Question

Let A = {$xϵZ:0\le x\le 12$ }. Show that

R = {(a,b) : a, b $ϵA,|a-b|$ is divisible by 4} is an equivalence relation.

Find the set of all elements related to 1. Also write the equivalence class [2].

OR Show that the function $f:R\to R$ defined by $f\left(x\right)=\frac{x}{x^2+1}\forall xϵR$ is neither one-one nor onto. Also, if $g:R\to R$ is defined as g(x) = 2x-1, find fog(x).

Answer 1

We have $R=\left\{\right(a,b):a,bϵA|a-b|\text{is divisible by 4}\}$ and $A=\{xϵZ:0\le x\le 12\}$ i.e., A = {0,1,2,....,12}

Reflexivity : For any $aϵA$ we have |a-a| = 0 which is divisible by 4. That is , (a,a) $ϵ$ R so,

R is reflexive

Symmetry: Let (a,b) $ϵR\forall a,bϵA$ We have |a-b| is divisible by 4.

As |a-b| = |b-a|. So |b-a| is also divisible by 4. that is,(b,a) $ϵ$ R so, R is symmetric.

Transitivity: Let a,b,c $ϵ$ A such that (a, b ) $ϵ$ R and (b,c) $ϵ$ R then , |a-b| is divisible by 4 and |b-c| is divisible by 4.

This implies that |a-b| = 4m and |b-c| = 4p where m, p $ϵ$ Z.

As $(a-b)+(b-c)=\pm 4(m+p)\Rightarrow |a-c|=4(m+p)$

That is |a-c| is also divisible by 4 so R is transitive.

Since the relation R is reflexive, symmetric and transitive so, it is an equivalence relation.

Also the set of all elements related to 1 in relation R is {1, 5, 9}.

Now let (2, x) $ϵ$ R so that |2-x| is divisible by $4\forall xϵA$

That is , x = 2,6,10. Hence [2] = {2,6,10}.

OR

We have $f:R\to R$ defined by $f\left(x\right)=\frac{x}{x^2+1},\forall xϵR$

Let m, n $ϵ$ R so that f(m)= f(n).

That implies, $\frac{m}{m^2+1}=\frac{n}{n^2+1}\Rightarrow m{n}^2+m=n{m}^2+n\Rightarrow m-n=mn(m-n)\to m-n=mn(m-n)\Rightarrow (mn-1)(m-n)=0$

Either m -n = 0 or mn-1 = 0 i.e., m = n or , m = $\frac{1}{n}$

So, f(m) = f(n) does not necessarily imply m = n for all m, n $ϵ$ R. Hence f(x) is not one-one.

Let y = f(x) = $\frac{x}{x^2+1}$ $\Rightarrow y{x}^2-x-y=0$ $\Rightarrow =\frac{1\pm \sqrt{1-4y^2}}{2y}$

Now for x to be real, we must have $1-4y^2\ge 0$ i.e., $(1-2y)(1+2y)\ge 0$ ie., $yϵ[-\frac{1}{2},\frac{1}{2}]$

That is for all x $ϵ$ R (domain), we do not have y $ϵ$ R (codomain) i.e., range $\ne$ codomain.

Hence f(x) is not onto.

Now fog(x) = f(g(x)) = f(2x-1) = $\frac{2x-1}{(2x-1{)}^2+1}=\frac{2x-1}{4x^2-4x+2}=\frac{2x-1}{2(2x^2-2x+1)}$