MATH 308 — Lec 13 — Relations

Lecture 13 — Relations
MATH 308  Bridge to Advanced Mathematics  Chartrand §9.1–9.2

Goals for today. By the end of this lecture you should be able to:

• state that a relation from $A$A to $B$B is a subset $R \subseteq A \times B$R⊆A×B, and use the notation $a\,R\,b$aRb;

• write a given relation as an explicit set of ordered pairs and draw it (arrow diagram, grid);

• define reflexive, symmetric, and transitive for a relation on a set;

• prove or disprove that a specific relation has each property — a universal claim by argument, its failure by one counterexample.

The prove-or-disprove mindset (Ch. 8)

Chapter 8 hands you a statement and asks a sharper question than “prove it”: it asks you to first decide whether it is true. This is the working stance for everything below.

• To establish a universal claim (“for all $x$x …”), you must argue for an arbitrary element — one counterexample is fatal, so you cannot check cases.

• To disprove a universal claim, you need exactly one counterexample: a single $x$x where it fails.

Every property of relations in §3 is a universal statement, so each one is a little prove-or-disprove problem. Keep both moves — argue-for-all, or exhibit-one — in the same hand.

Relations (§9.1)

Recall from Lecture 2 that the Cartesian product $A \times B = \left\{\, (a,b) \;:\; a\in A,\ b\in B \,\right\}$A×B={(a,b):a∈A, b∈B} collects all ordered pairs. A relation simply selects some of them: the pairs that “are related.”

Definition 1. A relation $R$R from a set $A$A to a set $B$B is a subset $R \subseteq A \times B$R⊆A×B. We write

$a\,R\,b \quad\text{to mean}\quad (a,b) \in R, \qquad\text{and}\qquad a \not\mathrel{R} b \ \ \text{to mean } (a,b) \notin R.$aRbto mean(a,b)∈R,andaRb  to mean (a,b)∈/R.

When $A = B$A=B we call $R$R a relation on $A$A; then $R \subseteq A \times A$R⊆A×A.

Nothing here is new machinery: a relation is a subset (Lecture 1) of a product (Lecture 2). Everything else is vocabulary laid on top of that one idea.

Example 1. Let $A = \{1,2,3\}$A={1,2,3} and $B = \{a,b,c\}$B={a,b,c}, and let

$R = \{(1,a),\ (1,c),\ (2,b),\ (3,b)\} \ \subseteq\ A \times B.$R={(1,a), (1,c), (2,b), (3,b)} ⊆ A×B.

Then $1\,R\,a$1Ra and $2\,R\,b$2Rb are true, while $2 \not\mathrel{R} a$2Ra (the pair $(2,a)$(2,a) is not in $R$R). There are $\left\lvert A\times B \right\rvert = 3\cdot 3 = 9$∣A×B∣=3⋅3=9 possible pairs; $R$R chose $4$4 of them. In fact any of the $2^{9}$29 subsets of $A \times B$A×B is a legal relation — most have no tidy description at all.

The relation $R=\{(1,a),(1,c),(2,b),(3,b)\}$R={(1,a),(1,c),(2,b),(3,b)} as an arrow diagram: draw $a\to b$a→b exactly when $a\,R\,b$aRb. A relation is its set of arrows.

Example 2 (Relations you already know). Each of these is a relation on a familiar set, i.e. a subset of a square product.

$\text{``}\le\text{'' on }\mathbb{R}:\ \left\{\, (x,y) \;:\; x \le y \,\right\} \subseteq \mathbb{R}^2, \qquad \text{``divides'' on }\mathbb{N}:\ \left\{\, (a,b) \;:\; a \mid b \,\right\} \subseteq \mathbb{N}^2.$“≤” on R: {(x,y):x≤y}⊆R2,“divides” on N: {(a,b):a∣b}⊆N2.

Recall $a \mid b$a∣b (Lecture 7) means $b = ak$b=ak for some $k \in \mathbb{Z}$k∈Z. So $2 \mid 6$2∣6 gives $2\,R\,6$2R6, while $4 \not\mid 6$4∣6 gives $4 \not\mathrel{R} 6$4R6.

Properties of a relation on a set

Fix a relation $R$R on a set $A$A. The three properties below are the vocabulary that organizes all of Chapter 9. Notice the quantifiers (Lecture 3–4): each is a “for all” statement.

Definition 2. The relation $R$R on $A$A is

1. reflexive  if  $x\,R\,x$xRx  for every $x \in A$x∈A;

2. symmetric  if  $x\,R\,y \Rightarrow y\,R\,x$xRy⇒yRx  for all $x,y \in A$x,y∈A;

3. transitive  if  $\bigl(x\,R\,y \ \text{and}\ y\,R\,z\bigr) \Rightarrow x\,R\,z$(xRy and yRz)⇒xRz  for all $x,y,z \in A$x,y,z∈A.

It pays to see these on the grid $A \times A$A×A, exactly the picture of a Cartesian product from Lecture 2. Plot a dot at $(x,y)$(x,y) precisely when $x\,R\,y$xRy. Then reflexive means the relation contains the whole diagonal, and symmetric means the picture is unchanged when reflected across that diagonal.

The relation “$\le$≤” on $\{1,2,3,4\}${1,2,3,4} as a subset of the grid $\{1,2,3,4\}^2${1,2,3,4}2. Red = the diagonal ($x\,R\,x$xRx): every diagonal cell is present, so $\le$≤ is reflexive. Blue = strictly above. The open cells below the diagonal are missing, so the picture is not symmetric about the diagonal — $\le$≤ is not symmetric.

Example 3 (Concrete first). Let $A = \{1,2,3\}$A={1,2,3} and let $R = \left\{\, (x,y) \in A^2 \;:\; x+y \text{ is even} \,\right\}$R={(x,y)∈A2:x+y is even}. Listing the pairs,

$R = \{(1,1),(1,3),(2,2),(3,1),(3,3)\}.$R={(1,1),(1,3),(2,2),(3,1),(3,3)}.

Reflexive? Yes: $x+x = 2x$x+x=2x is even, so $(x,x) \in R$(x,x)∈R for every $x$x. Symmetric? Yes: $x+y = y+x$x+y=y+x, so $x+y$x+y even $\Leftrightarrow y+x$⇔y+x even. Transitive? Yes: if $x+y$x+y and $y+z$y+z are both even then their sum $x+2y+z$x+2y+z is even, hence $x+z = (x+2y+z) - 2y$x+z=(x+2y+z)−2y is even. This relation is really “$x$x and $y$y have the same parity” — the partition of $\mathbb{Z}$Z into evens and odds from Lecture 2, now seen as a relation. Such all-three relations are equivalence relations, the whole topic of Lecture 14.

Example 4 (Non-examples, via one counterexample each). Take the relation “$<$<” on $\mathbb{Z}$Z.

• Not reflexive: $1 < 1$1<1 is false, so $(1,1)\notin{<}$(1,1)∈/<. One witness kills the “for all.”

• Not symmetric: $1 < 2$1<2 but $2 \not< 1$2<1.

• Transitive: this one is true — but now we owe an argument, not an example: if $x<y$x<y and $y<z$y<z then $x<z$x<z for all integers.

So “$<$<” is transitive only; “divides” below is reflexive and transitive but not symmetric. Different relations sit in different boxes — that is the point of having names.

Proposition 1. The relation “divides” on $\mathbb{N}$N, namely $R = \left\{\, (a,b) \in \mathbb{N}^2 \;:\; a \mid b \,\right\}$R={(a,b)∈N2:a∣b}, is reflexive and transitive, but not symmetric.

Proof. Reflexive. Let $a \in \mathbb{N}$a∈N be arbitrary. Since $a = a\cdot 1$a=a⋅1, we have $a \mid a$a∣a, so $(a,a) \in R$(a,a)∈R. As $a$a was arbitrary, $R$R is reflexive.

Transitive. Let $a,b,c \in \mathbb{N}$a,b,c∈N with $a \mid b$a∣b and $b \mid c$b∣c. Then $b = ak$b=ak and $c = b\ell$c=bℓ for some $k,\ell \in \mathbb{Z}$k,ℓ∈Z. Substituting, $c = (ak)\ell = a(k\ell)$c=(ak)ℓ=a(kℓ), and $k\ell \in \mathbb{Z}$kℓ∈Z, so $a \mid c$a∣c. Hence $R$R is transitive.

Not symmetric. We disprove the universal claim with one counterexample: $2 \mid 4$2∣4, so $(2,4)\in R$(2,4)∈R, but $4 \nmid 2$4∤2, so $(4,2)\notin R$(4,2)∈/R. Thus $R$R is not symmetric. ◻

Remark 1. A property can fail in one direction and hold in another — “divides” is reflexive but not symmetric — so you must check each of the three separately. They are independent questions about the same subset of $A \times A$A×A.

Homework

1. Let $A = \{2,3,4\}$A={2,3,4} and $B = \{4,6,9,12\}$B={4,6,9,12}. Write the relation “$a$a divides $b$b” from $A$A to $B$B as an explicit set of ordered pairs, and draw its arrow diagram.

2. On $A=\{1,2,3,4\}$A={1,2,3,4} define $x\,R\,y$xRy iff $x - y$x−y is even. List the pairs of $R$R, then decide (with proof or counterexample) whether $R$R is reflexive, symmetric, transitive.

3. Prove or disprove each property (reflexive, symmetric, transitive) for the relation “$\le$≤” on $\mathbb{Z}$Z.

4. Draw the grid picture (the subset of $A \times A$A×A) for the relation “$\le$≤” on $A=\{1,2,3\}$A={1,2,3}. Which cells encode reflexivity? Which reflection encodes symmetry?

5. Give a relation on $\{1,2,3\}${1,2,3} that is symmetric but not transitive, and a different one that is reflexive and transitive but not symmetric. (One counterexample settles each “not.”)

Suggested textbook exercises (Chartrand §9.1): 9.2, 9.4, 9.6, 9.10, 9.14, 9.18.