Q.Let $A = \{1, 2, 3, 4\}$, $B = \{1, 5, 9, 11, 15, 16\}$ and $f = \{(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)\}$. Are the following true?

Concept: Relation vs Function

A relation from $A$ to $B$ is any subset of $A \times B$. A function requires each element of $A$ to be paired with exactly one element of $B$.

Step 1 – Check if $f$ is a relation

Every ordered pair in $f$ has its first element in $A$ and second in $B$:

$(1,5), (2,9), (3,1), (4,5), (2,11)$ — all satisfy $A \times B$. So $f \subseteq A \times B$.

Step 2 – Check if $f$ is a function …

We need to check two things: whether $f$ is a relation from $A$ to $B$, and whether it is a function from $A$ to $B$. These are different conditions, and the second is stricter.

1. Is $f$ a relation from $A$ to $B$?

A relation from $A$ to $B$ is simply any subset of the Cartesian product $A \times B$. That means every ordered pair in $f$ must have its first element from $A$ and its second element from $B$.

Let’s check each pair:

• $(1, 5)$: $1 \in A$, $5 \in B$ ✓
• $(2, 9)$: $2 \in A$, $9 \in B$ ✓
• $(3, 1)$: $3 \in A$, $1 \in B$ ✓
• $(4, 5)$: $4 \in A$, $5 \in B$ ✓
• $(2, 11)$: $2 \in A$, $11 \in B$ ✓

All pairs satisfy the condition. So $f \subseteq A \times B$, meaning $f$ is indeed a relation from $A$ to $B$.

2. Is $f$ a function from $A$ to $B$?

A function is a special kind of relation. For $f$ to be a function from $A$ to $B$, every element of $A$ must appear exactly once as the first component of a pair in $f$. That is:

• Each $x \in A$ must have some pair $(x, y)$ in $f$ (no element of $A$ is left out).
• No $x \in A$ can appear in more than one pair (each input has a unique output).

Now look at $f$: …