Question

(a) What is the minimum number of ordered pairs to form a non-zero reflexive relation on a set of n elements?
(b) On the set R of real numbers S is a relation defined as
$S$={$(x,y)|xϵR,yϵR,x+y=xy$}.
Find $aϵR$ such that 'a' is never the first element of an ordered pair in S. Also find $bϵR$ such that 'b' is never the second element of an ordered pair in S.
(c) Consider the function $f\left(x\right)=\frac{3x+4}{x-2},x\ne 2$. Find a function $g\left(x\right)$ on a suitable domain such that $\left(gof\right)\left(x\right)=x=\left(fog\right)\left(x\right)$.

Answer 1

(a)

If we are given a set $S$ of n elements and a relation $R$ on the set, then we say the relation is reflexive if for every $x\in S$, we have $xRx$.

In other words, we have the ordered pair $(x,x)\in R$, where $R$ is the relation set. It relates each elemts of the set to itself.

Hence, if we have the set of all ordered pairs of the form $(x,x)$ where each $x\in S$, then we form a relation on $S$ that is reflexive.

And thus, the number of ordered pairs will be $n$.

(b)

$S=\left\{\right(x,y)|x\in R,y\in R,x+y=xy\}$

$x+y=xy$

$⟹(1-y)x+y=0$

$⟹x=\frac{y}{y-1}$

So, $y$ can't be $1$

Similarly, $y=\frac{x}{x-1}$ then $x$ can't be $1$.

Thus, $1\in R$ is never the first element and also, $1\in R$ is never the second element.

(c)

$f\left(x\right)=\frac{3x+4}{x-2}$

$fog\left(x\right)=x$ ........ [Given]

$⟹f\left(g\right(x\left)\right)=x$

$⟹\frac{3g\left(x\right)+4}{g\left(x\right)-2}=x$

$⟹3g\left(x\right)+4=xg\left(x\right)-2x$

$⟹(3-x)g\left(x\right)=-2x-4$

$⟹g\left(x\right)=\frac{-2x-4}{3-x}$

$⟹g\left(x\right)=\frac{2x+4}{x-3}$

We can verify $\left(gof\right)\left(x\right)=x=\left(fog\right)\left(x\right)$

$\left(gof\right)\left(x\right)=g\left(f\right(x\left)\right)$

$=g\left(\frac{3x+4}{x-2}\right)$

$=\frac{2\left(\frac{3x+4}{x-2}\right)+4}{\frac{3x+4}{x-2}-3}$

$=\frac{6x+8+4x-8}{3x+4-3x+6}$

$=x$

Similarly, $\left(fog\right)\left(x\right)=x$

Hence, $g:R\to R$ is defined as $g\left(x\right)=\frac{2x+4}{x-3}$