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 some