) \cdot (w_1, \ldots, w_n) = v_1w_1 + \cdots + v_n w_n\text{.} \end{equation*}Given \(M,N \in \operatorname{M}_n(\R)\text{,}\) let \(\mathbf{v}_1, \ldots, \mathbf{v}_n \in \R^n\) be the rows of \(M\) and let \(\mathbf{w}_1, \ldots, \mathbf{w}_n \in \R^n\) be the columns of \(N\text{.}\) Recall the product of \(M\) and \(N\) is defined to be the \(n \times n\) matrix with entry \(\mathbf{v}_i \cdot \mathbf{w}_j\) in row \(i\) and column \(j\text{,}\)\begin{equation*} MN = \begin{bmatrix} \mathbf{v}_1 \cdot \mathbf{w}_1 \amp \mathbf{v}_1 \cdot \mathbf{w}_2 \amp \cdots \amp \mathbf{v}_1 \amp \mathbf{w}_n\\ \mathbf{v}_2 \cdot \mathbf{w}_1 \amp \mathbf{v}_2 \cdot \mathbf{w}_2 \amp \cdots \amp \mathbf{v}_2 \amp \mathbf{w}_n\\ \vdots \amp \vdots \amp \ddots \amp \vdots\\ \mathbf{w}_n \cdot \mathbf{w}_1 \amp \mathbf{w}_n \cdot \mathbf{w}_2 \amp \cdots \amp \mathbf{w}_n \amp \mathbf{w}_n\\ \end{bmatrix} \end{equation*}This is a binary operation (called multiplication) on \(\operatorname{M}_n(\R)\text{.}\)
Warning 1.10.
While we can define the product of matrices more generally, this product is not a binary operation unless our matrices are square. In order for the matrix product to be well-defined, we must have three positive integers \(m,n,p\) such that the left matrix has dimension \(m \times n\) and the right matrix has dimension \(n \times p\text{.}\) The result of the multiplication is a matrix of dimension \(m \times p\text{.}\) In general, this says that matrix multiplication is a function
\begin{align*}
\operatorname{M}_{m \times n}(\R) \times \operatorname{M}_{n \times p}(\R) \amp\to \operatorname{M}_{m \times p}(\R)\\
(M,N) \amp \mapsto MN
\end{align*}
which does not conform to Definition 1.1.
Example 1.11. Some More Operations.
We can concoct all sorts of operations on a given set that are not necessarily arithmetic. The only requirement for a binary operation is that it maps a pair of elements from the set to some element of the same set.
-
On the integers, we could define the uninteresting operation\begin{align*} \Z \times \Z \amp\to \Z\\ (m,n)\amp \mapsto 1234 \end{align*}that maps every pair to the number \(1234\text{.}\)
-
We could also define an operation such as\begin{align*} \Z \times \Z \amp \to \Z\\ (m,n) \amp \mapsto m \end{align*}that simply maps a pair to the entry in the first slot.
-
For a finite set like \(X = \left\{\bigcirc, \Box, \bigtriangleup,\diamond\right\}\text{,}\) we could define a binary operation, \(\ast\text{,}\) by simply listing where each pair gets mapped. The most concise way to do this is to use a table.
Figure 1.12. A table defining a binary operation on \(X\)
Section 1.2 Basic Properties of Binary Operations
Subsection 1.2.1 Closure
Definition 1.13. Subsets Closed Under an Operation.
Suppose \(X\) is a set equipped with an operation, \(\ast\text{.}\) We say \(Y \subseteq X\) is closed under \(\ast\) if for all \(y_1, y_2 \in Y\text{,}\) \(y_1 \ast y_2 \in Y\text{.}\) In this case, we refer to the restriction of \(\ast\) to the subset \(Y\) as the binary operation on \(Y\) induced by \(\ast\).
Example 1.14. A Subset Closed under Multiplication.
Consider the set \(\R\) equipped with the operation \(\times\text{.}\) The set \(\R^\times = \R\setminus\{0\}\) of non-zero real numbers is closed under \(\times\) because the product of two non-zero numbers is again a non-zero number.
Example 1.15. A Subset Not Closed under Addition.
Consider the set \(\R\) equipped with the operation \(+\text{.}\) The set \(\R^\times = \R\setminus\{0\}\) of non-zero real numbers is not closed under \(+\) because, for example, \(1 + -1 = 0 \not\in \R^\times\text{.}\)
Subsection 1.2.2 Associative Binary Operations
Definition 1.16. Associative Binary Operation.
Assume \(X\) is a set equipped with a binary operation. We say this operation is associative if for all \(x,y,z \in X\text{,}\)
\begin{equation*}
(x y) z = x (y z)\text{.}
\end{equation*}
Theorem 1.17. Associativity of Composition.
For all sets \(X\text{,}\) function composition is an associative binary operation on \(\End_{\Set}(X)\text{.}\)
Proof.
Let \(f,g,h \colon X \to X\) be given. We must show that \((f \circ g) \circ h = f \circ (g \circ h)\text{.}\) Recall that these two functions are equal if and only if for every \(x \in X\text{,}\) \((f \circ g) \circ h (x) = f \circ (g \circ h) (x)\text{.}\) This equality holds by definition of function composition:
\begin{align*}
(f \circ g) \circ h (x) \amp= f \circ g \left(h \left(x\right)\right)\\
\amp= f\left(g\left(h\left(x\right)\right)\right)\\
\amp= f\left(\left(g \circ h\right) \left(x\right)\right)\\
\amp= f \circ (g \circ h) (x)
\end{align*}
Subsection 1.2.3 Identity Elements
Definition 1.18. Identity.
Assume \(X\) is a set equipped with a binary operation.
-
We say \(\ast\) has an identity if there exists \(e \in X\) such that \(e\) is a left identity and a right identity for \(\ast\text{.}\)
Example 1.19. Familiar Identities.
-
For any number system (e.g. \(\N, \Z, \Q, \R, \C\)), the number \(1\) is the multiplicative identity because \(1 \times x = x = x\times 1\text{.}\)
-
For any number system except \(\N\) (e.g. \(\Z, \Q, \R, \C\)), the number \(0\) is the additive identity because \(0 + x = x = x + 0\text{.}\) Note that we exclude \(\N\) because \(0\) is not a natural number.
-
For a set \(X\text{,}\) the identity function\begin{align*} \operatorname{id}_X \colon X \amp\to X\\ x \amp \mapsto x \end{align*}is the identity for composition of functions on \(\End_{\Set}(X)\text{.}\)
-
The identity element for addition on \(\operatorname{M}_n(\R)\) is \(m \times n\) zero matrix,\begin{equation*} \mathbf{0} = \begin{bmatrix} 0 \amp 0 \amp \cdots \amp 0\\ 0 \amp 0 \amp \cdots \amp 0\\ \vdots \amp \vdots \amp \ddots \amp \vdots\\ 0 \amp 0 \amp \cdots \amp 0 \end{bmatrix} \end{equation*}that we recall satisfies the property for all \(M \in \operatorname{M}_{m \times n}(\R)\text{,}\)\begin{equation*} \mathbf{0} + M = M = M + \mathbf{0}\text{.} \end{equation*}
-
The identity element for multiplication on \(\operatorname{M}_n(\R)\) is the \(n \times n\) identity matrix,\begin{equation*} I_n = \begin{bmatrix} 1 \amp 0 \amp \cdots \amp 0\\ 0 \amp 1 \amp \cdots \amp 0\\ \vdots \amp \vdots \amp \ddots \amp \vdots\\ 0 \amp 0 \amp \cdots \amp 1 \end{bmatrix} \end{equation*}that we recall satisfies the condition for all \(M \in \operatorname{M}_n(\R)\text{,}\)\begin{equation*} M I_n = M = I_n M\text{.} \end{equation*}
Example 1.20. Left Identity.
Consider the binary operation on the set \(\{a,b,c\}\) given by the following table.
The element \(a\) is a left identity because
\begin{align*}
aa \amp= a \amp ab \amp= b \amp ac \amp= c
\end{align*}
but not a right identity because \(ba = a \neq b\text{.}\)
Example 1.22. Right Identity.
Consider the binary operation on the set \(\{a,b,c\}\) given by the following table.
The element \(a\) is a right identity because
\begin{align*}
aa \amp= a \amp ba \amp= b \amp ca \amp= c
\end{align*}
but not a left identity because \(ab = a \neq b\text{.}\)
Warning 1.24.
Left identities and right identities are not guaranteed to be unique! In Example 1.20, we saw that \(a\) is a left identity. However, \(b\) is also a left identity because
\begin{align*}
ba \amp= a \amp bb \amp= b \amp bc \amp= c\text{.}
\end{align*}
Similarly, in Example 1.22, we saw that \(a\) and \(b\) are both right identities. This problem vanishes when the binary operation has both a left and right identity element.
Theorem 1.25.
Assume \(X\) is a set equipped with a binary operation. This binary operation has an identity if and only if it has a left and right identity.
Proof.
Assume \(e \in X\) is the identity for the binary operation. Then \(e\) is a left and right identity by definition.
Conversely, assume there exist \(e,f \in X\) such that \(e\) is a left identity and \(f\) is a right identity. Then \(f = e f\) because \(e\) is a left identity and \(e = ef\) because \(f\) is a right identity. Therefore \(e = e f = f\) is an identity for \(\ast\text{.}\)
Corollary 1.26. Uniqueness of Identity.
Assume \(X\) is a set equipped with a binary operation \(\ast \colon X \times X \to X\text{.}\) Then \(\ast\) has at most one identity element.
Proof.
Assume \(e,f \in X\) are both identities for \(\ast\text{.}\) By definition, \(e\) is a left identity for \(\ast\) and \(f\) is a right identity for \(\ast\text{.}\) Therefore \(e = f\) by Theorem 1.25.
When we are working with a general binary operation, we have already adopted the convention that \(xy\) is the result of performing the binary operation on \(x,y \in X\text{.}\) To further the analogy with multiplication, we will typically refer to the identity element (when it exists!) as \(1\text{.}\) In the specific case of the binary operation \(+\text{,}\) we write \(0\) for the identity element. We will frequently refer to the element \(1\) as the multiplicative identity and to the element \(0\) as the additive identity.
Subsection 1.2.4 Inverses
Definition 1.27. Inverse.
Assume \(X\) is a set equipped with a binary operation \(\ast \colon X \times X \to X\) with identity.
-
We say \(x \in X\) has an inverse if there exists an element of \(X\) that is both a left and right inverse for \(x\text{.}\) We write \(x^{-1}\) for the inverse of \(x\) under a general binary operation and we write \(-x\) when the binary operation is addition.
Example 1.28. Familiar Inverses.
-
For any number system (e.g. \(\N, \Z, \Q, \R, \C\)), the inverse of a number \(x\) under addition is just \(-x\) because\begin{equation*} x + (-x) = 0 = -x + x\text{.} \end{equation*}We call this the additive inverse of \(x\text{.}\)
-
For \(\Q\) or \(\R\text{,}\) the inverse of a non-zero number \(x\) under multiplication is \(x^{-1} = 1/x\) because\begin{equation*} x\left(\frac{1}{x}\right) = 1 = \left(\frac{1}{x}\right)x\text{.} \end{equation*}We call this the multiplicative inverse of \(x\text{.}\)
-
For a set \(X\text{,}\) we say a function, \(f \colon X \to X\text{,}\) has an inverse if there exists a function \(g \colon X \to X\) such that\begin{equation*} f \circ g = \id_X \quad\text{and}\quad g \circ f = \id_X\text{.} \end{equation*}In this situation, we typically write \(g = f^{-1}\) and we say that \(f\) is invertible.A concrete example of this is the function\begin{align*} f \colon \R \amp\to \R\\ x \amp \mapsto x + 2 \end{align*}whose inverse is the function\begin{align*} f^{-1} \colon R \amp\to \R\\ x \amp\mapsto x - 2 \end{align*}because\begin{align*} f^{-1} \circ f(x) \amp= f^{-1}(x + 2) = (x+2)-2 = x,\,\text{and}\\ f \circ f^{-1}(x) \amp= f(x - 2) = (x-2) + 2 \end{align*}
Warning 1.29.
It is important to remember that the inverse of an element under a binary operation must be an element of the set on which the binary operation is defined. One may be tempted to say that every non-zero integer, \(n\text{,}\) has a multiplicative inverse, \(1/n\text{.}\) This is true if we are working with multiplication defined on \(\Q\) and we identify \(n = n/1\text{.}\) However, this is false when we are working with multiplication defined on \(\Z\text{!}\) For example, the multiplicative inverse of \(2\) is \(1/2\text{,}\) but \(1/2\) is not an integer.
Example 1.30. Left/Right Inverse.
Consider the function
\begin{align*}
f \colon \N \amp\to \N\\
x \amp\mapsto x + 1
\end{align*}
The function
\begin{align*}
g \colon \N \amp\to \N\\
x \amp\mapsto
\begin{cases}
1 \amp\text{if}\ x = 1\\
x - 1 \amp\text{else}
\end{cases}
\end{align*}
is a left inverse for \(f\) because for all \(n \in \N\text{,}\) \(n + 1 \geq 2\) implies
\begin{equation*}
g \circ f(n) = g(n+1) = (n+1) - 1 = n = \id_{\N}(n)\text{.}
\end{equation*}
However, \(g\) is not a right inverse for \(f\) because
\begin{equation*}
f \circ g(1) = f(1) = 2 \neq 1\text{.}
\end{equation*}
Symmetrically, \(f\) is a right inverse for \(g\) and not a left inverse for \(g\text{.}\)
Theorem 1.31.
Assume \(X\) is a set equipped with an associative binary operation with identity. For all \(x \in X\text{,}\) \(x\) has an inverse if and only if \(x\) has a left and right inverse.
Proof.
Let \(x \in X\) be given. If \(x \in X\) has an inverse, then \(x^{-1}\) is a left inverse and a right inverse by definition.
Conversely, assume that \(y \in X\) is a left inverse for \(x\) and \(z \in X\) is a right inverse for \(x\text{.}\) Then
\begin{align*}
y \amp= y1\\
\amp= y(xz)\\
\amp= (yx)z\\
\amp= 1z\\
\amp= z
\end{align*}
by associativity. Therefore \(x^{-1} = y = z\text{.}\)
Corollary 1.32. Uniqueness of Inverses.
Assume \(X\) is a set equipped with an associative binary operation with identity. For all \(x \in X\text{,}\) the inverse of \(x\) is unique when it exists.
Proof.
Assume \(x \in X\) has an inverse, \(x^{-1}\text{.}\) If \(y\) satisfies \(xy = 1 = yx\text{,}\) then \(x^{-1}\) is a left inverse of \(x\) and \(y\) is a right inverse of \(x\text{.}\) Therefore \(x^{-1} = y\) by Theorem 1.31.
Warning 1.33.
The previous result requires the associativity assumption. Consider the set \(\{1,a,b,c\}\) equipped with the binary operation given by the following table.
It’s easy to see that \(1\) is the identity, this binary operation is not associative because
\begin{equation*}
a(bc) = a1 = a \neq c = 1c = (ab)c\text{,}
\end{equation*}
and the element \(a\) has three distinct inverses—\(a\text{,}\) \(b\text{,}\) and \(c\)—because
\begin{align*}
aa \amp = 1\\
ab \amp = ba = 1\\
ac \amp = ca = 1
\end{align*}
Subsection 1.2.5 Commutative Binary Operations
Definition 1.35. Commutative.
Assume \(X\) is a set equipped with a binary operation. We say this binary operation is commutative if for all \(x,y \in X\text{,}\) \(xy = yx\text{.}\)
If we wish to emphasize that a binary operation is not commutative, we will say it is noncommutative
Example 1.36. Familiar Commutative Binary Operations.
-
In nearly every context in which we encounter addition, it will always be a commutative binary operation.
-
Multiplication of numbers is always commutative
Example 1.37. Noncommutative Example.
Recall from linear algebra that matrix multiplication is noncommutative in general. For example,
\begin{equation*}
\begin{pmatrix}
1 \amp 2 \\
3 \amp 4
\end{pmatrix}
\begin{pmatrix}
4 \amp 3 \\
2 \amp 1
\end{pmatrix}
=
\begin{pmatrix}
8 \amp 5 \\
20 \amp 13
\end{pmatrix}
\end{equation*}
is not the same as
\begin{equation*}
\begin{pmatrix}
4 \amp 3 \\
2 \amp 1
\end{pmatrix}
\begin{pmatrix}
1 \amp 2 \\
3 \amp 4
\end{pmatrix}
=
\begin{pmatrix}
13 \amp 20 \\
5 \amp 8
\end{pmatrix}
\end{equation*}
Exercises 1.3 Exercises For Undergrads & Grads
Pointwise Binary Operations.
Assume \(Y\) is a set equipped with a binary operation, \(\ast\text{.}\) For any nonempty set \(X\text{,}\) define the binary operation pointwise \(\ast\) on \(Y^X\) as follows. For all \(f, g \in Y^X\text{,}\)
\begin{align*}
f \ast g \colon X \amp\to Y\\
x \amp \mapsto f(x) \ast g(x)
\end{align*}
or, equivalently, \((f\ast g)(x) = f(x) \ast g(x)\text{.}\)
1.
Prove that if \(\ast\) is an associative binary operation on \(Y\text{,}\) then for every nonempty set \(X\text{,}\) pointwise \(\ast\) is an associative binary operation on \(Y^X\text{.}\)
Solution.
Assume \(\ast\) is an associative binary operation on \(Y\text{.}\) Assume \(X\) is any nonempty set. Let \(f,g,h \in Y^X\) be given. For all \(x \in X\text{,}\) \(f(x), g(x), h(x) \in Y\) implies
\begin{equation*}
f(x) \ast \left(g(x) \ast h(x)\right) = \left(f(x) \ast g(x)\right) \ast h(x)
\end{equation*}
because \(\ast\) is associative. Hence
\begin{align*}
\left(f\ast\left(g\ast h\right)\right)(x) \amp= f(x) \ast \left(g \ast h\right)(x)\\
\amp= f(x) \ast \left(g(x) \ast h(x)\right)\\
\amp= \left(f(x) \ast g(x)\right) \ast h(x)\\
\amp= \left(f \ast g\right)(x) \ast h(x)\\
\amp= \left(\left(f \ast g\right) \ast h\right)(x)\text{.}
\end{align*}
Therefore pointwise \(\ast\) is an associative binary operation on \(Y^X\text{.}\)
2.
Prove that if \(e \in Y\) is the identity element for \(\ast\text{,}\) then for any nonempty set \(X\text{,}\) the function
\begin{align*}
\epsilon \colon X \amp\to Y\\
x \amp \mapsto e
\end{align*}
is the identity element for pointwise \(\ast\) on \(Y^X\text{.}\)
Solution.
Assume \(e \in Y\) is the identity element for \(\ast\text{.}\) Assume \(X\) is any nonempty set. Let \(f \in Y^X\) be given. For every \(x \in X\text{,}\) \(f(x) \in Y\) implies
\begin{align*}
\left(\epsilon \ast f\right)(x) \amp = \epsilon(x) \ast f(x)\\
\amp= e \ast f(x)\\
\amp = f(x)\\
\amp= f(x) \ast e\\
\amp= f(x) \ast \epsilon(x)\\
\amp= (f\ast \epsilon)(x)\text{.}
\end{align*}
Therefore \(\epsilon \in Y^X\) is the identity element for pointwise \(\ast\text{.}\)
3.
Assume \(e \in Y\) is the identity for \(\ast\) and for every \(y \in Y\text{,}\) \(y\) has an inverse under \(\ast\text{,}\) \(y^{-1} \in Y\text{.}\) Prove that for every nonempty set \(X\text{,}\) the inverse of \(f \in Y^X\) under pointwise \(\ast\) is the function
\begin{align*}
f^{-1} \colon X \amp\to Y\\
x \amp\mapsto f(x)^{-1}
\end{align*}
Solution.
Assume \(X\) is any nonempty set. Let \(f \in Y^X\) be given. For all \(x \in X\text{,}\)
\begin{align*}
\left(f \ast f^{-1}\right)(x) \amp= f(x) \ast f^{-1}(x)\\
\amp= f(x) \ast \left(f(x)\right)^{-1}\\
\amp= e\\
\amp= \left(f(x)\right)^{-1} \ast f(x)\\
\amp= f^{-1}(x) \ast f(x)\\
\amp= \left(f^{-1} \ast f\right)(x)
\end{align*}
Therefore \(f^{-1}\) is the inverse of \(f\text{.}\)
4.
Assume \(\ast\) is a commutative binary operation on \(Y\text{.}\) Prove that for every nonempty set \(X\text{,}\) pointwise \(\ast\) is a commutative binary operation on \(Y^X\text{.}\)
Solution.
Assume \(X\) is any nonempty set. Let \(f,g \in Y^X\) be given. For all \(x \in X\text{,}\) \(f(x), g(x) \in Y\) implies
\begin{equation*}
f(x) \ast g(x) = g(x) \ast f(x)\text{.}
\end{equation*}
Hence for all \(x \in X\text{,}\)
\begin{align*}
(f \ast g)(x) \amp= f(x) \ast g(x)\\
\amp= g(x) \ast f(x)\\
\amp= (g \ast f)(x)\text{.}
\end{align*}
Therefore pointwise \(\ast\) is a commutative binary operation on \(Y^X\text{.}\)
Exercises 1.4 Exercises For Grads
1.
Let \(S\) be a set and let \(\ast\) be a binary operation on \(S\) satisfying the two laws
Show that \(\ast\) is associative and commutative
Solution.
Observe that it suffices to prove that \(\ast\) is commutative, for then the second axiom guarantees
\begin{equation*}
(xy)z = (yz)x = x(yz)\text{.}
\end{equation*}
Towards that end, let \(x,y \in S\) be given. Write
\begin{align*}
yx \amp= (yx)(yx)\\
\amp= x(yx)y\\
\amp= [(yx)y]x\\
\amp= [(xy)y]x\\
\amp= [(yy)x]x\\
\amp= (yx)x\\
\amp = (xx)y\\
\amp= xy
\end{align*}
