CSCI 301 L15 Notes

Lecture 15 - Notes

Relations

• Be able to recognize when a relation is an equivalence relation
• Know how to identify the equivalence classes in a set given an equivalence relation on that set.

Functions

• Know the definition of a function.
• Know how to identify a function’s domain, codomain, and range.

Announcements

• A4 due Friday

Equivalence Relations and Equivalence Classes

Example:

Show that \(\equiv \pmod{3}\) (that is, congruence mod 3 is an equivalence relation.

Recall: \(a \equiv b \pmod{n}\) means \(n \mid (a-b)\).

• Reflexive: is \(a \equiv a \pmod{n}\)?

  • Yes: \(n \mid (a-a)\), or \(n \mid 0\) because \(0 = kn\) where the integer \(k = 0\).

• Symmetric: \(a \equiv b \pmod{n} \Rightarrow b \equiv a \pmod{n}\)?

  • Yes: By definition of modular congruence, \(n \mid (a-b)\) means \((a-b) = kn\) for some integer \(k\). Negating both sides we get \((b-a) = -kn\). Therefore \(n \mid (b-a)\), so \(b \equiv a \pmod{n}\).

• Transitive: \(\left(a \equiv b \pmod{n} \land b \equiv c \pmod{n}\right) \Rightarrow b \equiv a \pmod{n}\)

  • Yes: Suppose \(a \equiv b \pmod{n}\) and \(b \equiv c \pmod{n}\). Then \(n \mid (a-b)\), so \((a-b) = pn\) for some integer \(p\), and \((n \mid b-c)\), so \((b-c) = qn\) for some integer \(q\). Add these two equations together: \[ \begin{align*} a - b &= pn\\ b - c &= qn \end{align*} \] to yield $$ \[\begin{align*} (a - c) &= pn + qn\\ (a - c) &= n(p+q) \end{align*}\] $$ which satisfies the definition of \(n \mid (a - c)\), and therefore \(a \equiv c \pmod{n}\).

Equivalence Classes

Example: Consider the relation on \(A = \{-1, 1, 2, 3, 4\}\) that means “has the same sign as”:

Do Exercises Part A

Relations Between Sets

You can have a relation between two sets. In this case, \(R \subseteq A \times B\). Otherwise, everything’s more or less the same.

Example: Let \(A = \{1, 2\}\) and \(B = \mathcal{P}(A)\). Define \(R = \{1, \{1\}), (2, \{2\}), (1, \{1, 2\}), (2, \{1, 2\})\}\)

In this case, the relation means \(\in\).

In this case, if we draw a diagram it will have two groups of points, one for \(A\) and one for \(B\), and arrows can only go from a member of the \(A\) group to a member of the \(B\) group.

Functions

We’re used to seeing functions like: \(f(x) = x^2\). Here we’ll define them more rigorously and ground them in our existing understanding of sets as the foundation of mathematics.

Definition: Suppose \(A\) and \(B\) are sets. A function from \(A\) to \(B\), written \(f: A \rightarrow B\), is a relation \(f \subseteq A \times B\) that has the property that for each \(a \in A\), the relation \(f\) contains exactly one ordered pair of the form \((a, b)\). We abbreviate the statement \((a, b) \in f\) as \(f(a) = b\).

Intuition: A function is a special kind of relation that relates all elements of \(A\) to exactly one element of \(B\). There cannot be any \(a \in A\) that doesn’t map to any \(b\), and there can’t be any \(a \in A\) that maps to more than one \(b \in B\).

Example: Let \(f: \mathbb{Z} \rightarrow \mathbb{N} = \{(n, |n| + 2) : n \in \mathbb{Z}\}\)

Definition: If \(f : A \rightarrow B\), then

• \(A\) is the domain (the set of possible inputs for \(f\))
• \(B\) is the codomain (the set of things \(f\) might map elements of \(A\) to)
• \(\{f(a): a \in A\}\) is the range (i.e., the set of things \(f\) actually does map elements of \(A\) to)

Example: For \(f: \mathbb{Z} \rightarrow \mathbb{N} = \{(n, |n| + 2) : n \in \mathbb{Z}\}\) from above:

• The domain is \(\mathbb{Z}\).
• The codomain is \(\mathbb{N}\)
• The range is \(\{n \in \mathbb{N} : n \ge 2\}\)

Do Exercises Part B

Properties of Functions

Definition: A function \(f: A \rightarrow B\) is:

• injective, or one-to-one, if for all \(a, a' \in A\), \(a \ne a'\) implies \(f(a) \ne f(a')\).
  • intuition: no two elements of \(a\) map to the same element of \(b\).
• surjective, or onto, if for all \(b \in B\) there is an \(a \in A\) with \(f(a) = b\)
  • intuition: every element of \(b\) is mapped to by some element of \(a\)
• bijective if \(f\) is both injective and surjective

Example: \(f: \mathbb{Z} \rightarrow \mathbb{N} = \{(n, |n| + 2) : n \in \mathbb{Z}\}\) is:

• not injective because \(f(-2) = f(2) = 4\).
• not surjective because there is no \(a\) such that \(f(a) = 1\).
• not bijective

Example: If \(f : \mathbb{Z} \rightarrow \mathbb{N}\) is defined as \(f(n) = |n|+ 1\), then \(f\) is:

• not injective, for the same reason as the above example
• surjective because for all \(n \in \mathbb{N}\), it is the case that \(f(n-1) = n\)
• not bijective

Here are two ways to prove that a function is injective:

And here is an approach for proving that a function is surjective: