Relations

Given: $S = \{1, 2, 3, 4, 5\}$, $R = \{(x, y) : x \mid y\}$.

First, list the ordered pairs in R:
$$R = \{(1,1),(1,2),(1,3),(1,4),(1,5),(2,2),(2,4),(3,3),(4,4),(5,5)\}$$

Reflexive: For every $x \in S$, $x$ divides $x$, so $(x, x) \in R$ for all $x \in S$. ✓ Hence R is reflexive.

Symmetric: Check whether $(x,y) \in R \Rightarrow (y,x) \in R$.
Counter-example: $(1, 2) \in R$ because 2 is divisible by 1, but $(2, 1) \notin R$ because 1 is not divisible by 2. ✗ Hence R is not symmetric.

Transitive: Suppose $(x, y) \in R$ and $(y, z) \in R$, i.e., $x \mid y$ and $y \mid z$.
Then $y = kx$ and $z = my$ for some integers $k, m$, so $z = mkx$, meaning $x \mid z$, i.e., $(x, z) \in R$. ✓ Hence R is transitive.

Conclusion: R is reflexive and transitive but not symmetric.

Given: $L$ = set of all lines in a plane, $R = \{(L_1, L_2) : L_1 \perp L_2\}$.

Reflexive: A line cannot be perpendicular to itself. So $(L, L) \notin R$ for any line $L$. ✗ Hence R is not reflexive.

Symmetric: Suppose $(L_1, L_2) \in R$, i.e., $L_1 \perp L_2$. Then $L_2 \perp L_1$ as well, so $(L_2, L_1) \in R$. ✓ Hence R is symmetric.

Transitive: Suppose $(L_1, L_2) \in R$ and $(L_2, L_3) \in R$, i.e., $L_1 \perp L_2$ and $L_2 \perp L_3$.
If $L_1 \perp L_2$ and $L_2 \perp L_3$, then $L_1 \parallel L_3$ (both perpendicular to $L_2$), so $L_1$ is not perpendicular to $L_3$, meaning $(L_1, L_3) \notin R$. ✗ Hence R is not transitive.

Conclusion: R is symmetric only; it is neither reflexive nor transitive.

Given: $R = \{(a, b) : a < b^2\}$ on $\mathbb{R}$.

Not Reflexive: We need $(a, a) \in R$ for all $a \in \mathbb{R}$, i.e., $a < a^2$ for all $a$.
Counter-example: Take $a = \dfrac{1}{2}$. Then $a^2 = \dfrac{1}{4}$ and $a = \dfrac{1}{2} > \dfrac{1}{4} = a^2$.
So $\left(\dfrac{1}{2}, \dfrac{1}{2}\right) \notin R$. Hence R is not reflexive.

Not Symmetric: We need $(a,b) \in R \Rightarrow (b,a) \in R$, i.e., $a < b^2 \Rightarrow b < a^2$.
Counter-example: Take $a = 1,\ b = 3$. Then $a < b^2 \Rightarrow 1 < 9$ ✓, so $(1,3) \in R$.
But $b < a^2 \Rightarrow 3 < 1$ ✗, so $(3,1) \notin R$. Hence R is not symmetric.

Not Transitive: We need $(a,b) \in R$ and $(b,c) \in R \Rightarrow (a,c) \in R$.
Counter-example: Take $a = 2,\ b = -2,\ c = \dfrac{1}{2}$.
- $(a,b)$: $2 < (-2)^2 = 4$ ✓, so $(2,-2) \in R$.
- $(b,c)$: $-2 < \left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4}$ ✓, so $\left(-2, \dfrac{1}{2}\right) \in R$.
- $(a,c)$: $2 < \left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4}$? No, $2 > \dfrac{1}{4}$. ✗, so $\left(2, \dfrac{1}{2}\right) \notin R$.

Hence R is not transitive.

Conclusion: R is neither reflexive, nor symmetric, nor transitive. $\blacksquare$

Given: $R = \{(a, b) : 2 \mid (a - b)\}$ on $\mathbb{Z}$.

To show R is an equivalence relation, we verify reflexivity, symmetry, and transitivity.

1. Reflexive: For any $a \in \mathbb{Z}$,
$$a - a = 0 = 2 \times 0$$
So $2 \mid (a - a)$, hence $(a, a) \in R$ for all $a \in \mathbb{Z}$. ✓ R is reflexive.

2. Symmetric: Let $(a, b) \in R$, so $2 \mid (a - b)$, i.e., $a - b = 2k$ for some integer $k$.
Then $b - a = -2k = 2(-k)$, and $-k \in \mathbb{Z}$, so $2 \mid (b - a)$, giving $(b, a) \in R$. ✓ R is symmetric.

3. Transitive: Let $(a, b) \in R$ and $(b, c) \in R$.
Then $a - b = 2k$ and $b - c = 2m$ for integers $k, m$.
$$a - c = (a - b) + (b - c) = 2k + 2m = 2(k + m)$$
Since $k + m \in \mathbb{Z}$, we have $2 \mid (a - c)$, so $(a, c) \in R$. ✓ R is transitive.

Conclusion: Since R is reflexive, symmetric, and transitive, R is an equivalence relation on $\mathbb{Z}$. $\blacksquare$