$(a,b)\in S\circ R$ if there is some $c$ such that $(a,c)\in R$ and $(c,b)\in S$. That is, $a$ is a parent of $c$ and $c$ is a sibling of $b$--equivalently, $a$ is a parent of $b$ and $b$ has a sibling. $(a,b)\in R\circ S$ if there is some $c$ such that $(a,c)\in S$ and $(c,b)\in R$. That is, $a$ is a sibling of $c$ and $c$ is a parent of $b$--equivalently, $a$ is an aunt or uncle of $b$.
To determine if $R$ is reflexive, we need to determine if $x^2\geq 1$ for all integers $x$; symmetric, if $yx\geq 1$ whenever $xy\geq 1$; antisymmetric, if $xy\geq 1$ and $yx\geq 1$ together imply $x=y$; transitive, if $xy\geq 1$ and $yz\geq 1$ together implies $xz\geq 1$.
The first three are almost trivial to prove or find counterexamples for. For the last, note that $xy\geq 1$ if and only if $x,y$ are non-zero and have the same sign. Does it follow then that $xy\geq 1$ and $yz\geq 1$ implies $xz\geq 1$?
For the last, we need to look at each pair of nodes $(x,y)$--where $x$ and $y$ may be the same--and determine the fewest arrows we need to travel to get from $x$ to $y$. Call this number $\ell(x,y)$. The maximum of $\ell(x,y)$ over all $25$ pairs $(x,y)$ will give you the least $n$ that you'll need to get $R^*=\bigcup_{k=1}^nR^k$. But why is this? Let me dig into the definitions of $R^k$ and $R^*$ to make this clearer.
In general, $R^k$ is alternative notation for repeated composition--e.g.: $R^1=R$, $R^2=R\circ R$, $R^3=R\circ R\circ R$, etc. In this circumstance, $R^k$ is the set of all vertex pairs $(x,y)$ such that we can travel from $x$ to $y$ along a path composed of exactly $k$ of the arrows in the given diagram. (Do you see why?) For example, $$R^1=\bigl\{(A,D),(B,A),(B,C),(C,B),(C,E),(D,C),(D,E),(E,B)\bigr\}$$ $$R^2=\bigl\{(A,C),(A,E),(B,B),(B,D),(B,E),(C,A),(C,B),(C,C),(D,B),(E,A),(E,B)\bigr\}$$
In general, $R^*$ is the least reflexive transitive relation containing $R$. In this circumstance, any transitive relation containing $R$ will contain all pairs of nodes $(x,y)$ such that there is some path from $x$ to $y$ in the given diagram. Observe that we can start at any node and follow a circuit to get back to the starting node, and we can start at any node and travel to any other node. The least transitive relation containing $R$, then, will be the set of all vertex pairs $(x,y)$ (including those with $x=y$). Then $R^*=\bigcup_{k=1}^\infty R^k$, but we don't need arbitrarily long paths to get the job done. In fact, none of the paths we need are any longer than $4$, so we can rule out E, but the only simple circuit starting at and returning to A has length $4$, so the answer is D.
The emptyset is indeed a relation, and it's the only one you're missing. Remember that a "binary relation on $X$" is just "a subset of $X^2$" - the emptyset is definitely a subset of $X^2$!
As a relation, the emptyset is not reflexive, but it is symmetric and transitive; do you see why? (HINT: universal statements about the emptyset are "vacuously" true - e.g. "Every flying elephant is purple" is a true statement.)
As to antisymmetry, just look at the definition! A relation $R$ is antisymmetric iff $xRy$ and $yRx$ implies $x=y$. So:
$\{(a, b), (b, a)\}$ is not antisymmetric: we have $aRb$ and $bRa$ but $a\not=b$.
$\{(a, b)\}$ is antisymmetric - we never have $xRy$ and $yRx$ hold, so it trivially satisfies antisymmetry.
$\{(a, a)\}$ is antisymmetric - we only have $xRy$ and $yRx$ when $x, y=a$, and this means $x=y$.
Do you see how to check antisymmetry for the others, now?
Best Answer
Defininitions
A binary relation $\sim$ on a set $X$ is reflexive if and only if, for all $a$ in $X$
I binary relation on a set $X$ is said to be irreflexive (or antireflexive) if and only if for all $a$ in $X$,
A binary relation $\sim$ on a set $X$ is symmetric if and only if, for all $a$ and $b$ in $X$,
A binary relation $\sim$ on a set $X$ is antisymmetric if and only if, for all $a$ and $b$ in $X$,
$\quad$ or, equivalently
A binary relation on a set $X$ is said to be asymmetric if and only if it is NOT symmetric, that is,
A binary relation $\sim$ on a set $X$ is transitive if and only if, for all $a, b, c$ in $X$,
For application to the relation "is a sibling of" (which we can denote using the notation $\sim_{s}$) on the set of all people:
Can anyone be a sibling to oneself? No. Hence reflexivity fails. Even more, since no one is a sibling of oneself, the relation is irreflexive.
If $a$ is the sibling of $b$, then $b$ is certainly a sibling of $a$.
If $a$ is a sibling of $b$ we know $b$ is a sibling of $a$ (because the relation is symmetric). But we know from the fact that the relation “is a sibling of” is not reflexive, if $a$ is a sibling of $b$, then $a$ cannot be the same person as $b$, since no one can be a sibling to oneself. So “is a sibling of" is NOT anti-symmetric. This provides you with an example of a relation that is symmetric but not anti-symmetric.
No. Why not? Can it every occur that $a$ is a sibling of $b$, but $b$ is not a sibling of $a$?
Suppose $a$ and $b$ are siblings, and that $b$ and $c$ are siblings. Is it necessarily the case that $a$ and $c$ are siblings? If we define "siblinghood" strictly (where siblings have both parents in common), then the relation is transitive. If we include in our definition of "siblinghood" the relation of being a half-sibling, then the relation is not transitive, because if $a$ and $b$ share only the same mother, and $b$ and $c$ share only the same father, it will not be the case that $a$ and $c$ are siblings.
How a relation is defined, and knowing the elements of the set on which it is defined is crucial.
I hope the example helps. Try to work with the definitions given above and reason through your other relations. "is a sister of" models this example. The same reasoning applies in that relation as it does for "siblinghood."