Problem 53 - Discrete Mathematics with Applications

Question

Let \(f: A \rightarrow B, g: B \rightarrow C,\) and \(h: C \rightarrow D .\) Prove that \(h \circ(g \circ f)=\) \((h \circ g) \circ f\) (associative property). [Hint: Verify that \((h \circ(g \circ f))(x)=((h \circ g) \circ f)(x) \text { for every } x \text { in } A .]\)

Step-by-Step Solution

Answer

In order to prove the associative property of function composition, we compute \((h \circ(g \circ f))(x)\) and \(((h \circ g) \circ f)(x)\) for given functions \(f : A \rightarrow B\), \(g : B \rightarrow C\), and \(h : C \rightarrow D\). We find that \((h \circ(g \circ f))(x) = h(g(f(x)))\) and \(((h \circ g) \circ f)(x) = h(g(f(x)))\), which are equal for every \(x \in A\). Thus, we can conclude that function composition is associative, and \(h \circ(g \circ f) = (h \circ g) \circ f\).

Step 1: Define the problem and the terms

We are given three functions, \(f : A \rightarrow B\), \(g : B \rightarrow C\), and \(h : C \rightarrow D\). The function composition is represented by the symbol \(\circ\). For two functions \(f\) and \(g\), the composition is denoted by \((g \circ f)(x) = g(f(x))\). To prove the associative property of function composition, we need to show that \(h \circ(g \circ f) = (h \circ g) \circ f\), which means \((h \circ(g \circ f))(x) = ((h \circ g) \circ f)(x)\) for every \(x \in A\).

Step 2: Compute \(h \circ(g \circ f)(x)\)

First, we will compute \((h \circ(g \circ f))(x)\), which, by the definition of function composition, is equal to \(h((g \circ f)(x))\). Since \((g \circ f)(x) = g(f(x))\), we can substitute it back into the expression to get \(h(g(f(x)))\). So, \((h \circ(g \circ f))(x) = h(g(f(x)))\).

Step 3: Compute \(((h \circ g) \circ f)(x)\)

Now, we will compute \(((h \circ g) \circ f)(x)\), which, by the definition of function composition, is equal to \(((h \circ g))(f(x))\). Since \((h \circ g)(x) = h(g(x))\), we can substitute it back into the expression with \(x\) replaced by \(f(x)\), to get \(h(g(f(x)))\). So, \(((h \circ g) \circ f)(x) = h(g(f(x)))\).

Step 4: Compare the results

Comparing the two results, we have: \((h \circ(g \circ f))(x) = h(g(f(x)))\) and \(((h \circ g) \circ f)(x) = h(g(f(x)))\) Since the two expressions are equal for every \(x \in A\), this proves that function composition is associative: \(h \circ(g \circ f) = (h \circ g) \circ f\).

Key Concepts

Function Composition

Function composition involves applying one function to the outcomes of another function. Suppose we have the functions \(f: A \rightarrow B\), \(g: B \rightarrow C\), and \(h: C \rightarrow D\). When we talk about composing functions, we denote this by \((h \circ g \circ f)(x)\), which means applying \(f\) to \(x\), \(g\) to the result of \(f(x)\), and finally applying \(h\) to \(g(f(x))\). The composition of functions relies heavily on the correct ordering of these operations, and it is notated from right to left. This means you first apply \(f\) to \(x\), then \(g\) to \(f(x)\), and \(h\) to \(g(f(x))\), resulting in \(h(g(f(x)))\). In our exercise, verifying that \((h \circ (g \circ f))(x) = ((h \circ g) \circ f)(x)\) helps in understanding the associative property of function composition.

• Order matters: function composition is applied right to left.
• Skip no steps: ensure each composition step is correctly applied.

Mathematical Proof

A mathematical proof is a logical argument that demonstrates the truth of a mathematical statement. It is essential in mathematics to ensure that results are valid and reliable. In proving the associative property for function composition, our goal is to show equality between \(h \circ(g \circ f)\) and \((h \circ g) \circ f\) for any input \(x\).The proof often starts by defining expressions or manipulations of the given components:

• Define each composed function separately; for example, \((h \circ(g \circ f))(x)\) simplifies to \(h(g(f(x)))\).
• Similarly, simplify \(((h \circ g) \circ f)(x)\) to \(h(g(f(x)))\).
• Compare both reduced expressions to show they are identical for any \(x\) in \(A\).

Engaging with these steps ensures that the assertion of the associative property isn’t a mere assumption but a confirmed mathematical truth.

Set Theory

Set theory is the foundation of modern mathematics and helps to understand functions and mappings. In our scenario, it allows us to clearly define the domains and codomains of the functions \(f\), \(g\), and \(h\).

• \(f: A \rightarrow B\) denotes that \(f\) maps elements from set \(A\) to set \(B\).
• Likewise, \(g: B \rightarrow C\) maps elements of \(B\) to \(C\), and \(h: C \rightarrow D\) maps \(C\) to \(D\).

Set theory ensures functions have clear inputs and outputs, supporting the logical flow in verifying associativity. It provides a structural background that is crucial when discussing subsets, elements, and ensuring that elements are appropriately transported across function compositions.

Functions and Mappings

Functions and mappings are crucial in understanding how elements from one set correspond to elements in another set. Each function is a relation where each input is related to exactly one output. In our example:

• \(f: A \rightarrow B\) associates each element in \(A\) with one in \(B\).
• \(g: B \rightarrow C\) continues this mapping from \(B\) to \(C\).
• \(h: C \rightarrow D\) similarly maps \(C\) to \(D\).

These mappings are not just simple relations but function-specific, meaning the association must be consistent and well-defined across the domains and codomains indicated. Understanding functions in terms of mappings helps clarify why the compositions work seamlessly and underscore mathematical structures like associativity in our exercise framework.