Chapter 1 · Relations and Functions
Mathematics is full of connections: one number divides another, two lines are parallel, two students are in the same section. A relation describes such connections precisely, and a function is a special relation that gives every input exactly one output. In this chapter you will classify relations as reflexive, symmetric and transitive, study equivalence relations and the classes they create, and decide when a function is one-one (injective), onto (surjective) or both (bijective).
\[n(A)=m,\ n(B)=n \;\Rightarrow\; 2^{mn} \text{ relations from } A \text{ to } B\]
On a set with \(n\) elements there are \(2^{n^2}\) relations, \(2^{n^2-n}\) reflexive relations and \(2^{n(n+1)/2}\) symmetric relations.
\[a\equiv b \pmod n \iff n\mid(a-b)\]
An equivalence relation on \(\mathbb{Z}\) with exactly \(n\) classes: \([0],[1],\dots,[n-1]\).
\[\#(A\to B)=n^m,\qquad \#\text{one-one}=\frac{n!}{(n-m)!}\ (m\le n)\]
Bijections from an \(n\)-element set to itself: \(n!\). Onto functions from an \(m\)-set onto a 2-element set: \(2^m-2\).
Prove a property with arbitrary elements; disprove it with a single concrete counter-example.
On a non-empty set, the empty relation is symmetric and transitive but not reflexive; the universal relation \(A\times A\) is an equivalence relation.
Equivalence classes are either identical or disjoint, and their union is the whole set.
\(x\mapsto x^2\): neither on \(\mathbb{R}\to\mathbb{R}\); one-one but not onto on \(\mathbb{N}\to\mathbb{N}\); bijective on \([0,\infty)\to[0,\infty)\).
One-one: every horizontal line cuts the graph at most once. Onto: every horizontal line at a codomain height cuts it at least once.
For \(R=\{(1,1),(2,2)\}\) on \(A=\{1,2,3\}\), \(R\) is not reflexive because \((3,3)\) is missing.
In testing transitivity, \((1,2)\) and \((2,1)\) form a chain that requires \((1,1)\).
Onto depends on the codomain: \(e^x\) is not onto \(\mathbb{R}\) because negative numbers and 0 are never outputs.
Showing two particular inputs give different outputs proves nothing; you must start from f(x₁) = f(x₂) in general.
On \(A=\{1,2,\dots,9\}\), \(a\,R\,b \iff 4\mid(a-b)\).
Classes: \([1]=\{1,5,9\}\), \([2]=\{2,6\}\), \([3]=\{3,7\}\), \([4]=\{4,8\}\). They are disjoint and cover \(A\).
Show \(f:\mathbb{R}\to\mathbb{R},\ f(x)=5-2x\) is bijective.
One-one: \(5-2x_1=5-2x_2\Rightarrow x_1=x_2\). Onto: for any \(y\), \(x=\frac{5-y}{2}\in\mathbb{R}\) gives \(f(x)=y\). Hence bijective.
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