1. Problem: Determine which ordered pairs belong to the given relations. **a)** Relation $R = \{(x, y) \mid y < x\}$ - Check each pair: - $(2, 3)$: $3 < 2$? No. - $(2, 5)$: $5 < 2$? No. - $(0, 2)$: $2 < 0$? No. - $(2, 3)$: repeated, no. - $(5, -9)$: $-9 < 5$? Yes. - $(3, 3)$: $3 < 3$? No. - $(2, 5)$: repeated, no. **Pairs that belong:** $(5, -9)$ **b)** Relation $R = \{(x, y) \mid y \text{ is a multiple of } x\}$ - Check each pair: - $(2, 7)$: Is 7 multiple of 2? No. - $(1, 5)$: 5 is multiple of 1? Yes. - $(3, 9)$: 9 is multiple of 3? Yes. - $(16, 4)$: 4 multiple of 16? No. - $(\frac{1}{2}, 2)$: 2 multiple of 0.5? Yes, since $2 = 0.5 \times 4$. - $(4, -8)$: -8 multiple of 4? Yes. **Pairs that belong:** $(1, 5), (3, 9), (\frac{1}{2}, 2), (4, -8)$ **c)** Relation $R = \{(x, y) \mid y = 3x + 1\}$ - Check each pair: - $(-1, 2)$: $3(-1)+1 = -3+1 = -2 \neq 2$ no. - $(3, 10)$: $3(3)+1=9+1=10$ yes. - $(\frac{1}{2}, \frac{5}{2})$: $3(\frac{1}{2})+1=\frac{3}{2}+1=\frac{5}{2}$ yes. - $(-7, 20)$: $3(-7)+1=-21+1=-20 \neq 20$ no. - $(4, 12)$: $3(4)+1=12+1=13 \neq 12$ no. - $(-4, -11)$: $3(-4)+1=-12+1=-11$ yes. **Pairs that belong:** $(3, 10), (\frac{1}{2}, \frac{5}{2}), (-4, -11)$ --- 2. Problem: Relation $R$ on integers where $x - y$ is an integral multiple of 3. - Check each pair $(x,y)$ if $x - y = 3k$ for some integer $k$: - $(9, 3)$: $9 - 3 = 6 = 3 \times 2$ yes. - $(3, 9)$: $3 - 9 = -6 = 3 \times (-2)$ yes. - $(-1, 2)$: $-1 - 2 = -3 = 3 \times (-1)$ yes. - $(1, 5)$: $1 - 5 = -4$ no. - $(-7, -2)$: $-7 - (-2) = -5$ no. - $(0, 4)$: $0 - 4 = -4$ no. - $(-4, -7)$: $-4 - (-7) = 3 = 3 \times 1$ yes. **Pairs that belong:** $(9, 3), (3, 9), (-1, 2), (-4, -7)$