Let a, b and c be sets. prove that a ∪ (b \ c) = (a ∪ b) \ (c \ a).

prove in full grammatical sentences.

A ∪ (B \ C) = (A ∪ B) \ (C \ A).

Step-by-step explanation:

Set B \ C represents the set of all elements of B except the element of C. It means B \ C = B - C.

Let A, B and C be sets.

To prove: A∪(B\C) = (A∪B) \ (C\A).

Proof :

First we need to prove that A ∪ (B\C)  ⊆ (A∪B) \ (C\A).

Let x∈A∪(B\C) , then x∈A or x∈(B\C) .

Case 1: If x∈A, then

x∈A then x ∉ C\A and  x∈A then  x∈A∪B

Since x∈A∪B but  x∉C\A, therefore x∈(A∪B) \ (C\A).

Case 2: If x∈B\C , then

If x∈B\C then x∈B but x ∉ C

x∈B then x∈A∪B and x ∉ C then  x ∉ C\A

Since x∈A ∪ B but x ∉ C\A , therefore x∈(A∪B) \ (C\A).

From case 1 and case 2 we can say that  x∈ (A ∪ B) \ (C \ A).  So,

A ∪ (B \ C)  ⊆ (A ∪ B) \ (C \ A).                         (1)

Now we need to prove that (A ∪ B) \ (C \ A). ⊆  A ∪ (B \ C)

Let x∈(A ∪ B) \ (C \ A) , then   x∈A ∪ B but  x∉C\A

x∈ A∪B then x∈ A or x∈ B

Case 1: If x∈ A

If x∈A then  x∈A∪(B \ C)

Case 2: If x∈ B

Suppose that x ∉ A and x∈ C, then since  x ∉ A we have that  x ∈ C\A, a contradiction.  Therefore  x ∉ C.

Since  x∈ B and  x ∉ C. then  x ∈ B\C, hence  x∈ A ∪ (B \ C)

From case 1 and case 2 we can say that x∈ A ∪ (B \ C) . So,

(A ∪ B) \ (C \ A). ⊆  A ∪ (B \ C)                       (2)

Using (1) and (2) we get

A ∪ (B \ C) = (A ∪ B) \ (C \ A).

Hence proved.

hope this .

r3t40

1: -2 3/4, -2 1/2, -14, 0, 3, 8 1/3

2: -4, -3, -2 1/2, -2 1/4, 3 1/4,

step-by-step explanation: