Relations and Functions

It is given that $f:R\to \left\{x\in R:-1<x<1\right\}$ is defined as $f\left(x\right)=\frac{x}{1+\left|x\right|},x\in R.$

Suppose $f\left(x\right)=f\left(y\right)$ , where $x,y\in R.$

$\Rightarrow \frac{x}{1+x}=\frac{y}{1-y}\Rightarrow 2xy=x-y$

Since x is positive and y is negative:

$x>y\Rightarrow x-y>0$

But, 2xy is negative.

Then, $2xy\ne x-y$ .

Thus, the case of x being positive and y being negative can be ruled out.

Under a similar argument, x being negative and y being positive can also be ruled out

$\therefore$ x and y have to be either positive or negative.

When x and y are both positive, we have:

$f\left(x\right)=f\left(y\right)\Rightarrow \frac{x}{1+x}=\frac{y}{1+y}\Rightarrow x+xy=y+xy\Rightarrow x=y$

When x and y are both negative, we have:

$f\left(x\right)=f\left(y\right)\Rightarrow \frac{x}{1-x}=\frac{y}{1-y}\Rightarrow x-xy=y-yx\Rightarrow x=y$

$\therefore f$ is one-one.

Now, let $y\in R$ such that $-1<y<1$ .

If x is negative, then there exists $x=\frac{y}{1+y}\in R$ such that

$f\left(x\right)=f\left(\frac{y}{1+y}\right)=\frac{\left(\frac{y}{1+y}\right)}{1+\frac{y}{1+y}}=\frac{\frac{y}{1+y}}{1+\frac{-y}{1+y}}=\frac{y}{1+y-y}=y$

If x is positive, then there exists $x=\frac{y}{1-y}\in R$ such that

$f\left(x\right)=f\left(\frac{y}{1-y}\right)=\frac{\left(\frac{y}{1-y}\right)}{1+\left(\frac{y}{1-y}\right)}=\frac{\frac{y}{1-y}}{1+\frac{y}{1-y}}=\frac{y}{1-y+y}=y$

$\therefore f$ is onto.

Hence, f is

It is given that:

$f:W\to W$ is defined as $f\left(n\right)=\{\begin{array}{l}n-1,if.n,odd\\ n+1,if.n,even\end{array}$

One-one:

Let, $f\left(n\right)=f\left(m\right).$

It can be observed that if n is odd and m is even, then we will have n-1=m+1.

$\Rightarrow n-m=2$

However, the possibility of n being even and m being odd can also be ignored under a similar argument.

$\therefore$ Both n and m must be either odd or even.

Now, if both n and m are odd, then we have:

$f\left(n\right)=f\left(m\right)\Rightarrow n-1=m-1\Rightarrow n=m$

Again, if both n and m are even, then we have:

$f\left(n\right)=f\left(m\right)\Rightarrow n+1=m+1\Rightarrow n=m$

$\therefore f$ is one-one.

It is clear that any odd number 2r+1 in co-domain N is the image of 2r in domain N and any even 2r in co-domain N is the image of 2r+1 in domain N.

$\therefore f$ is onto.

Hence, f is an invertible function.

Let us define $g:W\to W$ as:

$g\left(m\right)=\{\begin{array}{l}m+1,if.n,even\\ m-1,if.n,odd\end{array}$

Now, when n is odd:

$gof\left(n\right)=g\left(f\left(n\right)\right)=g\left(n-1\right)=n-1+1=n$

And, when

A = N * N

* is a binary operation on A and is defined by:

(a, b) * (c, d) = (a + c, b + d)

Let (a, b), (c, d) ∈ A

Then, a, b, c, d ∈ N

We have:

(a, b) * (c, d) = (a + c, b + d)

(c, d) * (a, b) = (c + a, d + b) = (a + c, b + d) 

[Addition is commutative in the set of natural numbers]

∴(a, b) * (c, d) = (c, d) * (a, b)

Therefore, the operation * is commutative.

Now, let (a, b), (c, d), (e, f) ∈A

Then, a, b, c, d, e, f ∈ N

We have:

$\begin{array}{l}\left(\left(a,b\right).\left(c,d\right)\right).\left(e,f\right)=\left(a+c,b+d\right).\left(e,f\right)=\left(a+c+e,b+d+f\right)\\ \left(a,b\right).\left(\left(c,d\right).\left(e,f\right)\right)=\left(a,b\right).\left(c+e,d+f\right)=\left(a+c+e,b+d+f\right)\\ \therefore \left(\left(a,b\right).\left(c,d\right)\right).\left(e,f\right)=\left(a,b\right).\left(\left(c,d\right).\left(e,f\right)\right)\end{array}$

Therefore, the operation* is associative.

An element $e=\left(e_1,e_2\right)$ will be an identity element for the operation* if

$a*e=a=e*a\forall a=\left(a_1,a_2\right)\in A$

$i.e.,\left(a_1+e_1,a_2+e_2\right)=\left(a_1,a_2\right)=\left(e_1+a_1,e_2+a_2\right)$ which is not true for any element in A.

Therefore, the operation* does not have any identity

(i) On Q the operation* is defined as a*b=a-b.

It can be observed that:

$\begin{array}{l}\frac{1}{2}.\frac{1}{3}=\frac{1}{2}-\frac{1}{3}=\frac{3-2}{6}=\frac{1}{6}"and"\frac{1}{3}.\frac{1}{2}=\frac{1}{3}-\frac{1}{2}=\frac{2-3}{6}=\frac{-1}{6}\\ \therefore \frac{1}{2}.\frac{1}{3}\ne \frac{1}{3}.\frac{1}{2}where,\frac{1}{2},\frac{1}{3}\in Q\end{array}$

Thus, the operation* is not commutative.

It can also be observed that:

$\begin{array}{l}\left(\frac{1}{2}.\frac{1}{3}\right).\frac{1}{4}=\left(\frac{1}{2}-\frac{1}{3}\right).\frac{1}{4}=\frac{1}{6}.\frac{1}{4}=\frac{1}{6}-\frac{1}{4}=\frac{2-3}{12}=\frac{-1}{12}\\ \frac{1}{2}.\left(\frac{1}{3}.\frac{1}{4}\right)=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{4}\right)=\frac{1}{2}.\frac{1}{12}=\frac{1}{2}-\frac{1}{12}=\frac{6-1}{12}=\frac{5}{12}\\ \therefore \left(\frac{1}{2}.\frac{1}{3}\right).\frac{1}{4}\ne \frac{1}{2}.\left(\frac{1}{3}.\frac{1}{4}\right)where,\frac{1}{2},\frac{1}{3},\frac{1}{4}\in Q\end{array}$

Thus, the operation* is not associative.

(ii) On Q the operation* is defined as a*b=a²+b.

For $a,b\in Q$ , we have:

$\begin{array}{l}a.b=a^2+b^2=b^2+a^2=b.a\\ \therefore a*b=b*a\end{array}$

Thus, the operation* is commutative.

It can be observed that:

$\begin{array}{l}\left(1*2\right)*3=\left(12+22\right)*3=\left(1+4\right)*3=5*3=52+32=25+9=34\\ 1*\left(2*3\right)=1*\left(22+32\right)=1*\left(4+9\right)=1*13=12+132=1+169=170\\ \therefore \left(1*2\right)*3\ne 1*\left(2*3\right),where,1,2,3\in Q\end{array}$

Thus, the operation* is not commutative.

(iii) On Q the operation* is defined as a*b=a+ab.

It can be observed that:

$\begin{array}{l}1.2=1+1*2=1+2=3\\ 2.1=2+2*1=2+2=4\\ \therefore 1.2\ne 2.1,where,1,2\in Q\end{array}$

Thus, the operation* is not commutative.

It can also be observed that:

$\begin{array}{l}\left(1.2\right).3=\left(1+1*2\right).3=3.3=3+3*3=3+9=12\\ 1.\left(2.3\right)=1.\left(2+2*3\right)=1+1*8=9\\ \therefore \left(1.2\right).3\ne 1.\left(2.3\right),where,1,2,3\in Q\end{array}$

Thus, the operation* is not associative.

(iv) On Q the operation* is defined as a*b=(a-b)²

For $a,b\in Q$ , we have:

$\begin{array}{l}a*b={\left(a-b\right)}^2\\ b*a={\left(b-a\right)}^2={\left[-\left(a-b\right)\right]}^2={\left(a-b\right)}^2\\ \therefore a*b=b*a\end{array}$

Thus, the operation* is commutative