Q.Given $A = \{1, 2, 3, 4, 5\}$, $S = \{(x, y) : x \in A,\ y \in A\}$. Find the ordered pairs which satisfy the conditions given below:

Concept: Relation Arrow Diagram — we list all pairs $(x, y)$ from $A \times A$ that satisfy each condition.

Step 1: For $x + y = 5$, possible pairs are $(1,4), (2,3), (3,2), (4,1)$.

Step 2: For $x + y < 5$, list all pairs where sum is $2, 3,$ or $4$:

$(1,1), (1,2), (1,3), (2,1), (2,2), (3,1)$.

Step 3: For $x + y > 8$, sum must be $9$ or $10$: …

First, let’s understand what $S$ actually is. The definition says $S = \{(x, y) : x \in A,\ y \in A\}$. That’s just the set of all ordered pairs where both coordinates come from $A = \{1,2,3,4,5\}$. So $S$ has $5 \times 5 = 25$ elements — every combination of a first number and a second number from 1 to 5.

The three conditions are simply filters on these 25 pairs. We’re not being asked to do anything fancy; we just need to list the pairs that satisfy each inequality or equation.

A good way to visualise this is to imagine a $5 \times 5$ grid. The rows are the $x$-values (1 to 5) and the columns are the $y$-values (1 to 5). Each cell is a pair $(x,y)$. The sum $x+y$ is constant along the diagonals running from bottom-left to top-right. For example, the diagonal where $x+y=5$ runs through $(1,4), (2,3), (3,2), (4,1)$. The region where $x+y<5$ is the set of cells above and to the left of that diagonal. The region where $x+y>8$ is the set of cells below and to the right of the diagonal $x+y=9$ (since 8 is the threshold, we want sums of 9 or 10).

Let’s work through each part systematically.

1. Condition (i): $x+y = 5$

   We need all pairs $(x,y)$ with $x,y \in \{1,2,3,4,5\}$ such that $x+y=5$.

   • If $x=1$, then $y=4$.
   • If $x=2$, then $y=3$.
   • If $x=3$, then $y=2$.
   • If $x=4$, then $y=1$.
   • If $x=5$, then $y=0$, but 0 is not in $A$, so discard. So the pairs are $(1,4), (2,3), (3,2), (4,1)$. That’s 4 pairs.

2. Condition (ii): $x+y < 5$

   The smallest possible sum is $1+1=2$. So we want sums of 2, 3, or 4.

   • Sum = 2: only $(1,1)$.
   • Sum = 3: $(1,2)$ and $(2,1)$.
   • Sum = 4: $(1,3), (2,2), (3,1)$. Adding them up: $(1,1), (1,2), (1,3), (2,1), (2,2), (3,1)$. That’s 6 pairs. …