Try saying this out loud. It's not only fun, it's also a concise definition. And accurate.

The Pythagorean Theorem: the sum of the squares of the lengths of the legs of a right triangle is equal to the square of the length of the hypotenuse.

This definition amuses me. I teach Pre-Algebra and Algebra; for the sake of integrating disciplines, I have my students count the prepositional phrases therein.

Pythagoras's theorem says that for a right angled triangle with sides of length a,b,c (with c the length of the hypotenuse) we have c2=a2+b2.

Here is one proof (of many). Start with a square of side length a+b, call it square 1. Put a square of side length c in the middle of square 1, call it square 2. Now rotate square 2 so that each vertex of the square 2 meets one of the edges of square 1.

    ____
   | /\ |
   |/  \|
 a |\c /|
   |_\/_|
    b

At this point our original right angled triangle occurs in each of the four corners of square 1. So the area of square 1 is the area of square 2 + the 4 x the area of our original triangle. That is square 1 has area c2+2ab. On the other hand square 1 has area (a+b)2. The result follows easily.

See also Pythagorean triple.

In eighth grade, after studying this theorem, my class was given the assignment to ask some adults about how often they used this theorem. It was expected that most people use this on a daily basis, and it was a good to have some examples to answer "When am I ever going to use this in real life?" But every adult I asked rarely used the theorem if at all. The results were similar for the rest of my class. Another unsuccessful attempt to make math class useful.

Knowing even just that the 3-4-5 triangle is a right triangle is enough to do geometry in the field. Apparently the ancient Egyptians knew this fact, and used it (all of Egyptian mathematics was devoted to tax inspection, for which being able to measure an area is crucial). Given a rope of length 12, knotted at every unit distance, you can build a right triangle by ensuring two of the edges have lengths 3 and 4.

It is a shame that people are unable to do the same, 3000 years later.

Useless Trivia: According to the Guinness Book Of World Records this is the most proved Mathematical Theorem, with over 145 different proofs known to exist. Notably, Prince Charles is credited with creating one of these proofs

The pythagorean theorem works in the third dimension as well. In other words, the diagonal of a rectangular prism equals the square root of the sum of the squares of the sides.
Less wordy: d2 = x2 + y2 + z2

I said that this works in the third dimension, but it isn't limited to it. I have little doubt It works in the fourth, fifth, sixth, etc. dimensions, but I can't verify it, because it's hard for me to visualize even the fourth dimension! I'm sure the theorem can be proven, or probably has been proven, for all dimensions, but that is beyond my knowledge right now.

The diagonal of a rectangular prism is a line that connects any two corners which do not share a face. It goes through and across the inside of the prism, not along a side.

Technically, Pythagoras did not think up the theorem, he made the first proof of it. In doing so, it lost the title postulate and gained the much-needed tag theorem.

Also, the theorem works in any number of dimensions, just tack on another whatever-squared and it works fine.
Technically, Pythagoras did not think up the theorem. Nor did he prove it. Nor did he exist in the first place. But: had he existed, and had he known about the theorem, he would probably have been last person on earth to realise it. Thus it is logical that it is named after him.
In fact the Pythagorean theorem was known by many people before the Greeks, eg the Chinese and the Babylonians.
There is no evidence that it was known to the Egyptians though, a popular myth that has also been noded here. It appears, however, that they did know that triangles have three sides, and in the early 20th century a Belgian historian concluded that they might have known the Pythogorean theorem. After a few references that nobody bothered to check this became an accepted truth.
There are indeed many proofs of the Pythagorean Theorem. However, the following one is particularly charming. This proof was given by the 11 year old Albert Einstein.

Jacob Einstein taught his nephew Albert the fundamentals of Euclidean geometry. The 11 year old youngster felt that some of Euclid's proofs were unnecessarily complicated. For instance, proof of the Pythagorean theorem required many additional lines, angles and squares. The young Einstein came up with an elegant proof that only required one additional line, the altitude above the hypotenuse.

         *
        **   *
       * *       *      b
   a  *  *           *
     *   Ec              *
    *    *                   *
   *     *                       *
  *   Ea *       Eb                  *
 *       *                               *
* * * * * * * * * * * * * * * * * * * * * * *
               c

The height divides the large triangle into two smaller triangles that are similar to each other and similar to the large triangle. In Euclidean geometry, the area ratio of two similar closed figures is equal to the square of the ratio of corresponding linear dimensions. Therefore, the areas of the triangles Ea, Eb, and the larger Ec (E as in German Ebene) are:

Ea= ma2
Eb= mb2
Ec= mc2

(The resemblance with Einstein's famous relationship E=mc2 is of course entirely coincidental).

The larger area Ec is the sum of the two smaller areas:

Ec=Ea+Eb

or:

mc2 = ma2+mb2
c2 = a2+ b2

Source: Fractals, Chaos, Power Laws, Minutes from an Infinite Paradise, Manfred Schroeder, New York, W.H. Freeman and Company, 1991

U.S. President Garfield's proof of the Pythagorean theorem

Consider a trapezoid like this:
     _b__
  a |  ⁄
    |⁄
  b |\ 
    |_\
     a
Where the trapezoid has bases of length a and length b , and a height of length a + b .
The area of the trapezoid can be found by the conventional formula:
½(base1 + base2) × height
= ½(a + b)(a + b)
The area of the trapezoid can also be found by adding the area of the triangles:
½ab + ½ab + ½cc
= ab + ½cc
Set these two equal, and solve:
ab + ½cc = ½(a + b)(a + b)
 2ab + cc = (a + b)(a + b)
 2ab + cc = aa + 2ab + bb
           cc = aa + bb
[]
In book 1 of Euclid's Elements, the pythagorean theorem appears as proposition 47, and its converse appears as proposition 48.

Probably one of the most general ways to state the Pythagorean theorem is in the context of an inner product space. A modicum of abstract nonsense follows (in the non-technical sense of the word).

So let V be your favourite inner product space or Hilbert space. Mine happens to be at the moment L2(X, dμ), the space of square integrable functions in the measure space X with respect to the measure μ, but if you prefer something more commonplace, you could think of Euclidean space Rn, or even of the plain ol' Euclidean plane R2 of elementary geometry. Take any two orthogonal vectors v and w in V. By definition of orthogonality, <v|w> = 0.

In V, the lengths of v and w are defined by

||v|| := <v|v>1/2

||w|| := <w|w>1/2.

So, if the Pythagorean theorem is translated to the language of vectors, inner products, norms, and such, it should read in this context as

||v - w||2 = ||v||2 + ||w||2.

In other symbols,

<v - w|v - w> = <v|v> + <w|w>.

Is this true? Of course it is! Just recall the axioms of an inner product, linearity and quasi-symmetry (or unqualified symmetry, in case our field of scalars is real) being all that we need. This allows us to write

<v - w|v - w> = <v|v>+ <w|w> - <w|v> - <v|w>,

but <w|v> = <v|w> = 0, since the two vectors are orthgonal. This establishes the Pythagorean theorem.

But wait! We have not yet squeezed all the juice out of this orange. What if the two vectors v and w were not initially orthogonal? Then the two terms being subtracted on the right do not vanish, but instead

<v - w|v - w> = <v|v> + <w|w> - 2Re<v|w>,

where Re denotes the real part. To translate this, recall that the inner product can also be written as Re<v|w> = ||v|| ||w|| cos θ, where θ is the angle between v and w. Thus, we may write,

||v - w||2 = ||v||2 + ||w||2 - 2||v|| ||w|| cos θ

which we recognize to be the Law of Cosines.

Of course, once all the definitions and axioms are understood, everything written above is utterly trivial (in the mathematical sense of the word). In a way, the axioms and definitions are built just so that the Pythagorean theorem holds in an inner product space. Unfortunately, why the definitions are the way they are is not something that can be readily explained, but only comes with experience. One of those things that cannot be taught. Or rather, something that I would not know how to teach.

For another interesting version of the Pythagorean theorem (that probably could be subsumed under the present interpretation if the correct definitions are made), look at the Bessel inequality and Parseval's theorem, sometimes known as Plancherel's theorem. Warning: in those two interpretations the abstract nonsense gets piled higher and deeper, yet the basic idea is still the same, despite initial appearances.

Here is one more of the 145 proofs of this theorem.

The reason I particularly like this proof is that you can actually get a piece of paper and a pair scissors and do it. (My old maths teacher had a set of pretty wooden blocks to demonstrate it.)

Ok here goes:

  1. Draw a square.
  2. Call the vertices A, B, C, D.
  3. Inside the square draw two identical right angle triangles with the lines AB and AD as their hypotenuses. The shorter side of one should lie on the longer side of the other.

You should end up with something looking like this:



                      A_
                      /|`-._
                     / |    `-._
                    /  |        `-._
                   /   |____________`-._ B        
                  /    |F               /
                 /c    |               /
                /      |b             /
               /       |             /
              /________|            / 
             D`-._ a    E          /
                  `-._            /
                      `-._       /
                          `-._  /
                              `'
                                C

The area of the square is c2

Now rotate triangle ADE 90o clockwise about point D. This should map line AD onto line DC.

Now rotate triangle ABF 90o anticlockwise about point B mapping AB onto BC.

The resulting shape should look like this:

                               b
                        _________________ 
                       |                /|
                       |               / | 
                       |              /  |
                   a   |             /   |
              _________|            /    | b
              |`-.               c /     |  
              |    `-.            /      |
            a |      c `-.       /       |
              |            `-.  /        |  
              |_______________`'_________| 


This shape consists of two squares with sides of length a and b. Since all the operations performed on the original square are area preserving the area of the new shape must have the same as the area of the square.

Therefore a2=b2+c2

OK. It's 1978, and I've decided to not go to college. This is applauded by my peers (most of whom have Ivy League parents, trust funds, and Marxist leanings) and treated as a failure by my parents, who are Republican lower-middle-class strivers who figured that my turning 18 would take me off their hands so that at last they could enjoy the Playboy lifestyle....

Athem. I digress.

So it was that I was placed as Electrician's Apprentice, Mark 1, in the Building Trades Division of the Wallingford Board of Education, with a free ten-mile cab drive to and fro my workplace, a uniform, and a million in prizes....that last was a lie, but we were paid twice minimum wage, enough to finance my (a) high life in NYC and partici-pation in the Rocky Horror Picture Show (for which I was given a crew cut, so I could work all week in boy drag and wear wigs as a female drag queen on the weekends), (b)insatiable love for beauty products to pull the above off (self-same crew cut was dyed pink), (c) a similar love for rare and imported recordings of progressive rock, jazz, blues and similar genres, and even (d) more s/f, modern, and classic literature than a person should be allowed to read. (As I said, my life is complex.)Now, all of us had a Project, which was to Maintain the Buildings that we have been charged with (I got a new definition of nipple, in the bargain, and an everloving appreciation of Nikola Tesla in the form of me maintaining many flourescent lights), do a New Project (in which service of, I mixed mortar and moved many cement blocks)...and a Yearly Appreciation of our Progress.

I don't know how I did on the first round.We went to the local community college, spent a few hours listening to self-important types telling us How Much this Meant to the Community, and much more doing Dolce Far Niente, meaning that I got to sit around in the library doing pentomino puzzles and checking up on the latest short stories by John Updike. At the end of this I passed and was sent to Storrs, at the end of the month, with some of the other kids, and Roger, my right hand man. (Let's memorialize Roger, for a second. I liked him. He looked a lot like Clark Gable, except that he didn't tan well, and wasn't as slim. He liked to do Edward G. Robinson imitations, which I found comical. Only I knew his true soul, and although we pretty much owned the turf, I never made a move on him, for good or worse.So there.)

And so, to Storrs...

Storrs, Connecticut is on the Massachusetts border of the state. It's also where the University of Connecticut is, which features a cafeteria with incredible food, all grown on a University-operated model farm (even meat and dairy) and a showplace garden of pharmaceutical herbs. Of the second I remember taking many samples, and of the former, that most of the participants (and myself) routinely took three and four times the quantity of food we'd be eating... at a time of high food prices, it felt exhilarating to be able to load up one's plate and stuff one's self for free, and even toss some away, just because we could! Now, to the Conference....

We were to hear Secrets of the Builder's Trade, and in service of which, I was to hear "...now, when we have a line four units in one direction, and another, three units in a perpendicular direction...we have a diagonal line..."

"....which is five, sir." I said, rising, from the back of the classroom.

"Please stay after class." I did so.

"Please explain your comment."

First, I gave the proof of the Pythagorean Theorem. Not the proof, just a cut-it-out-with-scissors-and-reassemble proof I'd found easy to memorize, a variant of Behold! Then I showed how, given parameters in a grid, one could find the distance between two points on a WW-II era military map, something I'd learned in a book by George Gamow. (I gritted my teeth on this one: most wargame maps I'd seen used hexes, not squares.) Then two points in three-space (without the curvature of the Earth, I added, as an afterthought). With a final flourish, I said, "Now, there isn't any law against using a fourth term here..." I did so. "...but that would lead us into the theory of Relativity, and.."

"Look, we thought you were just a dumb girl. How do you know this?"

"Tenth grade geometry, some books...your point?"

"Squaring up a building...the theory of relativity, this isn't high school, isn't it?" Somehow, I got the feeling that I'd dropped The True Name of God (or more accurately, the Mason Word) in casual conversation, and much to everyone's wonder, lightning had failed to strike.

"Well, I kind of like math and physics...I figured electrical work would tap into this."

"So, we've all had nothing beyond sixth grade. What in the hell are you doing here?"

"I love my family."

I brought home the trophy for Second Place, as the only female builder in the State, to the tune of ELO's Mr. Blue Sky, which Roger played in my honor ion his car radio, all the way home.

Note: the Theorem is also the short speech given by the Scarecrow in the movie version of the Wizard of Oz after he recieves his diploma. (Thanks, Mouse!)

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