Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination, 04-BS-16 Discrete Mathematics, undated sitting (May 2019). Closed book; approved Casio or Sharp calculator only. The exam instructs "answer 10 of the 12 questions"; every question is answered below as a complete study resource.
Source note: This paper is the May 2019 sitting (every page footer reads "04-BS-16/May 2019"). Two printed statements are defective as set and are flagged where they occur: Question 7(a) prints the last term of $\{1,5,9,\dots\}$ as $4n-1$ (the pattern and the stated sum require $4n-3$), and Question 8(b) prints "$n>2$" although $4^n>n^4$ fails at $n=3,4$. Question 12(c)'s parameters also make a connected graph impossible; this is noted at that part.
Reference texts: Rosen, Discrete Mathematics and Its Applications, 7th ed. (Pearson) — used throughout for logic, set theory, functions, combinatorics, probability, induction, asymptotic (Big-O) notation, and graph theory.
Given. $A=\{-2,-1,0,1,2\}$; $(x,y)\in R\iff x=-y+1$ or $x=-y-1$, i.e. $x+y=\pm1$.
Find. (a) all pairs of $R$. (b) which of the four properties hold. (c) all pairs of $R^2=R\circ R$. (d) a relation on $A$ that is both symmetric and antisymmetric.
Approach. Rewrite the condition as $x+y\in\{1,-1\}$ and enumerate; test each property against the definition; compose by finding, for each $(x,y)\in R$, every $(y,z)\in R$; for (d) use the fact that symmetric-and-antisymmetric relations are exactly the subsets of the identity relation.
a) Enumerate $R$. $x+y=1$: $(-1,2),(0,1),(1,0),(2,-1)$. $x+y=-1$: $(-2,1),(-1,0),(0,-1),(1,-2)$. $$\boxed{R=\{(-2,1),(-1,0),(-1,2),(0,-1),(0,1),(1,-2),(1,0),(2,-1)\}}\ (8\text{ pairs}).$$
b) Properties.Reflexive: $(x,x)\in R$ needs $2x=\pm1$, impossible for an integer, so no element is related to itself — $\boxed{\text{not reflexive}}$ (in fact irreflexive). Symmetric: $x+y=\pm1\iff y+x=\pm1$, so $(x,y)\in R\Rightarrow(y,x)\in R$ — $\boxed{\text{symmetric}}$. Antisymmetric: $(0,1)\in R$ and $(1,0)\in R$ with $0\ne1$ — $\boxed{\text{not antisymmetric}}$. Transitive: $(0,1)\in R$ and $(1,0)\in R$ but $(0,0)\notin R$ — $\boxed{\text{not transitive}}$.
c) $R^2=R\circ R$. $(x,z)\in R^2$ iff some $y\in A$ has $x+y=\pm1$ and $y+z=\pm1$; subtracting, $x-z\in\{0,\pm2\}$, but the middle vertex $y$ must also lie in $A$. Working through each $x$: $x=-2$ reaches only $y=1$, then $z\in\{0,-2\}$; $x=-1$ reaches $y\in\{0,2\}$, then $z\in\{1,-1\}$; $x=0$ reaches $y\in\{1,-1\}$, then $z\in\{0,-2,2\}$; $x=1$ reaches $y\in\{0,-2\}$, then $z\in\{1,-1\}$; $x=2$ reaches only $y=-1$, then $z\in\{0,2\}$. $$\boxed{R^2=\{(-2,-2),(-2,0),(-1,-1),(-1,1),(0,-2),(0,0),(0,2),(1,-1),(1,1),(2,0),(2,2)\}}\ (11\text{ pairs}).$$ These are exactly the 11 pairs of $A\times A$ with $x-z\in\{0,\pm2\}$ (5 with $x=z$, 6 with $|x-z|=2$). Note that $R^2$ contains all five loops $(x,x)$ even though $R$ has none: each $x$ goes to a neighbour and straight back.
d) A relation on $A$ that is symmetric and antisymmetric. Symmetry gives $(x,y)\in S\Rightarrow(y,x)\in S$, and antisymmetry then forces $x=y$; so exactly the subsets of the identity relation qualify. Example: $\boxed{S=\{(-2,-2),(-1,-1),(0,0),(1,1),(2,2)\}}$ (any subset of it, including $\varnothing$, also works).
