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hram777 [196]
3 years ago
13

Let A = {a, b, c}, B = {b, c, d}, and C = {b, c, e}. (a) Find A ∪ (B ∩ C), (A ∪ B) ∩ C, and (A ∪ B) ∩ (A ∪ C). (Enter your answe

r in set-roster notation.) A ∪ (B ∩ C) = (A ∪ B) ∩ C = (A ∪ B) ∩ (A ∪ C) = Which of these sets are equal? A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) (A ∪ B) ∩ C = (A ∪ B) ∩ (A ∪ C) A ∪ (B ∩ C) = (A ∪ B) ∩ C (b) Find A ∩ (B ∪ C), (A ∩ B) ∪ C, and (A ∩ B) ∪ (A ∩ C). (Enter your answer in set-roster notation.) A ∩ (B ∪ C) = (A ∩ B) ∪ C = (A ∩ B) ∪ (A ∩ C) = Which of these sets are equal? A ∩ (B ∪ C) = (A ∩ B) ∪ C (A ∩ B) ∪ C = (A ∩ B) ∪ (A ∩ C) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) (c) Find (A − B) − C and A − (B − C). (Enter your answer in set-roster notation.) (A − B) − C = A − (B − C) = Are these sets equal? Yes No
Mathematics
1 answer:
wariber [46]3 years ago
4 0

Answer:

(a)

A\ u\ (B\ n\ C) = \{a,b,c\}

(A\ u\ B)\ n\ C = \{b,c\}

(A\ u\ B)\ n\ (A\ u\ C) = \{b,c\}

(A\ u\ B)\ n\ C = (A\ u\ B)\ n\ (A\ u\ C)

(b)

A\ n\ (B\ u\ C) = \{b,c\}

(A\ n\ B)\ u\ C = \{b,c,e\}

(A\ n\ B)\ u\ (A\ n\ C) = \{b,c\}

A\ n\ (B\ u\ C) = (A\ n\ B)\ u\ (A\ n\ C)

(c)

(A - B) - C = \{a\}

A - (B - C) = \{a,b,c\}

<em>They are not equal</em>

<em></em>

Step-by-step explanation:

Given

A= \{a,b,c\}

B =\{b,c,d\}

C = \{b,c,e\}

Solving (a):

A\ u\ (B\ n\ C)

(A\ u\ B)\ n\ C

(A\ u\ B)\ n\ (A\ u\ C)

A\ u\ (B\ n\ C)

B n C means common elements between B and C;

So:

B\ n\ C = \{b,c,d\}\ n\ \{b,c,e\}

B\ n\ C = \{b,c\}

So:

A\ u\ (B\ n\ C) = \{a,b,c\}\ u\ \{b,c\}

u means union (without repetition)

So:

A\ u\ (B\ n\ C) = \{a,b,c\}

Using the illustrations of u and n, we have:

(A\ u\ B)\ n\ C

(A\ u\ B)\ n\ C = (\{a,b,c\}\ u\ \{b,c,d\})\ n\ C

Solve the bracket

(A\ u\ B)\ n\ C = (\{a,b,c,d\})\ n\ C

Substitute the value of set C

(A\ u\ B)\ n\ C = \{a,b,c,d\}\ n\ \{b,c,e\}

Apply intersection rule

(A\ u\ B)\ n\ C = \{b,c\}

(A\ u\ B)\ n\ (A\ u\ C)

In above:

A\ u\ B = \{a,b,c,d\}

Solving A u C, we have:

A\ u\ C = \{a,b,c\}\ u\ \{b,c,e\}

Apply union rule

A\ u\ C = \{b,c\}

So:

(A\ u\ B)\ n\ (A\ u\ C) = \{a,b,c,d\}\ n\ \{b,c\}

(A\ u\ B)\ n\ (A\ u\ C) = \{b,c\}

<u>The equal sets</u>

We have:

A\ u\ (B\ n\ C) = \{a,b,c\}

(A\ u\ B)\ n\ C = \{b,c\}

(A\ u\ B)\ n\ (A\ u\ C) = \{b,c\}

So, the equal sets are:

(A\ u\ B)\ n\ C and (A\ u\ B)\ n\ (A\ u\ C)

They both equal to \{b,c\}

So:

(A\ u\ B)\ n\ C = (A\ u\ B)\ n\ (A\ u\ C)

Solving (b):

A\ n\ (B\ u\ C)

(A\ n\ B)\ u\ C

(A\ n\ B)\ u\ (A\ n\ C)

So, we have:

A\ n\ (B\ u\ C) = \{a,b,c\}\ n\ (\{b,c,d\}\ u\ \{b,c,e\})

Solve the bracket

A\ n\ (B\ u\ C) = \{a,b,c\}\ n\ (\{b,c,d,e\})

Apply intersection rule

A\ n\ (B\ u\ C) = \{b,c\}

(A\ n\ B)\ u\ C = (\{a,b,c\}\ n\ \{b,c,d\})\ u\ \{b,c,e\}

Solve the bracket

(A\ n\ B)\ u\ C = \{b,c\}\ u\ \{b,c,e\}

Apply union rule

(A\ n\ B)\ u\ C = \{b,c,e\}

(A\ n\ B)\ u\ (A\ n\ C) = (\{a,b,c\}\ n\ \{b,c,d\})\ u\ (\{a,b,c\}\ n\ \{b,c,e\})

Solve each bracket

(A\ n\ B)\ u\ (A\ n\ C) = \{b,c\}\ u\ \{b,c\}

Apply union rule

(A\ n\ B)\ u\ (A\ n\ C) = \{b,c\}

<u>The equal set</u>

We have:

A\ n\ (B\ u\ C) = \{b,c\}

(A\ n\ B)\ u\ C = \{b,c,e\}

(A\ n\ B)\ u\ (A\ n\ C) = \{b,c\}

So, the equal sets are:

A\ n\ (B\ u\ C) and (A\ n\ B)\ u\ (A\ n\ C)

They both equal to \{b,c\}

So:

A\ n\ (B\ u\ C) = (A\ n\ B)\ u\ (A\ n\ C)

Solving (c):

(A - B) - C

A - (B - C)

This illustrates difference.

A - B returns the elements in A and not B

Using that illustration, we have:

(A - B) - C = (\{a,b,c\} - \{b,c,d\}) - \{b,c,e\}

Solve the bracket

(A - B) - C = \{a\} - \{b,c,e\}

(A - B) - C = \{a\}

Similarly:

A - (B - C) = \{a,b,c\} - (\{b,c,d\} - \{b,c,e\})

A - (B - C) = \{a,b,c\} - \{d\}

A - (B - C) = \{a,b,c\}

<em>They are not equal</em>

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