Note that the characteristic of sets {1, 4, 7}, {2, 5, 8} and {3, 6, 9} is that difference between any two elements of these sets is a multiple of 3.
Therefore, (x, y) \(\in\) R₁ \(\Rightarrow\) x – y is a multiple of 3 \(\Rightarrow\) {x, y} \(\subset\) {1, 4, 7} or {x, y} \(\subset\) {2, 5, 8} or {x, y} \(\subset\) {3, 6, 9} \(\Rightarrow\) (x, y) \(\in\) R₂.
Hence, R₁ \(\subset\) R₂.
Similarly, {x, y} \(\in\) R₂ Þ {x, y}\(\subset\) {1, 4, 7} or {x, y} \(\subset\) {2, 5, 8} or {x, y} \(\subset\) {3, 6, 9} \(\Rightarrow\) x – y is divisible by 3 \(\Rightarrow\) {x, y} \(\in\) R₁.
This shows that R₂ \(\subset\) R₁.
Hence, R₁ = R₂

As we studied in earlier class that for sets A, B,C
\(A\bigcup B=B\bigcup A\ and\ (A\bigcup B)\bigcup C\)
\(=A\bigcup (B\bigcup C)\)
Therefore, for any A, B,C E P(X), we have
\(A\bigcup B=B\bigcup A\ and\ (A\bigcup B)\bigcup C\)
\(=A\bigcup (B\bigcup C)\)
\(\Rightarrow A*B=B*A\ and\ (A*B)*C\)
= A*B(B*C)
Thus, is both commutative and associative on
P(X)
Now,  \(A\bigcup \phi =\phi \bigcup A\ for\ all\ A\in \ P(X)\)
\(A\bigcup \phi =\phi \bigcup *A\ for\ all\ A\in P(X)\)
So is \(\phi \) the identity element for '>t' on P(X).
Let A P(X) be an invertible element. The, there exists P(X) such that
\(A*S=\phi =S*A\)
\(\Rightarrow A\bigcup S=\phi =S\bigcup A\)
\(\Rightarrow S=\phi =A\)
Hence \(\phi \)  is the only invertible element.

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
\(y\neq [-5,\infty )\)
\(x=\frac { \sqrt { y+6 } -1 }{ 3 } \)
\(f^{ -1 }[-5,\infty )\rightarrow [0,\infty ),f^{ -1 }(y)\)
\(=\frac { \sqrt { y+6 } -1 }{ 3 } \)

\(f(x_{ 1 })=f(x_{ 2 })\)
\(\Rightarrow \left( \frac { x_{ 1 }-2 }{ x_{ 1 }-3 } \right) =\left( \frac { x_{ 2 }-2 }{ x_{ 2 }-3 } \right) \)
\(\Rightarrow x_{ 1 }=x_{ 2 }\)
f is one-one function
Let, y = f(x) = \(\frac { x-2 }{ x-3 } \)
\(\Rightarrow x=\frac { 3y-2 }{ y-1 } ,y\neq 1,x\neq 3\)
For each \(y\in B\) there exists \(x\in A\) such that \(f(x)=y=\Rightarrow f\) is onto
f is one-one and onto function
f is bijective function
\(f^{ 1 }:B\rightarrow A\quad f^{ -1 }(x)=\frac { 3x-2 }{ x-1 } \)

f(x) = y = 10x + 7
y = 10x + 7
10x = y - 7
\(x=\frac { y-7 }{ 10 } \)
\(g\left( x \right) =\frac { x-7 }{ 10 } \)