Exercise 1
Determine whether each of the following relations are reflexive, symmetric and transitive:
(i)
Relation R in the set \(\displaystyle \mathrm{A}=\{1,2,3, \ldots, 13,14\}\) defined as \[\mathrm{R}=\{(x, y): 3 x-y=0\} \]
(ii)
Relation R in the set N of natural numbers defined as \[\mathrm{R}=\{(x, y): y=x+5 \text { and } x<4\} \]
(iii)
Relation R in the set \(\displaystyle \mathrm{A}=\{1,2,3,4,5,6\}\) as \[\mathrm{R}=\{(x, y): y \text { is divisible by } x\} \]
(iv)
Relation R in the set \(\displaystyle \mathbf{Z}\) of all integers defined as \[\mathrm{R}=\{(x, y): x-y \text { is an integer }\} \]
(v)
Relation R in the set A of human beings in a town at a particular time given by
(a)
\(\displaystyle \mathrm{R}=\{(x, y): x\) and \(\displaystyle y\) work at the same place \(\displaystyle \}\)
(b)
\(\displaystyle \mathrm{R}=\{(x, y): x\) and \(\displaystyle y\) live in the same locality \(\displaystyle \}\)
(c)
\(\displaystyle \mathrm{R}=\{(x, y): x\) is exactly $\displaystyle 7$ cm taller than \(\displaystyle y\}\)
(d)
\(\displaystyle \mathrm{R}=\{(x, y): x\) is wife of \(\displaystyle y\}\)
(e)
\(\displaystyle \mathrm{R}=\{(x, y): x\) is father of \(\displaystyle y\}\)
Not cross-checked
This solution has not been cross-checked against the answer printed in NCERT.
NCERT’s answer
(i)
Neither reflexive nor symmetric nor transitive. (ii) Neither reflexive nor symmetric but transitive. (iii) Reflexive and transitive but not symmetric. (iv) Reflexive, symmetric and transitive. (v) (a) Reflexive, symmetric and transitive. (b) Reflexive, symmetric and transitive. (c) Neither reflexive nor symmetric nor transitive. (d) Neither reflexive nor symmetric but transitive. (e) Neither reflexive nor symmetric nor transitive.
Definitions used throughout: a relation \(\displaystyle \mathrm{R}\) on a set \(\displaystyle \mathrm{A}\) is reflexive if \(\displaystyle (a,a)\in\mathrm{R}\) for every \(\displaystyle a\in\mathrm{A}\); symmetric if \(\displaystyle (a,b)\in\mathrm{R}\Rightarrow(b,a)\in\mathrm{R}\); transitive if \(\displaystyle (a,b)\in\mathrm{R}\) and \(\displaystyle (b,c)\in\mathrm{R}\) together force \(\displaystyle (a,c)\in\mathrm{R}\). A single counterexample denies a property.
(i)
\(\displaystyle 3x-y=0\) means \(\displaystyle y=3x\). Keeping only those pairs with both entries in \(\displaystyle \mathrm{A}=\{1,2,\ldots,14\}\),
\[\mathrm{R}=\{(1,3),(2,6),(3,9),(4,12)\}. \]
Reflexive: \(\displaystyle (1,1)\) would need \(\displaystyle 3(1)-1=0\), which is false, so \(\displaystyle (1,1)\notin\mathrm{R}\). Not reflexive.
Symmetric: \(\displaystyle (1,3)\in\mathrm{R}\), but \(\displaystyle (3,1)\) needs \(\displaystyle 3(3)-1=8=0\), false. Not symmetric.
Transitive: \(\displaystyle (1,3)\in\mathrm{R}\) and \(\displaystyle (3,9)\in\mathrm{R}\), but \(\displaystyle (1,9)\) needs \(\displaystyle 3(1)-9=-6=0\), false. Not transitive.
(ii)
In \(\displaystyle \mathbf{N}\) the condition \(\displaystyle x<4\) allows only \(\displaystyle x=1,2,3\), so
\[\mathrm{R}=\{(1,6),(2,7),(3,8)\}. \]
Reflexive: \(\displaystyle (1,1)\notin\mathrm{R}\) since \(\displaystyle 1\ne 1+5\). Not reflexive.
Symmetric: \(\displaystyle (1,6)\in\mathrm{R}\) but \(\displaystyle (6,1)\notin\mathrm{R}\) (it fails both \(\displaystyle 1=6+5\) and \(\displaystyle 6<4\)). Not symmetric.
Transitive: the second entries \(\displaystyle 6,7,8\) never occur as a first entry, so there is no pair \(\displaystyle (x,y),(y,z)\) both lying in \(\displaystyle \mathrm{R}\); the transitivity requirement is never triggered and is therefore satisfied vacuously. R is transitive.
(iii)
\(\displaystyle \mathrm{R}=\{(x,y): y\text{ is divisible by }x\}\) on \(\displaystyle \mathrm{A}=\{1,2,3,4,5,6\}\).
Reflexive: every \(\displaystyle x\) divides itself, \(\displaystyle x=1\cdot x\), so \(\displaystyle (x,x)\in\mathrm{R}\) for all \(\displaystyle x\in\mathrm{A}\). Reflexive.
Symmetric: \(\displaystyle 2\) is divisible by \(\displaystyle 1\), so \(\displaystyle (1,2)\in\mathrm{R}\); but \(\displaystyle 1\) is not divisible by \(\displaystyle 2\), so \(\displaystyle (2,1)\notin\mathrm{R}\). Not symmetric.
Transitive: if \(\displaystyle y=px\) and \(\displaystyle z=qy\) with \(\displaystyle p,q\) integers, then \(\displaystyle z=(pq)x\), so \(\displaystyle x\) divides \(\displaystyle z\). Transitive.
(iv)
\(\displaystyle \mathrm{R}=\{(x,y):x-y\text{ is an integer}\}\) on \(\displaystyle \mathbf{Z}\). Since a difference of two integers is always an integer, \(\displaystyle \mathrm{R}=\mathbf{Z}\times\mathbf{Z}\) — every ordered pair belongs to R.
Reflexive: \(\displaystyle x-x=0\in\mathbf{Z}\). Yes.
Symmetric: \(\displaystyle y-x=-(x-y)\in\mathbf{Z}\). Yes.
Transitive: \(\displaystyle x-z=(x-y)+(y-z)\in\mathbf{Z}\). Yes.
So R is reflexive, symmetric and transitive, i.e. an equivalence relation.
(v)
A is the set of human beings in a town at a fixed instant.
(a)
\(\displaystyle x\) and \(\displaystyle y\) work at the same place: \(\displaystyle x\) works where \(\displaystyle x\) works, so reflexive; if \(\displaystyle x,y\) share a workplace so do \(\displaystyle y,x\), so symmetric; if \(\displaystyle x,y\) share a workplace and \(\displaystyle y,z\) share a workplace, then \(\displaystyle x,z\) share that same workplace, so transitive. Reflexive, symmetric and transitive (an equivalence relation).
(b)
\(\displaystyle x\) and \(\displaystyle y\) live in the same locality: the same three arguments apply verbatim. Reflexive, symmetric and transitive.
(c)
\(\displaystyle x\) is exactly $\displaystyle 7$ cm taller than \(\displaystyle y\): a person is not $\displaystyle 7$ cm taller than himself, so not reflexive; if \(\displaystyle x\) is $\displaystyle 7$ cm taller than \(\displaystyle y\) then \(\displaystyle y\) is $\displaystyle 7$ cm shorter than \(\displaystyle x\), so not symmetric; if \(\displaystyle x\) is $\displaystyle 7$ cm taller than \(\displaystyle y\) and \(\displaystyle y\) is $\displaystyle 7$ cm taller than \(\displaystyle z\), then \(\displaystyle x\) is $\displaystyle 14$ cm taller than \(\displaystyle z\), so not transitive.
(d)
\(\displaystyle x\) is wife of \(\displaystyle y\): nobody is her own wife, so not reflexive; if \(\displaystyle x\) is the wife of \(\displaystyle y\) then \(\displaystyle y\) is the husband of \(\displaystyle x\), not the wife, so not symmetric. For transitivity, \(\displaystyle (x,y)\in\mathrm{R}\) forces \(\displaystyle y\) to be a husband and \(\displaystyle (y,z)\in\mathrm{R}\) forces \(\displaystyle y\) to be a wife, so no such chain \(\displaystyle (x,y),(y,z)\) exists in R; the transitivity condition is never tested and holds vacuously. So R is transitive, but neither reflexive nor symmetric.
(e)
\(\displaystyle x\) is father of \(\displaystyle y\): nobody is his own father, so not reflexive; if \(\displaystyle x\) is the father of \(\displaystyle y\), then \(\displaystyle y\) is a child of \(\displaystyle x\), not the father, so not symmetric; if \(\displaystyle x\) is the father of \(\displaystyle y\) and \(\displaystyle y\) is the father of \(\displaystyle z\), then \(\displaystyle x\) is the grandfather of \(\displaystyle z\) and \(\displaystyle (x,z)\notin\mathrm{R}\), so not transitive.