Question

The following three relation are defined on the set of natural numbers N:
$R=\left\{\right(x,y):x<y,x\in N,y\in N\}$
$S=\left\{\right(x,y):x+y=10,x\in N,y\in N\}$
$T=\left\{\right(x,y):x=yorx-y=1,x\in N,y\in N\}$
Explain clearly which of the following options are correct:

• A. R is not reflexive, symmetric, transitive
• B. S is not reflexive, but symmetric and transitive
• C. T is reflexive, not symmetric, not transitive.
• D. T is reflexive, symmetric, not transitive.

Answer 1

Answer: (C)

T is reflexive, not symmetric, not transitive.

(i) $R=\left\{\right(x,y):x<y,x\in N,y\in N\}$
We cannot say that a number is less than itself.
So, $\text{less than}(<)$ is not reflexive.
Also, it is not symmetric as $2<3$ does not implies $3<2$
Hence, $x<y$ does not implies $y<x$ .
R is transitive as $x<y,y<z\Rightarrow x<z$
Example $2<3,3<4$ implies $2<4$
Hence, R is not reflexive, not symmetric but transitive.
(ii)$S=\left\{\right(x,y):x+y=10,x\in N,y\in N\}$
i.e. $S=\left\{\right(1,9),(9,1),(2,8),(8,2),(3,7),(7,3),(4,6),(6,4),(5,5\left)\right\}$
Clearly, S is not reflexive as (1,1)(2,2) , ... are not in S
For every $(x,y)\in S$, there is a $(y,x)\in S$
So, S is symmetric.
Clearly, S is not transitive as $(1,9),(9,1)\in S$ but $(1,1)\notin S$
Hence, S is not reflexive and transitive but symmetric.
(iii) $T=\left\{\right(x,y):x=yorx-y=1,x\in N,y\in N\}$
So, T will have the elements of the form $\left\{\right(1,1),(2,2)......\}$ or of the form $(1+y,y);y\in N$
Clearly, T is not symmetric and not transitive.