Question

For each of the following relations, indicate whether they are reflexive, symmetric, and transitive (for example select reflexive and transitive if the relation is reflexive and transitive, but not symmetric).

(a) Let R be the relation on N given by xRy if and only if x divides y.
(b) Let X be a set and let R be the relation "⊆" on . In other words, SRT if and only if S⊆T.

Answer 1

Conditions

For each of the following relations, indicate whether they are reflexive, symmetric, and transitive (for example select reflexive and transitive if the relation is reflexive and transitive, but not symmetric).

(a) Let $R$ be the relation on $N$ given by $xRy$ if and only if $x$ divides $y$.

(b) Let $X$ be a set and let $R$ be the relation "⊆" on

. In other words, SRT if and only if $S \subseteq T$. # Solution The relation $R$ is called reflexive on a set $X$, if:

\forall a \in X \ aRa

The relation $R$ is called symmetric on a set $X$, if:

\forall a, b \in X \ aRb \rightarrow bRa

The relation $R$ is called transitive on a set $X$, if:

\forall a, b, c \in X \ aRb \text{ and } bRc \rightarrow aRc

a) First of all, we must notice, that every number of $N$ can divide every number of $N$. But we will check these properties, meaning integer division.

R: \forall x, y \in N: xRy \leftrightarrow \frac{y}{x} \in N

The relation $R$ is reflexive:

\forall x \in N: xRx \leftrightarrow \frac{x}{x} = 1 \in N

The relation $R$ is not symmetric:

\exists x = 5, y = 10, x, y \in N: xRy = \frac{10}{5} = 2 \in N, \text{ but } yRx = \frac{5}{10} = 0.5 \in Q \setminus N

The relation $R$ is transitive:

\forall a, b, c \in N \ aRb \text{ and } bRc \rightarrow b = t a, t \in N, c = k b, k \in N

$$Consider$$

aRc = \frac{c}{a} = \frac{k b}{\frac{b}{t}} = k t.

As $k, t \in N, \rightarrow kt \in N$. And $aRc$ b)

R: \forall S, T \in X S R T \leftrightarrow S \subseteq T

$$The relation R is reflexive:$$

\forall S \in X S R S = S \subseteq S

$$The relation R is not symmetric:$$

\exists S, T \in X, M \in X, M \cap S = \emptyset : T = S \cup M. \text{ Then } S \subseteq T, \text{ but } T \text{ is not } \subseteq S, \text{ as it consist } M

$$The relation R is transitive:$$

\forall A, B, C \in X A R B \text{ and } B R C \rightarrow A \subseteq B, B \subseteq C

$$

Each element from A is in B, and each element from B is in C. Then each element of A is in C.