Number of Elements in Relation on Set Product

To find the number of elements in the relation $R$ defined on $A \times A$, we need to find the number of 4-tuples $(x, y, z, w) \in A \times A \times A \times A$ such that:

1. $x$ divides $z$ (written as $x \mid z$)
2. $y \le w$

where $A = \{2, 3, 4, 5, 6\}$.

Since the condition for $x$ and $z$ is independent of the condition for $y$ and $w$, the total number of elements in $R$ is given by: $|R| = S_1 \times S_2$ where $S_1$ is the number of ordered pairs $(x, z) \in A \times A$ such that $x \mid z$, and $S_2$ is the number of ordered pairs $(y, w) \in A \times A$ such that $y \le w$.

Step 1: Calculate $S_1$ (Number of pairs $(x, z)$ such that $x \mid z$)

We list the possible values for $z$ corresponding to each $x \in A$:

• For $x = 2$, $z \in \{2, 4, 6\}$ (3 pairs: $(2,2), (2,4), (2,6)$)
• For $x = 3$, $z \in \{3, 6\}$ (2 pairs: $(3,3), (3,6)$)
• For $x = 4$, $z \in \{4\}$ (1 pair: $(4,4)$)
• For $x = 5$, $z \in \{5\}$ (1 pair: $(5,5)$)
• For $x = 6$, $z \in \{6\}$ (1 pair: $(6,6)$)

Summing these up: $S_1 = 3 + 2 + 1 + 1 + 1 = 8$

Step 2: Calculate $S_2$ (Number of pairs $(y, w)$ such that $y \le w$)

We list the possible values for $w$ corresponding to each $y \in A$:

• For $y = 2$, $w \in \{2, 3, 4, 5, 6\}$ (5 pairs)
• For $y = 3$, $w \in \{3, 4, 5, 6\}$ (4 pairs)
• For $y = 4$, $w \in \{4, 5, 6\}$ (3 pairs)
• For $y = 5$, $w \in \{5, 6\}$ (2 pairs)
• For $y = 6$, $w \in \{6\}$ (1 pair)

Summing these up: $S_2 = 5 + 4 + 3 + 2 + 1 = 15$

Step 3: Calculate the total number of elements in $R$

$|R| = S_1 \times S_2 = 8 \times 15 = 120$

Thus, the number of elements in $R$ is $120$.