Question

Let $A=Q\times Q$, where Q is the set of all rational numbers, and * be a binary operation on A defined by (a, b) * (c, d) = (ac, b + ad) for $(a.b),(c,d)ϵA$. Then find

(i) The identity element of * in A.

(ii) Invertible elements of A, and hence write the inverse of elements (5, 3) and. $(\frac{1}{2},4)$.

OR

Let f : $W\to W$ be defined as

$f\left(n\right)\{\begin{array}{cc}n-1,& \text{if n is odd}\\ n+1,& \text{if n is even}\end{array}$
Show that f is invertible and find the inverse of f. Here, W is the set of all whole numbers.

Answer 1

Let $A=Q\times Q$, where Q is the set of rational numbers.

Given that * is the binary operation on A defined by (a, b) * (c, d) = (ac, b + ad) for

$(a,b),(c,d)\hat{i}A.$

(i)

We need to find the identity element of the operation * in A.

Let (x, y) be the identity element in A.

Thus,

(a, b) * (x, y) = (x, y) * (a, b) = (a, b), for all $(a,b)ϵA$

$\Rightarrow (ax,b+ay)=(a,b)$

$\Rightarrow ax=aandb+ay=b$

$\Rightarrow y=0andx=1$

Therefore, $(1,0)ϵA$ is the identity element in A with respect to the operation *.

(ii)

We need to find the invertible elements of A.

Let (p, q) be the inverse of the element (a, b)

Thus,

$(a,b)\ast (p,q)=(1,0)$

$\Rightarrow (ap,b+aq)=(1,0)$

$\Rightarrow ap=1andb+aq=0$

$\Rightarrow p=\frac{1}{a}andq=-\frac{b}{a}$

Thus the inverse elements of (a, b) is $(\frac{1}{a},-\frac{b}{a})$

Now let us find the inverse of (5, 3) and $(\frac{1}{2},4)$

Hence, inverse of (5, 3) is $(\frac{1}{2},-\frac{3}{5})$

And inverse of $(\frac{1}{2},4)is(2,\frac{-4}{\frac{1}{2}})=(2,-8)$

OR

Let f: $W\to W$ be defined as

$f\left(n\right)\{\begin{array}{cc}n-1,& \text{if n is odd}\\ n+1,& \text{if n is even}\end{array}$

We need to prove that 'f' is invertible.

In order to prove that 'f' is invertible it is sufficient to prove that f is a bijection.

A function f: $A\to B$ is a one-one function or an injection, if

$f\left(x\right)=f\left(y\right)\Rightarrow x=yforallx,yϵA.$

Case i:

If x and y are odd.

Let f(x) = f(y)

$\Rightarrow x-1=y-1$

$\Rightarrow x=y$

Case ii:

If x and y are even,

Let f(x) = f(y)

$\Rightarrow x+1=y+1$

$\Rightarrow x=y$

Thus, in both the cases, we have,

$f\left(x\right)=f\left(y\right)\to x=yforallx,yϵW.$

Hence f is an injection.

Let n be an arbitrary element of W.

If n is an odd whole number, there exists an even whole number $n-1ϵW$ such that

f(n - 1) = n − 1 + 1 = n.

If n is an even whole number, then there exists an odd whole number $n+1ϵW$

such that f(n + 1) = n + 1 − 1 = n.

Also, f(1) = 0 and f(0) = 1

Thus, every element of W (co-domain) has its pre-image in W (domain).

So f is an onto function.

Thus, it is proved that f is an invertible function.

Thus, a function $g:B\to A$ which associates each element $yϵB$ to a unique element $xϵA$

such that f(x) = y is called the inverse of f.

That is, $f\left(x\right)=y\iff g\left(y\right)=x$

The inverse of f is generally denoted by $f^{-1}$.

Now let us find the inverse of f.

Let $x,yϵW$ such that f(x) = y

$\Rightarrow x+1=y$, if x is even

And

x − 1 = y, if x is odd

$\Rightarrow x=\{\begin{array}{cc}y-1,& \text{if y is odd}\\ y+1,& \text{if y is even}\end{array}$

$\Rightarrow f^{-1}\left(y\right)=\{\begin{array}{cc}y-1,& \text{if y is odd}\\ y+1,& \text{if y is even}\end{array}$

Interchange, x and y, we have,

$\Rightarrow f^{-1}\left(x\right)=\{\begin{array}{cc}x-1,& \text{if x is odd}\\ x+1,& \text{if x is even}\end{array}144$

Rewriting the above we have,

$\Rightarrow f^{-1}\left(x\right)=\{\begin{array}{cc}x-1,& \text{if x is even}\\ x+1,& \text{if x is odd}\end{array}Thus,f^{-1}\left(x\right)=f\left(x\right)$