For Problem V-16:

All of them are equivalence relations:

(v) is reflexive since $x - x = 0$, which is even; symmetric since $x-y$ is even implies $y-x$ is even; transitive since $x-y$ and $y-z$ both even imply $x-z = (x-y) + (y-z)$ is even.

(vi) is reflexive since $f(x) = f(x)$; symmetric since $f(x _ 1) = f(x _ 2)$ implies $f(x _ 2) = f(x _ 1)$; transitive since if $f$ has equal values at $x _ 1$ and $x _ 2$ and equal values at $x _ 2$ and $x _ 3$ then $f$ has equal values at $x _ 1$ and $x _ 3$.

(vii) is reflexive since $x - x =$   0$\in W$; symmetric since $x-y \in W$ implies $y-x = - (x-y) \in W$; transitive since $x-y \in W$ and $y-z \in W$ imply $x-z = (x-y) + (y-z) \in W$.

(viii) is reflexive since $A$ has the same row space as itself; symmetric since if $A$ and $B$ have the same row space then $B$ and $A$ have the same row space; transitive since if $A$ and $B$ have the same row space and $B$ and $C$ have the same row space then $A$ and $C$ have the same row space.

(ix) is reflexive since $A = I^{-1} A I$ so $A \sim A$; symmetric since if $A \sim B$ using $P^{-1} \dots P$ then $B \sim A$ using $P \dots P^{-1}$; transitive since $A \sim B$ and $B \sim C$ imply $P^{-1} A P = B$ and $Q^{-1} B Q = C$ for some invertible $P$ and $Q$, so that $Q^{-1}(P^{-1} A P) Q = C$, which is the same as $(PQ)^{-1} A (PQ) = C$, so $A \sim C$.

(x) is all three since any assertion that two elements are related is true, and the three properties consist of such assertions.

(xi) is reflexive since any element is in the same block as itself; symmetric since if $x$ and $y$ are in the same block then $y$ and $x$ are in the same block; transitive since if $x$ and $y$ are in the same block and $y$ and $z$ are in the same block then $x$ and $z$ are in the same block.

For Problem V-17: The ``proof'' gives the impression that it is talking about all possible $x \in X$, but what it actually says is that if $x$ is related to $y$ something is true. So what the proof actually proves is that if $x$ is related to $y$ then $x$ is related to itself. It doesn't say anything about an $x$ that is related to no elements.

For a counterexample to the assertion, then, let $X = \mathbb{R}$ and say $x \rho y$ when $x$ and $y$ are both integers. This $\rho$ is symmetric and transitive but not symmetric. For a simpler counterexample, take any nonempty set $S$ and let $\rho$ be the ``empty relation''--no two elements are related.

For Problem V-18:

For (i): Let's rephrase the three properties of equivalence relations using the definition of $B _ x$: Reflexivity says that (1) $x \in B _ x$; symmetry says that (2) $y \in B _ x \Rightarrow x \in B _ y$; transitivity says that (3) $y \in B _ x$ and $z \in B _ y$ imply $z \in B _ x$.

Notice that we also have
(4) if $x \in B _ y$ then $B _ x = B _ y$.

The reason: Suppose $x \in B _ y$. To show $B _ x \subseteq B _ y$: If $z \in B _ x$ then by $x \in B _ y$ and (3) we have $z \in B _ y$. To show $B _ y \subseteq B _ z$ notice that by (2) we can turn $x \in B _ y$ around to get $y \in B _ x$.

Now to show the distinct blocks make a partition: By (1), each $x$ is in its own block, so the union of all the blocks is all of $X$. To show that if two blocks are distinct then they are pairwise disjoint, let's do the contrapositive: If two blocks have an element $x$ in common, i.e., if $x \in B _ z$ and $x \in B _ z$ then $B _ y = B _ z$. But that's true by (4), since $B _ z = B _ x = B _ y$.

For (ii): We are asked to start with $\rho$, make the partition described in (i), and then show that it gives $\rho$ back. In other words if the relation $\sigma$ is defined by $x \sigma y$ when $x$ and $y$ are in the same block, then $\sigma = \rho$. We can do this in two parts: First, does $x \sigma y \Rightarrow x \rho y$? Yes: If $x \sigma y$, so that $x$ and $y$ are in the same block, then since $x \in B _ x$ that block must be $B _ x$, so $y \in B _ x$, which by definition says $x \rho y$. Next, does $x \rho y \Rightarrow x \sigma y$? Yes: If $x \rho y$ then by (4) $B _ x = B _ y$ and by (1) both $x$ and $y$ are in this block.

For Problem V-19:

Of (i)-(iv) only (iii) (equality) is an equivalence relation; it corresponds to (g) (singleton blocks).

All of (v)-(xi) are equivalence relations.

(v) corresponds to (e) in Q-2 (even/odd), which is the same as congruence blocks modulo 2.

(vi) corresponds to (h).

(vii) corresponds to (b)

(viii)-(xi) have no correspondence to examples in Q-2.

For Problem V-20: Congruence modulo $n$ is reflexive, since $n \vert 0$ so $n \vert x - x$. The relation is symmetric, since $n \vert y-x$ means $y - x = kn$ for some $k$, so $x - y = (-k)n$, giving $n \vert x-y$. The relation is transitive since $n \vert y-x$ and $n\vert z-y$ say $y - x = kn$ and $z-y = mn$ for some $k,m$ and then $z-x = (z-y)+(y-x) = jn + kn = (m+k)n$ so $n \vert z-x$. Therefore congruence modulo $n$ is an equivalence relation.

Notice that Example (i) in Problem V- is congruence modulo 2.

The blocks of a congruence relation are called ``congruence classes modulo $n$''. Again, it would be better if people called these ``congruence blocks'' instead.