My Dear Dear students from last year! It's been a while and I needed to know that you haven't stopped thinking and started memorizing random facts. I miss you all terribly and to show how much I miss you I shall pose a small question to you that should surely look like a "mole hill" to you when compared to your prior achievements.
Is there a geometric proof of the algebraic statement below?
a2 – b2 = (a+b)(a-b)
If so, share your suggestion or proof by commenting to this post and remember, you'll have to communicate a visual idea through words because it'll be hard to share images when commenting. Goodluck!
10 comments:
I'm sure the Greeks or someone showed this to be true way before algebra was invented... It's a brilliantly simple little proof...
I'm back in smartboard class with mrs spitzer this year. This kind of thought is simplly out of my league ;)
Mr. Kerai, how's it going... what's going on
Yes, but alas, I cannot write it in this margin.
Oh come on Michael. If you're gonna use Fermat's line, you should do it after writing a whole book!
As for Slaughter's question, it's going well. Don't have a class quite like we had but oh well....
Sam, do try and avoid the damaging rays of the smartboard!!!
I think I've gone down the same path as sam. yay memorizing random things!
Hi Mr. Kerai!
I haven't used the geometric side of my brain in months, so I might be oversimplifying things... but here are my thoughts on this problem:
Okay, say we have a square ABCD with side lengths 'a'. Inside square ABCD, we have square APQR (in which P and R are on lines AB and DA respectively) with side lengths 'b'.
We know the area of ABCD is a^2 because it is a square. We know the area of APQR is b^2, because it is also a sqare.
We are trying to prove a^2-b^2=(a+b)(a-b).
So we need to prove that the area of ABCD (a^2) minus the area of APQR (b^2) is (a+b)(a-b).
Let's draw an auxiliary line that is an extension of line PQ down to line CD, creating point S. Now we can see that inside of ABCD are three shapes: APQR, and two rectangles. If we add the areas of these two rectangles, then that is the answer we are looking for!
Let's look at rectangle PBCS. AB=AP+PB. AB, as given, is 'a'. AP, as given, is 'b'. thus, a=b+PB. by subtracting b from both sides, we see that PB=a-b. Because PBCS is a rectangle and it's oppposite sides are equal, then we know by the transitive property that PB=CS=(a-b). We know that BC is 'a', as given. The area of a rectangle is length x width, which in this case is a(a-b).
Now we look at rectangle RQSD. AD=AR+RD. AD=a, as given, and AR=b, as given. Thus, a=b+RD. Again, subtract b from both sides and we see that RD=a-b. Therefore, RD=QS=(a-b). RQ is b, as given. The area of the rectangle is thus b(a-b).
Then we add the areas to find the are left after subtracting APQR from ABCD. a(a-b)+b(a-b)=APQRarea-ABCDarea. Let's simplify this. (a-b)(a+b)=APQRarea-ABCDarea. Therefore (a-b)(a+b)=a^2-b^2.
SORRY. LONG POST IS LONG.
Lauren, don't memorize random things because the medical research that hasn't been done yet will most likely prove this to be damaging to the brain!
Emily, you rock! You have articulated the construction very well. I'm so completely proud of you!
Scott has another way of proving it, using the same setup as Emily's but the approach Scott uses involves relocating a piece of the left-over area to be determined; in essence, creating a rectangle of sides a+b and a-b. I shall implore him to put up his proof in the next comment post.
Await another challenging question next month....
Here is my proof (A diagram would be helpful since the proof is completely geometric, but here it is in words):
I proved it by drawing a large square with sides length a and another smaller square with sides length b inside of the large one. The L shaped area is the difference in the areas between the squares. By chopping off the section directly under the b square and turning it such that it is directly under the other difference in areas, that new rectangle has a short side equal to (a-b) because it is the length of a left over after the b square is drawn inside the larger square. The long side of the left over area has a length of (a+b) because the length of the one section inside the original square is a, and the length of the chopped off area is b because its side was formerly adjacent to the square with sides length b. Therefore, the area of the remaining area after b^2 is subtracted from a^2 is (a-b)(a+b).
No; I cannot provide a geometric proof. However I would like to propose an analogy. On Monday, Adam magnifies himself by use of a hall of mirrors in which he appears. Then Ben magnifies himself in the hall of mirrors on Tuesday. The difference is one day and one person and one name and one set of magnified images. Adam and Ben, however, when they stand together in front of a single gigantic single mirror on Wednesday that then falls upon them, crushing them so that all that remains of Adam is Ben's dead body over him, covered with mirror shards is equal in some ways to the marvel and miracle of a day without any people in it at all.
Post a Comment