Question 5 of 12

Question 5 (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Set $S=\{-1,0,1,2\}$; relation $R=\{(x,y)\in S\times S: x=y+1\text{ or }x=y-1\}$, i.e. $|x-y|=1$.

Find. (a) All ordered pairs in $R$. (b) The four standard relation properties. (c) All ordered pairs in $R^2=R\circ R$. (d) Whether $R^2$ is reflexive.

Approach. (a) test every ordered pair in $S\times S$ against $|x-y|=1$; (b) check each property against the pair list directly; (c) compose $R$ with itself: $(a,c)\in R^2$ iff $\exists b\in S$ with $(a,b)\in R$ and $(b,c)\in R$.

1. (a) Elements of $R$. $x=y\pm1$ means $|x-y|=1$: consecutive elements of $S=\{-1,0,1,2\}$ in both directions. $$R=\{(-1,0),(0,-1),(0,1),(1,0),(1,2),(2,1)\}$$ $\boxed{R=\{(-1,0),(0,-1),(0,1),(1,0),(1,2),(2,1)\}}$ (6 pairs)
2. (b) Properties of $R$. Reflexive? Would need $(x,x)\in R$ for every $x\in S$, i.e. $|x-x|=0=1$ — never true, so $R$ is NOT reflexive. Symmetric? $|x-y|=1\Leftrightarrow|y-x|=1$, so every pair's reverse is also in $R$ (confirmed by inspection: $(-1,0)$↔$(0,-1)$, $(0,1)$↔$(1,0)$, $(1,2)$↔$(2,1)$) — $R$ IS symmetric. Antisymmetric? Needs: if $(a,b)\in R$ and $(b,a)\in R$ then $a=b$. But $(0,1)\in R$ and $(1,0)\in R$ with $0\ne 1$ — $R$ is NOT antisymmetric. Transitive? Needs: if $(a,b),(b,c)\in R$ then $(a,c)\in R$. Take $(-1,0)\in R$ and $(0,1)\in R$: would need $(-1,1)\in R$, but $|-1-1|=2\ne1$, so it is not in $R$ — $R$ is NOT transitive. $$\boxed{R\text{: not reflexive, symmetric, not antisymmetric, not transitive}}$$
3. (c) Elements of $R^2=R\circ R$. For each $b\in S$, chain every incoming pair $(a,b)\in R$ with every outgoing pair $(b,c)\in R$: from $b=-1$: only $(0,-1)\in R$ incoming, $(-1,0)\in R$ outgoing $\to(0,0)$. From $b=0$: incoming $(-1,0),(1,0)$; outgoing $(0,-1),(0,1)$ $\to(-1,-1),(-1,1),(1,-1),(1,1)$. From $b=1$: incoming $(0,1),(2,1)$; outgoing $(1,0),(1,2)$ $\to(0,0),(0,2),(2,0),(2,2)$. From $b=2$: incoming $(1,2)$; outgoing $(2,1)$ $\to(1,1)$. Collecting distinct pairs: $$R^2=\{(-1,-1),(-1,1),(0,0),(0,2),(1,-1),(1,1),(2,0),(2,2)\}$$ $\boxed{R^2=\{(-1,-1),(-1,1),(0,0),(0,2),(1,-1),(1,1),(2,0),(2,2)\}}$ (8 pairs)
4. (d) Is $R^2$ reflexive? Every element of $S$ has an even displacement of $2$ available within $S$ (walk one step "out" then one step "back", or two steps in a valid direction and land back), and the pair list in (c) already contains $(-1,-1),(0,0),(1,1),(2,2)$ — all four diagonal pairs of $S$: $$\boxed{R^2\text{ IS reflexive}}$$