Invertible Functions and Binary Operations

Invertible Functions: Definition

The concept of an invertible function formalizes the idea of a function that can be "undone" or reversed. Not all functions have this property. An invertible function establishes a one-to-one correspondence between its domain and codomain, allowing for a unique mapping back from the codomain to the domain.

Definition of an Invertible Function

A function $f: A \to B$ is said to be invertible if there exists a function $g: B \to A$ such that the composition of $g$ with $f$ is the identity function on set $A$, and the composition of $f$ with $g$ is the identity function on set $B$.

Formally, $f: A \to B$ is invertible if there exists a function $g: B \to A$ such that:

1. The composite function $g \circ f$ maps every element in $A$ back to itself. This means $g(f(x)) = x$ for all $x \in A$. The function $g \circ f$ is the identity function on the set $A$, denoted $I_A$.

   $\mathbf{g \circ f = I_A}$

2. The composite function $f \circ g$ maps every element in $B$ back to itself. This means $f(g(y)) = y$ for all $y \in B$. The function $f \circ g$ is the identity function on the set $B$, denoted $I_B$.

   $\mathbf{f \circ g = I_B}$

If such a function $g$ exists, it is unique and is called the inverse function of $f$. The inverse function is denoted by $f^{-1}$.

So, if $f$ is invertible, its inverse $f^{-1}: B \to A$ satisfies:

• $f^{-1}(f(x)) = x$ for all $x \in A$.
• $f(f^{-1}(y)) = y$ for all $y \in B$.

The defining property of an inverse function can also be stated concisely: If $f(x) = y$ for some $x \in A$ and $y \in B$, then the inverse function maps $y$ back to $x$.

Important Note: The notation $f^{-1}(x)$ for the inverse function should not be confused with $\frac{1}{f(x)}$ (the reciprocal of $f(x)$), which is a completely different concept unless specifically noted by context (which is rare in this domain).

Condition for a Function to be Invertible (Bijective)

Not every function has an inverse. The existence of an inverse function is tied directly to the type of function it is (specifically, its injectivity and surjectivity).

Theorem: Invertibility and Bijectivity

A function $f: A \to B$ is invertible if and only if $f$ is bijective (meaning $f$ is both one-to-one/injective and onto/surjective).

This theorem provides a necessary and sufficient condition for a function to be invertible.

Proof of the Theorem:

We need to prove two implications:

1. If $f$ is invertible, then $f$ is bijective.

2. If $f$ is bijective, then $f$ is invertible.

Part 1: If $f$ is invertible, then $f$ is bijective.

Assume $f: A \to B$ is invertible. By definition, there exists a function $f^{-1}: B \to A$ such that $(f^{-1} \circ f)(x) = x$ for all $x \in A$ and $(f \circ f^{-1})(y) = y$ for all $y \in B$.

• Proof of Injectivity: Let $x_1, x_2 \in A$ such that $f(x_1) = f(x_2)$.

  $f(x_1) = f(x_2)$

  $\textsf{(Assume)}$

  Apply $f^{-1}$ to both sides of the equation:

  $f^{-1}(f(x_1)) = f^{-1}(f(x_2))$

  By the definition of the inverse function ($(f^{-1} \circ f)(x) = x$), the left side is $x_1$ and the right side is $x_2$.

  $x_1 = x_2$

  Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$, the function $f$ is one-to-one (injective).

• Proof of Surjectivity: Let $y$ be an arbitrary element in the codomain $B$. We need to show that there exists at least one element $x \in A$ such that $f(x) = y$.

  Let $y \in B$

  Since $f^{-1}: B \to A$, applying $f^{-1}$ to $y$ gives an element in $A$. Let $x = f^{-1}(y)$.

  Let $x = f^{-1}(y)$

  Now consider $f(x)$: $f(x) = f(f^{-1}(y))$.

  By the definition of the inverse function ($(f \circ f^{-1})(y) = y$), $f(f^{-1}(y)) = y$.

  $f(x) = y$

  We have found an element $x \in A$ (specifically, $x = f^{-1}(y)$) such that $f(x) = y$. Since $y$ was an arbitrary element in $B$, this holds for all $y \in B$. Therefore, $f$ is onto (surjective).

Thus, if $f$ is invertible, it is both injective and surjective, i.e., $f$ is bijective.

Part 2: If $f$ is bijective, then $f$ is invertible.

Assume $f: A \to B$ is bijective. This means $f$ is both one-to-one and onto.

We need to show that an inverse function $g: B \to A$ exists that satisfies the conditions $g \circ f = I_A$ and $f \circ g = I_B$.

Let $y$ be any element in the codomain $B$. Since $f$ is onto, there exists at least one element $x \in A$ such that $f(x) = y$.

Since $f$ is also one-to-one, this element $x$ in $A$ is unique for that particular $y \in B$. For every $y \in B$, there is one and only one $x \in A$ such that $f(x) = y$.

This unique correspondence allows us to define a rule for a function $g: B \to A$. For each $y \in B$, define $g(y)$ to be that unique element $x \in A$ such that $f(x) = y$.

Now let's verify the composition properties for this function $g$:

• Check $g \circ f = I_A$: Let $x \in A$. We need to show $(g \circ f)(x) = x$.

  $(g \circ f)(x) = g(f(x))$

  Let $f(x) = y$. By the way we defined $g$, $g(y)$ is the element in A that maps to $y$ under $f$. Since $f(x)=y$, this element is $x$.

  $g(f(x)) = g(y) = x$

  So, $(g \circ f)(x) = x$ for all $x \in A$. This means $g \circ f = I_A$.

• Check $f \circ g = I_B$: Let $y \in B$. We need to show $(f \circ g)(y) = y$.

  $(f \circ g)(y) = f(g(y))$

  Let $g(y) = x$. By the way we defined $g$, $x$ is the unique element in A such that $f(x) = y$.

  $f(g(y)) = f(x) = y$

  So, $(f \circ g)(y) = y$ for all $y \in B$. This means $f \circ g = I_B$.

Since we have defined a function $g: B \to A$ that satisfies both $g \circ f = I_A$ and $f \circ g = I_B$, the function $f$ is invertible, and $g$ is its inverse, $f^{-1}$.

Thus, if $f$ is bijective, it is invertible.

Combining Part 1 and Part 2, we conclude that a function $f$ is invertible if and only if it is bijective.

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Finding the Inverse of a Function

If a function $f: A \to B$ is bijective, we know it is invertible and its inverse $f^{-1}: B \to A$ exists. We can find the rule for this inverse function using an algebraic procedure, especially for real functions defined by an equation.

Steps to Find the Inverse Function $f^{-1}(x)$:

1. Verify Invertibility:

   First, confirm that the function $f: A \to B$ is indeed bijective (both one-to-one and onto) on the specified domain $A$ and codomain $B$. If it's not bijective, a strict inverse function does not exist (though sometimes invertible-like functions can be found by restricting the domain/codomain).

2. Set up the Equation:

   Write the equation for the function in the form $y = f(x)$. This represents the original mapping from an input $x$ in the domain to an output $y$ in the codomain.

3. Interchange Variables:

   Swap the roles of $x$ and $y$ in the equation. This step is crucial; it conceptually reverses the mapping. The equation becomes $x = f(y)$. Now, $x$ represents an input from the codomain of $f$ (which is the domain of $f^{-1}$), and $y$ represents the corresponding output in the domain of $f$ (which is the codomain of $f^{-1}$).

4. Solve for the New y:

   Algebraically solve the new equation ($x = f(y)$) to express $y$ in terms of $x$. Isolate $y$ on one side of the equation. This new expression for $y$ gives the rule for the inverse function, where the input is $x$ (from the original codomain) and the output is $y$ (mapping back to the original domain).

5. Replace y with $f^{-1}(x)$:

   Replace the variable $y$ in the final expression from step 4 with the standard notation for the inverse function, $f^{-1}(x)$. This gives the rule for the inverse function in terms of $x$.

6. Specify Domain and Codomain of $f^{-1}$:

   State the domain and codomain of the inverse function $f^{-1}$.
   • The domain of $f^{-1}$ is the codomain of the original function $f$. (Since $f$ is bijective, this is also the range of $f$).
   • The codomain of $f^{-1}$ is the domain of the original function $f$.
   So, if $f: A \to B$ is bijective, then $f^{-1}: B \to A$.

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Graphical Interpretation of Inverse Functions

The graph of the inverse function $f^{-1}$ is directly related to the graph of the original function $f$. If the point $(a, b)$ is on the graph of $f$, it means $f(a) = b$. Since $f$ is invertible, this implies $f^{-1}(b) = a$, so the point $(b, a)$ is on the graph of $f^{-1}$. The points $(a, b)$ and $(b, a)$ in the Cartesian plane are reflections of each other across the line $y = x$.

Therefore, the graph of $f^{-1}$ is the reflection of the graph of $f$ across the line $y = x$. This provides a visual way to understand and sometimes find inverse functions.

(Imagine a graph showing a curve for $y=f(x)$, the line $y=x$, and the curve for $y=f^{-1}(x)$ which is the mirror image of $y=f(x)$ across $y=x$).

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Binary Operations: Definition

In mathematics, operations like addition, subtraction, multiplication, and division are fundamental ways to combine two numbers to get a result. The concept of a binary operation generalizes this idea to any set. A binary operation on a set is a rule for combining any two elements of the set to produce a unique element that must also belong to the same set.

Definition of a Binary Operation

A binary operation $*$ on a non-empty set $S$ is a function that takes any ordered pair of elements from $S$ and assigns to it a unique element that is also in $S$.

Symbolically, a binary operation $*$ on a set $S$ is a function:

The domain of the function is the Cartesian product $S \times S$ (all possible ordered pairs of elements from $S$), and the codomain is the set $S$ itself. For an ordered pair $(a, b) \in S \times S$, the unique element assigned by the operation $*$ is usually denoted by $a * b$ instead of $*(a, b)$.

Closure Property

A key requirement in the definition of a binary operation $*$ on a set $S$ is that the result of combining any two elements from $S$ must also be an element of $S$. This property is called the closure property.

For a rule $*$ to be a binary operation *on a set S*, the set S must be closed under $*$. If there exists at least one pair of elements $(a, b)$ from $S$ such that $a * b$ is not in $S$, then $*$ is not a binary operation *on S*.

Examples of Binary Operations:

Let's examine some common operations on various sets to see if they are binary operations on those sets.

1. Addition (+) on the set of Natural Numbers $\mathbb{N} = \{1, 2, 3, ...\}$:

   Is $+$ a binary operation on $\mathbb{N}$? We need to check closure. Let $a, b \in \mathbb{N}$. Is $a+b$ always in $\mathbb{N}$? Yes, the sum of any two natural numbers is always a natural number.

   $1+2 = 3 \in \mathbb{N}$

   So, + is a binary operation on $\mathbb{N}$.

2. Subtraction (-) on the set of Natural Numbers $\mathbb{N}$:

   Is $-$ a binary operation on $\mathbb{N}$? We check closure. Let $a=3$ and $b=5$. Both $3 \in \mathbb{N}$ and $5 \in \mathbb{N}$. Calculate $a - b = 3 - 5 = -2$. Is $-2 \in \mathbb{N}$? No. Since we found at least one pair of elements from $\mathbb{N}$ whose difference is not in $\mathbb{N}$, subtraction is not a binary operation on $\mathbb{N}$.

   (However, subtraction is a binary operation on the set of Integers $\mathbb{Z}$, because the difference of any two integers is always an integer).

3. Multiplication ($\times$) on the set of Real Numbers $\mathbb{R}$:

   Is $\times$ a binary operation on $\mathbb{R}$? We check closure. Let $a, b \in \mathbb{R}$. Is $a \times b$ always in $\mathbb{R}$? Yes, the product of any two real numbers is always a real number.

   $\sqrt{2} \times \pi = \sqrt{2}\pi \in \mathbb{R}$

   So, $\times$ is a binary operation on $\mathbb{R}$.

4. Division ($\div$) on the set of Real Numbers $\mathbb{R}$:

   Is $\div$ a binary operation on $\mathbb{R}$? We check closure and whether it's defined for all pairs. Let $a=5 \in \mathbb{R}$ and $b=0 \in \mathbb{R}$. The ordered pair is $(5, 0) \in \mathbb{R} \times \mathbb{R}$. Calculate $a \div b = 5 \div 0$. This is undefined. Since the operation is not defined for all ordered pairs in $\mathbb{R} \times \mathbb{R}$, $\div$ is not a binary operation on $\mathbb{R}$.

   (However, $\div$ is a binary operation on the set of non-zero real numbers $\mathbb{R} \setminus \{0\}$, because for any $a, b \in \mathbb{R} \setminus \{0\}$, $a/b$ is defined and is also in $\mathbb{R} \setminus \{0\}$).

5. Maximum operation ($\max$) on $\mathbb{R}$:

   Define an operation $*$ by $a * b = \max(a, b)$. Is $*$ a binary operation on $\mathbb{R}$? We check closure. Let $a, b \in \mathbb{R}$. Is $\max(a, b)$ always in $\mathbb{R}$? Yes, the maximum of any two real numbers is a real number.

   $\max(7, -3) = 7 \in \mathbb{R}$

   So, $\max$ is a binary operation on $\mathbb{R}$.

6. Concatenation on the set of strings:

   Let $S$ be the set of all finite strings of characters (e.g., S = {"hello", "world", "a", "123", ""}). Define an operation $*$ by $s_1 * s_2 =$ concatenation of $s_1$ and $s_2$. Let $s_1, s_2 \in S$. Is $s_1 * s_2$ (the combined string) always a finite string? Yes. So, concatenation is a binary operation on the set of finite strings.

The definition of a binary operation requires that for *every* ordered pair of elements from the set, the operation produces a *unique* result *within* that set. This is strict.

Properties of Binary Operations (Closure, Commutativity, Associativity, Identity, Inverse)

Once we have established that a rule is a binary operation on a set $S$, we can further analyze the operation by examining several important properties it might possess. These properties reveal the algebraic structure of the set under the given operation.

Let $*$ be a binary operation on a set $S$ (meaning $S$ is closed under $*$ by definition).

Properties:

1. Closure Property:

   As stated in the definition, a binary operation $*$ on $S$ implies that $S$ is closed under $*$. This means for all $a, b \in S$, the result $a * b$ is always an element of $S$.

   $\forall a, b \in S, a * b \in S$

   When asked to check if a given operation is a binary operation *on a set S*, this closure property is the key thing to verify.

2. Commutativity:

   A binary operation $*$ on a set $S$ is commutative if the order in which the two elements are combined does not affect the result.

   Condition: $\mathbf{a * b = b * a}$ for all $a, b \in S$.

   If there exists even one pair of elements $a, b \in S$ such that $a * b \neq b * a$, then the operation is not commutative.

   Examples:

   • Standard addition (+) and multiplication ($\times$) on number sets like $\mathbb{Z}, \mathbb{Q}, \mathbb{R}$ are commutative ($a+b=b+a$, $a \times b = b \times a$).
   • The maximum operation ($\max(a, b)$) on $\mathbb{R}$ is commutative ($\max(a, b) = \max(b, a)$).
   • The operation $a * b = a+b-ab$ on $\mathbb{Q}$ (from Example 5 below) is commutative because $a+b-ab = b+a-ba$.

   Counter-examples:

   • Standard subtraction (-) on $\mathbb{Z}$: $3 - 5 = -2$, but $5 - 3 = 2$. Since $-2 \neq 2$, subtraction is not commutative on $\mathbb{Z}$.
   • Standard division ($\div$) on $\mathbb{Q}^+$: $2 \div 4 = 1/2$, but $4 \div 2 = 2$. Since $1/2 \neq 2$, division is not commutative on $\mathbb{Q}^+$.
   • Matrix multiplication is generally not commutative ($AB \neq BA$ for matrices A, B).

3. Associativity:

   A binary operation $*$ on a set $S$ is associative if the way elements are grouped does not affect the result when combining three elements.

   Condition: $\mathbf{(a * b) * c = a * (b * c)}$ for all $a, b, c \in S$.

   If there exists even one triplet of elements $a, b, c \in S$ such that $(a * b) * c \neq a * (b * c)$, then the operation is not associative.

   Examples:

   • Standard addition (+) and multiplication ($\times$) on number sets like $\mathbb{Z}, \mathbb{Q}, \mathbb{R}$ are associative ($(a+b)+c = a+(b+c)$, $(a \times b) \times c = a \times (b \times c)$).

   Counter-examples:

   • Standard subtraction (-) on $\mathbb{Z}$: $(8 - 5) - 2 = 3 - 2 = 1$. $8 - (5 - 2) = 8 - 3 = 5$. Since $1 \neq 5$, subtraction is not associative on $\mathbb{Z}$.
   • Standard division ($\div$) on $\mathbb{Q}^+$: $(8 \div 4) \div 2 = 2 \div 2 = 1$. $8 \div (4 \div 2) = 8 \div 2 = 4$. Since $1 \neq 4$, division is not associative on $\mathbb{Q}^+$.

4. Existence of Identity Element:

   An element $e \in S$ is called an identity element (or neutral element) for the binary operation $*$ on $S$ if, when combined with any element $a \in S$, it leaves $a$ unchanged. The identity element must work from both the left and the right.

   Condition: There exists an element $e \in S$ such that for all $a \in S$, $\mathbf{a * e = a}$ and $\mathbf{e * a = a}$.

   • If such an identity element exists for an operation on a set, it is unique.
   • Not every binary operation on every set has an identity element.

   Examples:

   • For standard addition (+) on $\mathbb{Z}$, the identity element is 0, because $a + 0 = a$ and $0 + a = a$ for all $a \in \mathbb{Z}$.
   • For standard multiplication ($\times$) on $\mathbb{R}$, the identity element is 1, because $a \times 1 = a$ and $1 \times a = a$ for all $a \in \mathbb{R}$.
   • For standard addition (+) on $\mathbb{N} = \{1, 2, 3, ...\}$, there is no identity element within the set $\mathbb{N}$. While $a+0=a$, $0 \notin \mathbb{N}$.
   • For subtraction (-) on $\mathbb{Z}$, there is no identity element. If $e$ were the identity, $a - e = a$ for all $a$, implying $e=0$. But $e - a = 0 - a = -a$, which is not equal to $a$ for all $a$ (only for $a=0$). So 0 is a right identity but not a left identity. For an identity, it must work on both sides.

5. Existence of Inverse Element:

   Suppose a binary operation $*$ on $S$ has an identity element $e$. An element $a \in S$ is said to have an inverse (or be invertible) with respect to $*$ if there exists an element $b \in S$ such that combining $a$ and $b$ (in either order) results in the identity element $e$.

   Condition: For a given $a \in S$, there exists an element $b \in S$ such that $\mathbf{a * b = e}$ and $\mathbf{b * a = e}$.

   • The element $b$ is called the inverse of $a$ and is often denoted by $a^{-1}$. In the case of additive operations, the inverse of $a$ is often denoted as $-a$.
   • The inverse of an element might not exist for all elements in the set $S$.
   • If an inverse exists for an element $a$ with respect to an associative operation, that inverse is unique.

   Examples:

   • For standard addition (+) on $\mathbb{Z}$ (identity $e=0$), the inverse of an element $a \in \mathbb{Z}$ is $-a$, because $a + (-a) = 0$ and $(-a) + a = 0$. Since $-a$ is always an integer when $a$ is an integer, every element in $\mathbb{Z}$ has an additive inverse in $\mathbb{Z}$.
   • For standard multiplication ($\times$) on $\mathbb{R}$ (identity $e=1$), the inverse of a non-zero element $a \in \mathbb{R}$ is $1/a$, because $a \times (1/a) = 1$ and $(1/a) \times a = 1$. If $a \neq 0$, then $1/a$ is always a real number. The element 0 does not have a multiplicative inverse in $\mathbb{R}$ (since $0 \times b = 0 \neq 1$).
   • For standard multiplication ($\times$) on $\mathbb{N} = \{1, 2, 3, ...\}$ (identity $e=1$), only the element 1 has a multiplicative inverse in $\mathbb{N}$ (which is $1$ itself, $1 \times 1 = 1$). For any other element $a > 1$, its multiplicative inverse $1/a$ is not a natural number.

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