Consider the following relations: $$R = \{(x, y) \mid x, y \text{ are real numbers and } x = wy \text{ for some rational number } w\}$$; $$S = \left\{\left(\frac{m}{n}, \frac{p}{q}\right) \mid m, n, p \text{ and } q \text{ are integers such that } n, q \ne 0 \text{ and } qm = pn\right\}$$. Then

1. Analyze Relation R

The relation $$R$$ states that two real numbers $$x$$ and $$y$$ are related if $$x$$ is a rational multiple of $$y$$ ($$x = wy$$ where $$w \in \mathbb{Q}$$).

Check Reflexivity:

For any real number $$x$$, we can write $$x = 1 \cdot x$$. Since $$1$$ is a rational number ($$1 \in \mathbb{Q}$$), the pair $$(x, x) \in R$$ for all real numbers. Thus, $$R$$ is reflexive.

Check Symmetry:

For a relation to be symmetric, if $$(x, y) \in R$$, then $$(y, x)$$ must also belong to $$R$$.

Consider a specific counterexample with $$x = 0$$ and $$y = 1$$.

The pair $$(0, 1) \in R$$ because $$0 = 0 \cdot 1$$, and $$0$$ is a rational number ($$0 \in \mathbb{Q}$$).

However, for the reverse pair $$(1, 0)$$, we would need to find a rational number $$w$$ such that $$1 = w \cdot 0$$. No such number exists because $$w \cdot 0 = 0 \neq 1$$.

Since $$(0, 1) \in R$$ but $$(1, 0) \notin R$$, the relation $$R$$ is not symmetric. Because it lacks symmetry, $$R$$ is not an equivalence relation.

2. Analyze Relation S

The relation $$S$$ acts on rational pairs of numbers defined as $$\left(\frac{m}{n}, \frac{p}{q}\right) \in S \iff qm = pn$$.

Dividing both sides of the condition by $$nq$$ gives the alternative equivalent condition:

$$\frac{m}{n} = \frac{p}{q}$$

This means relation $$S$$ is simply a statement of numerical equality between two rational fractions.

Check Reflexivity:

For any rational number $$\frac{m}{n}$$, it is always true that $$\frac{m}{n} = \frac{m}{n}$$ because $$nm = mn$$. Thus, $$S$$ is reflexive.

Check Symmetry:

If $$\left(\frac{m}{n}, \frac{p}{q}\right) \in S$$, then $$\frac{m}{n} = \frac{p}{q}$$. This directly implies that $$\frac{p}{q} = \frac{m}{n}$$, which means $$\left(\frac{p}{q}, \frac{m}{n}\right) \in S$$. Thus, $$S$$ is symmetric.

Check Transitivity:

If $$\left(\frac{m}{n}, \frac{p}{q}\right) \in S$$ and $$\left(\frac{p}{q}, \frac{r}{s}\right) \in S$$, then we have the equalities:

$$\frac{m}{n} = \frac{p}{q}$$

and

$$\frac{p}{q} = \frac{r}{s}$$

By the transitive property of numerical equality, it follows that:

$$\frac{m}{n} = \frac{r}{s} \implies sm = rn \implies \left(\frac{m}{n}, \frac{r}{s}\right) \in S$$

Thus, $$S$$ is transitive.

Since relation $$S$$ satisfies reflexivity, symmetry, and transitivity, it is a valid equivalence relation.

Final Answer

The relation $$S$$ is an equivalence relation, but $$R$$ is not an equivalence relation.