Given A = Set of all human beings in a town at a particular time and R = {(x, y); x is a wife of y; x, y \(\in \) A} 
(i) Since x is a wife of x, is not true (x, x) \(\notin \)  R
So, R is not reflexive. 
(ii) x is a wife of y, but y not wife of (y, x) \(\notin \) : R. So, R is not symmetric. 
(iii) x is a wife of y, and y is wife of z But this situation does not exist. So, R is not transitive.

A relation R on a non-empty set A is called a transitive relation if (a, b), (b, c) \(\in R\) then (a, c) \(\in R\) , i.e., aRb, bRc implies aRc.
Thus a relation R on a non empty set A is said to be transitive if there exist a, b, c \(\in A\) such that (a, b)(b, c) \(\in R\)  implies (a, c) . \(\in R\)
Example
Let A = (1, 2, 3, 6)
R = (3, 6) (6, 1) (3, 1)
3 R 6 and 6 R 1 \(\Rightarrow \)(3, 1)  \(\in R\)
\(\therefore\) A is transitive.

\(f o f(x)=f(f(x))=F\left\{\left(3-x^3\right)^{1 / 3}\right\}=\left[3-\left\{\left(3-x^3\right)^{1 / 3}\right\}^3\right]^{1 / 3}=\left(x^3\right)^{1 / 3}=x\)

\( a * b=2 a+b-a b \\ 3 * 4=2 \times 3+4-3 \times 4 \\ 3 * 4=-2 \)

22 * 4 = HCF (22,4) = 2

\( a * b=a+3 b^2 \\ 2 * 4=2+3(4)^2 \\ =2+3 \times 16=50 \)

$fog(x)=f[g(x)]$
$=f(x-7)$  ...................[ Since, $g(x)=x-7$ ]
$=x-7+7$ ..................[ Since, $f(x)=x+7$ ]           
$\Rightarrow$  $f$$o$$g(x)=x\Rightarrow$  $f$$o$$g(7)=7$

3*4 = 2(3) + 4 - 3 = 6 + 4 - 3 = 7

\(2*3=\frac { 2 }{ 2 } +\frac { 3 }{ 2 } =1+1=2\)

Proving" is commutative 
Proving" is associative 
Getting identity element as (0, 0) 
Getting inverse of (a, b) as (- a, - b)

\(\forall x\in [0,\infty ),y=9x^{ 2 }+6x-5\)
\(=(3x+1)^{ 2 }-6\ge -5\prec \)
\( f=\ [-5,\infty )\)
Co-domain f, hence f is not onto and hence not invertible
Let us take the modified co-domain
\(f=\ [-5,\infty )\)
Let us now check whether f is one-one
Let \(x_{ 1 },x_{ 2 }\neq [0,\infty )\)
\( f(x_{ 1 })=f(x_{ 2 })\)
\( \Rightarrow (3x_{ 1 }+1)^{ 2 }-6=(3x_{ 2 }+1)^{ 2 }-6\)
\( \Rightarrow 3x_{ 1 }+1=3x_{ 2 }+1\)
\(\Rightarrow x_{ 1 }=x_{ 2 }\)
Hence f is oe-one
Since with the modified co-domain = the range f, f is both o0ne-one and onto hence invertible
From (i) above for any
\(y\neq [-5,\infty )\)
\(x=\frac { \sqrt { y+6 } -1 }{ 3 } \)
\(f^{ -1 }[-5,\infty )\rightarrow [0,\infty ),f^{ -1 }(y)\)
\(=\frac { \sqrt { y+6 } -1 }{ 3 } \)