Help please its on unit 5 semester 2 a p e x math test

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Step-by-step explanation:

Math 311

Set Proofs Handout and Activity

Recall that a set A is a subset of a set B, written A ⊆ B if every element of the set A is also an element of the set B. To show

that one set is a subset of another set using a paragraph proof, we usually use what is called a “general element argument”.

Here is an example:

Example 1: We will prove that A ∩ B ⊂ A

Proof: Let x be an arbitrary element of A ∩ B. By definition of set intersection, since x ∈ A ∩ B, then x ∈ A and x ∈ B.

In particular, x ∈ A. Since every element of A ∩ B is also an element of A, A ∩ B ⊆ A ✷

Since that was a fairly straightforward example, let’s try another.

Example 2: We will prove that If A ⊆ B, then B ⊆ A.

Proof: Let x ∈ B. By definition of set complement, x /∈ B. Recall that since A ⊆ B, whenever y ∈ A, we also have y ∈ B.

Therefore, using contraposition, whenever y /∈ B, we must have y /∈ A. From this, since x /∈ B, then x /∈ A. Therefore x ∈ A.

Hence B ⊆ A. ✷

Lastly, in order to formally prove that two sets are equal, say S = T, we must show that S ⊆ T and that T ⊆ S. This will

require two general element arguments.

Example 3: We will prove that A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).

Proof:

“⊆” Let x ∈ A ∪ (B ∩ C). Then x ∈ A or x ∈ B ∩ C. We will consider these two cases separately.

Case 1: Suppose x ∈ A. Then, by definition of set union, x ∈ A ∪ B. Similarly, x ∈ A ∪ C. Thus. by definition of set

intersection, we must have x ∈ (A ∪ B) ∩ (A ∪ C).

Case 2: Suppose x ∈ B ∩ C. Then, by definition of set intersection, x ∈ B and x ∈ C. Since x ∈ B, then again by the

definition of set union, x ∈ A ∪ B. Similarly, since x ∈ C, then x ∈ A ∪ C. Hence, again by definition of set intersection,

x ∈ (A ∪ B) ∩ (A ∪ C).

Since these are the only possible cases, then A ∪ (B ∩ C) ⊆ (A ∪ B) ∩ (A ∪ C)

“⊇” Let x ∈ (A ∪ B) ∩ (A ∪ C). Then, by definition of set intersection, x ∈ A ∪ B and x ∈ A ∪ C. We will once again split

into cases.

Case 1: Suppose x ∈ A. Then, by definition of set union, x ∈ A ∪ (B ∩ C).

Case 2: Suppose x /∈ A. Since x ∈ A ∪ B, then, by definition of set union, we must have x ∈ B. Similarly, since x ∈ A ∪ C,

we must have x ∈ C. Therefore, by definition of set intersection, we have x ∈ B ∩ C. Hence, again by definition of set union,

A ∪ (B ∩ C).

Since these are the only possible cases, then A ∪ (B ∩ C) ⊇ (A ∪ B) ∩ (A ∪ C)

Thus A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C). ✷

Instructions: Use general element arguments to show that following (Note that only 3, 4, and 5 may be used as portfolio

proofs):

1. Proposition 1: B − A ⊆ B

2. Proposition 2: A − (A − B) ⊆ B

3. Proposition 3: A − (A − B) = A ∩ B

4. Proposition 4: (A − B) ∪ (B − A) = (A ∪ B) − (A ∩ B)

5. Proposition 5: A × (B ∩ C) = (A × B) ∩ (A × C)