Mathematical Proof/Appendix/Answer Key/Mathematical Proof/Methods of Proof/Constructive Proof

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Problem 1Edit

First, we wish to show that A\cup(B\cap C) \subset (A\cup B)\cap(A\cup C). Let x \in A\cup(B\cap C). Then x \in A or x \in B\cap C.

Case 1: x \in A

x \in A \subset A\cup B so that x \in A\cup B
x \in A \subset A\cup C so that x \in A\cup C
x \in A\cup B and x \in A\cup C so that x \in (A\cup B)\cap (A\cup C)

Case 2: x \in B\cap C

x \in B and x \in C
x \in B \subset A\cup Bso that x \in A\cup B
x \in C \subset A\cup C so that x \in A\cup C
x \in A\cup B and x \in A\cup C so that x \in (A\cup B)\cap (A\cup C)

Since in both cases, x \in (A\cup B)\cap (A\cup C), we know that A\cup(B\cap C) \subset (A\cup B)\cap(A\cup C)

Now we wish to show that (A\cup B)\cap(A\cup C) \subset A\cup(B\cap C). Let x \in (A\cup B)\cap (A\cup C). Then x \in A\cup B and x \in A\cup C.

Case 1a: x \in A

x \in A \subset A\cup(B\cap C), so x \in A\cup(B\cap C)

Case 1b: x \in B

We can't actually conclude anything we want with just this, so we have to also to consider the case x \in A\cup C.


Case 2a: x \in A : [see Case 1a]

Case 2b: x \in C

We now have x \in B and x \in C so that x \in B\cap C
Of course, since x \in B\cap C \subset A\cup(B\cap C), it follows that x \in A\cup(B\cap C).
Since both cases 2a and 2b yield x \in A\cup(B\cap C), we know that it follows from 1b.

Since in both cases 1a and 1b, x \in A\cup(B\cap C), we know that (A\cup B)\cap(A\cup C) \subset A\cup(B\cap C).

Since both A\cup(B\cap C) \subset (A\cup B)\cap(A\cup C) and (A\cup B)\cap(A\cup C) \subset A\cup(B\cap C), it follows (finally) that A\cup(B\cap C) = (A\cup B)\cap(A\cup C).


--will continue later, feel free to refine it if you feel it can be--

Problem 3Edit

  1. Because the question asks about the square of a number, you can substitute the definition of an odd number 2n + 1 into the number to be squared. So, say x is that number, then

    x^2 = (2n+1)^2 = (2n+1)(2n+1)

  2. Multiply both factors together

    (2n+1)(2n+1) = 4n^2 + 4n + 1

  3. Factor out a two for the first two terms

    4n^2 + 4n + 1 = 2(2n^2 + 2n) + 1

  4. The factor 2n^2 + 2n will always be a natural number. As such, it fits the definition of an odd number, 2n + 1
Problem solved!