Given a non-empty set X, let *: P(X) P(X) P(X) be defined as A*B =(A – B) (B – A), ,B (X).Show that the empty set is the identity for the operation * and all the elements A of P(X) are invertible with A^ –1 = A. (Hint: (A- ) ( -A)=A and (A-A) (A-A)=A*A= ).

It is given that *: P(X) P(X) P(X) is defined as A*B =(A – B) (B – A), ,B (X). Let A (X). Then, we have: A* =(A- ) ( -A)=A =A *A=( -A) (A- )= =A * =A= *A. (X) Thus, is the identity element for the given operation *. Now, an element A (X) will be invertible if there exists B (X) such that A*B= =B*A. (As is the identity element) Now, we observed that A*A=(A-A) (A-A)= = (X) Hence, all the elements A of P(X) are invertible with A^ -1 = A.

Given a non-empty set X, let *: P(X) P(X) P(X) be defined as A*B …

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Given a non-empty set $X,$ let $*: P(X) \times P(X) \rightarrow P(X)$ be defined as $\text{A}*\text{B} =(\text{A – B})\cup(\text{B – A}),\forall\text{A},\text{B}\in\text{P(X)}.$Show that the empty set $\phi$ is the identity for the operation $*$ and all the elements $A$ of $P(X)$ are invertible with $A^{–1} = A. (\text{Hint: }(\text{A}-\phi)\cup(\phi-\text{A})=\text{A}\ \text{and }(\text{A}-\text{A})\cup(\text{A}-\text{A})=\text{A}*\text{A}=\phi).$

$X = x_i:$	1	2	3	4
$P(X = x_i):$	2k	4k	3k	k

$o$	$a$	$b$	$c$	$d$
$a$	$a$	$a$	$a$	$a$
$b$	$a$	$b$	$c$	$d$
$c$	$a$	$c$	$d$	$b$
$d$	$a$	$d$	$b$	$c$

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$b$	$a$	$b$	$c$	$d$
$c$	$a$	$c$	$d$	$b$
$d$	$a$	$d$	$b$	$c$

$o$	$a$	$b$	$c$	$d$
$a$	$a$	$a$	$a$	$a$
$b$	$a$	$b$	$c$	$d$
$c$	$a$	$c$	$d$	$b$
$d$	$a$	$d$	$b$	$c$

RELATIONS AND FUNCTIONS — Maths STD 12 Science — Question

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$c$	$a$	$c$	$d$	$b$
$d$	$a$	$d$	$b$	$c$