How did Pythagoras come up with a^2+b^2=c^2?

[thharrimw]

I have wondered about this for a while now and can anyone tell me the proof behind A^2+B^2=C^2?

 

[HallsofIvy]

There have been many proofs of the Pythagorean theorem, including one by James Garfield, 20^th president of the United States. Several, but certainly not all, are given here:
http://jwilson.coe.uga.edu/EMT668/EMT668.Student.Folders/HeadAngela/essay1/Pythagorean.html

 

[LukeD]

I personally like the proof that uses the simple fact that if vectors u and v are orthagonal, then the dot product of u and v is 0. Pretty much gives the result immediately without needing any pretty pictures (as nice as those pretty pictures are)

Let u and w be the vectors corresponding to the 2 legs of your triangle. The definition of a right angle, is one that makes $$u \cdot w = 0$$. Also note that for any vector v $$\lvert v \rvert^2 = v \cdot v$$ where $$\lvert v \rvert$$ means the length of the vector v. Note that the hypotenuse is the vector u+w

You also need to know that the dot product is pretty much just like multiplication, so it distributes over addition, meaning that $$u \cdot (v + w) = u \cdot v + u \cdot w$$ and $$(u + v) \cdot w = u \cdot w + v \cdot w$$

Then $$\lvert u+w \rvert^2 = (u+w) \cdot (u+w) = u \cdot (u+w) + w \cdot (u+w) = u \cdot u + 2 (u \cdot w) + w \cdot w = v \cdot v + w \cdot w = \lvert v \rvert^2 + \lvert w \rvert^2$$

If you know how to work with vectors, that will probably be the shortest, most direct proof of the pythagorean theorem you will ever see

 

[CRGreathouse]

The Pythagoreans probably found the theorem (directly or indirectly) as a rule of thumb from the Egyptians, who were good practical geometers. Pythagoras was the first, I believe, to prove that the rule holds (exactly), unlike the various approximations of the day.

The theorem was probably first proved by dissection -- I can't remember, is this known?

 

[robert Ihnot]

How did Pythagoras come up with a^2+b^2 = c^2?

There are the books of Euclid, often you can find something right out of the big bookstores, and so if one is really interested that is a question easily researched.

The proofs by the Greeks involved geometry rather than algebra, so they are, to us anyway, harder than proofs by algebra. However, when I took high school Geometry I can remember we covered the Pythagorean theorem.

 

[robert Ihnot]

LukeD: I personally like the proof that uses the simple fact that if vectors u and v are orthagonal, then the dot product of u and v is 0. Pretty much gives the result immediately without needing any pretty pictures (as nice as those pretty pictures are)

I wonder how legitimate that is. Why not define a right angle such that the square of the sides equals the hypothesis?

 

[LukeD]

LukeD: I personally like the proof that uses the simple fact that if vectors u and v are orthagonal, then the dot product of u and v is 0. Pretty much gives the result immediately without needing any pretty pictures (as nice as those pretty pictures are)

I wonder how legitimate that is. Why not define a right angle such that the square of the sides equals the hypothesis?

Try out that definition, see what you get.
If $$\lvert u + w \rvert^2 = \lvert u \rvert^2 + \lvert w \rvert^2$$ is your definition of having a right angle between u and w, then what does that imply about $$u \cdot w$$?

 

[John Creighto]

In the book A profile of Mathematical Logic, Howard DeLong speculates that Pythagoras most likely used his theory of proportions.

http://en.wikipedia.org/wiki/Pythagorean_theorem#Proof_using_similar_triangles

The esiest proof is given here:

http://en.wikipedia.org/wiki/Pythagorean_theorem#Algebraic_proof

 

[dodo]

Personally I like this one, but I'm not sure if its origin is Indian or Babylonian.

The triangles, while in the red position, make a^2 + b^2 (the two black squares), while moved to the pink position they make c^2.

 

[M Grandin]

Simplest proof

Simplest possible proof I derived myself long ago:

According to similarity , if the big triangle area is k* c^2 then it contains two triangels of area k* a^2 resp k* b^2. Therefore k* c^2 = k* a^2 + k* b^2 and hence
c^2 = a^2 + b^2

 

[mathwonk]

it depends what you think the theorem says. you can regard it as a statement abiout numbers, i.e. a relation between the squares of the lengths of the sides of a right triangle, but euclid did not view it this way.

he viewed it as an equivalence between the geometric figure of square erected on the hypotenuse, and the two squares erected on the sides. And he had no theory of area at that point.

what he essentially proved was that it is possible to decompose the larger square into pieces which can be reassembled to form the other two squares. actually he did not quite prove this, but his same method of proof does do this.

i.e. one can define several different equivalence relations:

1) One figure can be decomposed into pieces which can be reassembled to form another figure.

2) two figures can be decomposed into pieces, and both sets of pieces can be reassembled together with the same third set of pieces, to form the same figure.

3) one can assign an area to every figure in such A WAY that figures which can be related in the ways above have the same area, and then the two figures have the same area. and so on. we tend to think of all these as the same, but actually defining area cREFULLY IS QUITE hard, and euclid did not do it. even showing the previous relations are transitive is some work, and eucloid did not do that either.

the proof that the numbers obtained by squaring the sides and adding are the same, does follow from the theory of similarity, but euclid did not have that theory in plACE AT THE TIME OF stating the pythagorean theorem.

so actually euclid did not prove the theorem at all, but gave only a partial proof.

i.e. a complete proof needs either a theory of similarity, or a theory of area, or the proof that the relations euclid used are transitive, none of which are in place fully in euclid, at the time he gave the theorem.

the treatment in hartshorne's geometry, euclid and beyond, is excellent.