Intro

This collection of Blogs will be an ongoing process. I will begin with examples of my math and education work from my second college experience--preparing to be a teacher. I plan on continuing adding to this blog throughout my teaching career, by adding reflections on classroom management, lesson plans, articles from scholarly journals, just my thoughts, and anything else I find pertinent to teaching.

I truly hope this is the beginning of an incredible journey through the mind of a beginning math teacher (even if this teacher is already 33 years old!) I hope I enjoy going back and reading my entries, listening to my Podcasts, watching my slideshows, etc. I hope my students enjoy visiting my blog as well.

This should also serve as an ongoing effort for me to constantly improve my knowledge and implementation of technology in the classroom. Let the journey begin!



Thursday, December 15, 2011

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!
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