Q.Traffic flows from 𝐷 to 𝐸 and 𝐷 to 𝐶. The department wants to represent and analyze this data using relations and functions. Use the given data to answer the following questions: I. Is the traffic flow reflexive? Justify. [1] II. Is the traffic flow transitive? Justify. [1] III
(A) Represent the relation describing the traffic flow as a set of ordered pairs. Also state the domain and range of the relation.

The traffic flow gives the relation $R=\{(D,E),(D,C)\}$.

Part (a)

I. Reflexive? No — $(D,D),(E,E),(C,C)$ are absent, so $R$ is not reflexive.

II. Transitive? There is no chain $(a,b),(b,c)$ (nothing starts with $E$ or $C$), so the condition holds vacuously — $R$ is transitive.

III(A). As ordered pairs / domain / range: …

The traffic data "flows from $D$ to $E$ and $D$ to $C$" becomes the relation $R=\{(D,E),(D,C)\}$ on the set of points $\{D,E,C\}$.

Part (a)

I. Reflexivity. $R$ is reflexive iff $(x,x)\in R$ for every element $x$. We would need $(D,D),(E,E),(C,C)$, none of which is present. Hence $R$ is not reflexive.

II. Transitivity. $R$ is transitive iff whenever $(a,b)\in R$ and $(b,c)\in R$ then $(a,c)\in R$. The only pairs are $(D,E)$ and $(D,C)$; no pair begins with $E$ or $C$, so the "if" part is never satisfied. An implication with a false hypothesis is true, so $R$ is vacuously transitive.

III(A). Ordered pairs, domain, range.

$R=\{(D,E),(D,C)\}.$ …

Method: Turning a scenario into a relation, then testing it

For "real-world data" questions, first translate the words into ordered pairs, then apply the standard property and function tests.

Steps

Step 1: Write the relation as ordered pairs and fix the ground set

"Flows from $D$ to $E$ and $D$ to $C$" becomes $R=\{(D,E),(D,C)\}$ on the set of nodes $\{D,E,C\}$.

Step 2: Reflexive over the whole node set

Check $(x,x)$ for every node, not just the ones that appear as a source. A missing self-loop on any node breaks reflexivity.

Step 3: Transitive by looking for chains

Only when some second coordinate reappears as a first coordinate can a chain form; otherwise transitivity is vacuously true. …

Common Mistakes

Mistake 1: Calling the relation reflexive because "it flows somewhere"

Why it's wrong: reflexive means every node relates to itself ($(D,D),(E,E),(C,C)$), which none do here. Correct approach: check self-loops on all nodes.

Mistake 2: Thinking it is a function because it "makes sense as a flow"

Why it's wrong: $D$ maps to two outputs $E$ and $C$, violating the one-input-one-output rule. Correct approach: a function needs a unique image per input; two outputs from $D$ rules it out. …