Q.Discuss the following relations for reflexivity, symmetricity and transitivity:

Step 1 (i) "$m$ divides $n$" on positive integers. Reflexive: $m\mid m$ always ($m/m=1$) $\Rightarrow$ yes. Symmetric: $2\mid4$ but $4\nmid2$ $\Rightarrow$ no. Transitive: $m\mid n,\ n\mid p\Rightarrow m\mid p$ $\Rightarrow$ yes. So: reflexive and transitive, not symmetric.

Step 2 (ii) "$\ell\perp m$" on lines in a plane. Reflexive: a line is never perpendicular to itself $\Rightarrow$ no. Symmetric: $\ell\perp m\Rightarrow m\perp\ell$ $\Rightarrow$ yes. Transitive: $\ell\perp m,\ m\perp n\Rightarrow\ell\parallel n$ (not perpendicular) $\Rightarrow$ no. So: symmetric only.

Step 3 (iii) "$a$ is not a sister of $b$" on ALL family members. Reflexive: nobody is their own sister, so "$a$ is not a sister of $a$" is always true $\Rightarrow$ yes. Symmetric: let $a$ be male, $b$ his actual sister -- $a$ is not a sister of $b$ (true, $a$ is male) but $b$ IS a sister of $a$, so "$b$ is not a sister of $a$" is false $\Rightarrow$ no. Transitive: let $a,c$ be actual sisters and $b$ an unrelated male; $a$-not-sister-of-$b$ (true) and $b$-not-sister-of-$c$ (true, $b$ male) but $a$ IS a sister of $c$, so "$a$-not-sister-of-$c$" is false $\Rightarrow$ no. So: reflexive only.

Step 4 (iv) same rule, restricted to FEMALE family members only. Reflexive: same reasoning as (iii) $\Rightarrow$ yes. Symmetric: among females, actual "is a sister of" is itself symmetric (siblinghood is mutual), so its negation is symmetric too $\Rightarrow$ yes. Transitive: reuse the same counterexample as (iii) with $a,b,c$ all female and $a,c$ actual sisters, $b$ unrelated $\Rightarrow$ still no. So: reflexive and symmetric, not transitive.

Step 5 (v) "$x+2y=1$" on $N$. For $x,y\ge1$, the minimum of $x+2y$ is $1+2=3>1$, so NO pair of natural numbers satisfies $x+2y=1$ -- the relation is the empty set. Reflexive: needs $(a,a)\in R$ for every $a\in N$, but $R=\varnothing$ has none $\Rightarrow$ no (fails, since $N$ is non-empty). Symmetric and transitive hold vacuously (there is no pair to ever violate either condition) $\Rightarrow$ yes to both.

Test reflexivity/symmetry/transitivity directly against the defining condition, using a counterexample to disprove and a short proof to confirm

• Treating the negated relation "not a sister of" the same as the original "is a sister of" relation -- negation flips which properties hold.
• Forgetting that an empty relation is vacuously symmetric and transitive (no pair exists to break either rule), while still failing reflexivity on a non-empty domain.