Beginning Proof of containment of sets.

Written in Words Version:

We will show [(the set A union B) take away (the set A intersect B)] equals [(the set A take away B) union (the set B take away A)].  You can see this represented visually in the imbedded image.

Proof:

First we will show [(the set A union B) take away (the set A intersect B)] is a subset of [(the set A take away B) union (the set B take away A)].  If x is an element of [(the set A union B) take away (the set A intersect B)], then x is an element of A or B, and X is not an element of A and B.  Therefore x is either an element of A and not B or x is an element of B and not A.  Hence X is an element of equals [(the set A take away B) union (the set B take away A)].  Thus we have shown [(the set A union B) take away (the set A intersect B)] is a subset of [(the set A take away B) union (the set B take away A)].

Second, we will show [(the set A take away B) union (the set B take away A)] is a subset of [(the set A union B) take away (the set A intersect B)].  If x is an element of [(the set A take away B) union (the set B take away A)], then x is an element of A and not B or X is an element of B and not A.  Thus x is an element of A or B, but X is not an element of A and B.  Then x is an element of [(the set A union B) take away (the set A intersect B)].  Thus we have shown show [(the set A take away B) union (the set B take away A)] is a subset of [(the set A union B) take away (the set A intersect B)].

Since we have shown containment in both directions (which means each set is a subset of the other), we have proven equality:  [(the set A union B) take away (the set A intersect B)] equals [(the set A take away B) union (the set B take away A)].  ■ 

(The little black box means the same as Q.E.D., which is Latin for Quod Erat Demonstrandum, or “which was to be demonstrated”).

And lastly, here is the visual version:

The visual representation succinctly shows how simple this proof really is.  I believe I could show the visual logic to high school students very quickly, then focus on the wordy proof (the one in the middle), and finally work on the succinct proof using mathematical symbols (the first one above).  Man, I love Math and Proofs!

M 307 Proof of the Well Ordering Principle of the Natural Numbers

Below is my proof of the Well Ordering Principle.  It is done via Induction and Contradiction.  Again, the box did not survive the file conversion.  Every time you see a box, there should be a large N, indicating the set of Natural Numbers (or positive Integers--counting numbers beginning with 1).

I really like this proof.  I find it relatively easy to understand, and one that I feel I could explain to high school students fairly simply.  It pretty much says every subset of the counting numbers must have a minimum, or smallest element.  The proof proceeds by contradiction by assuming there exists a subset of the Natural numbers which does not contain a minimum.  This subset must have a lower bound, or number outside of the set that is smaller than every number in the set.  The proof then proceeds by induction.  Is every number in the subset larger than 1?  Yes, since the subset does not contain a minimum and is a subset of the Natural numbers, every element in the subset must be larger than the lower bound 1.  Then we assume that our inductive step holds, and Q(2), Q(3), ..., Q(n) holds, which means that the numbers 1, 2, 3, ..., n are all lower bounds of our subset.  We must then show that Q(n+1) holds.  But this means that every natural number is a lower bound of our subset.  This is a contradiction, because our set is a subset of the natural numbers, and thus must contain natural numbers.  I think this makes fairly simple sense.  The big problem in understanding I see in high school students would be accepting the inductive step.  I still have some trouble with this--assuming that the assertion holds in order to prove the assertion seems logically incorrect.

Proof from Math 307 that the square root of 3 is irrational