Question

if A and B are any two non empty sets and A is proper subset of B.if n(A)=5 ,then minimum possible value of in (A∆B) is​

Answer 1

Answer:

A is a proper subset of B.

Minimum value of n -

(A × B) = $\dfrac{5}{5}$ = 1.

Step-by-step explanation:

Given -

A and B are any two non empty sets.

A is proper subset of B.

n(A) = 5

To Find -

The minimum possible value of in symmetric diff B.

Solution -

⇒ A∩B = A

⇒ AUB = B

⇒ A × B = $\sf \dfrac{A \cupB}{A \capB} = \dfrac{B}{A}$

Now,

⇒ n(A) = 5

⇒ n(B) = 5n(A × B) = $\dfrac{B}{A}$

As we know here,

Lowest possible value of n(B) = 5

So,

Minimum value of n -

⇒ (A × B) = $\dfrac{5}{5}$ = 1.

$\rule{300}{1.5}$

Remember -

Here I use -

⇒ × = Symmetric difference operation.

⇒ ∩ = Intersection.

⇒ U = Union

Answer 2

$\huge \red { \boxed{ \boxed{ \mathsf{ \mid \ulcorner Answer : </p><p>\urcorner \mid }}}}$

$\large{\sf{A \cap B \: = \: A}}$

$\large{\sf{A \cup B \: = \: B}}$

$\large{\sf{(A \times A) \: = \: {\frac{A}{A}} \: = \: {\frac{B}{A}}}}$

So,

$\sf{n(A) \: = \: 5}$

$\sf{n(B) \: = \: \frac{B}{A}}$

Minimum Value of n will be

(A × B) = 5/5 = 1

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