- The Greeks did have inventions, but they were stronger in the world of ideas, and we owe much to them for this. They developed geometry and studied astronomy, geography, and mechanics. These studies formed the basis of much science that followed. Their philosophers developed speculative philosophy which is the foundation of much of our speculation and a good portion of our Mathematics. Their art and architecture were very influential and set styles that are still popular and highly copied today. Museums around the world have much material from ancient Greece which is often the most valuable part of their collection. A list of inventions follow:
analog computer with clockwork mechanism
cyclorama
camera obscura
steam driven jet engine
Archimedian Screw
astrolabe
catapult
hydraulic music organ
coinage
parchment
trireme
Ancient Greek Scientists:
Alcmaeon of Croton
Anaxagorus of Clazomenae
Anaximander of Miletus
Apollnius of Perga
Archimedes of Syracuse
Archytas of Tarentum
Aristarchus of Samos
Aristotle
Callipus of Cyzicus
Chalcidius
Ctesibius of Alexandria
Pythagorean Theorem or Pythagoras TheoremIn mathematics, the Pythagorean theorem (American English) or Pythagoras' theorem (British English) is a relation in Euclidean geometry among the three sides of a right triangle (
right-angled triangle in British English). It states:
<BLOCKQUOTE>
In any right triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle).</BLOCKQUOTE>
The theorem can be written as an equation:
where
c represents the length of the hypotenuse, and
a and
b represent the lengths of the other two sides.
The Pythagorean theorem is named after the Greek mathematician Pythagoras, who by tradition is credited with its discovery and proof, although it is often argued that knowledge of the theory predates him. (There is much evidence that Babylonian mathematicians understood the principle, if not the mathematical significance).
If we let
c be the length of the hypotenuse and
a and
b be the lengths of the other two sides, the theorem can be expressed as the equation:
or, solved for
c:
If
c is already given, and the length of one of the legs must be found, the following equations (which are corollaries of the first) can be used:
or
This equation provides a simple relation among the three sides of a right triangle so that if the lengths of any two sides are known, the length of the third side can be found. A generalization of this theorem is the law of cosines, which allows the computation of the length of the third side of any triangle, given the lengths of two sides and the size of the angle between them. If the angle between the sides is a right angle it reduces to the Pythagorean theorem.
Proof using similar triangles Proof using similar triangles
Like most of the proofs of the Pythagorean theorem, this one is based on the proportionality of the sides of two similar triangles.
Let
ABC represent a right triangle, with the right angle located at
C, as shown on the figure. We draw the altitude from point
C, and call
H its intersection with the side
AB. The new triangle
ACH is similar to our triangle
ABC, because they both have a right angle (by definition of the altitude), and they share the angle at
A, meaning that the third angle will be the same in both triangles as well. By a similar reasoning, the triangle CBH is also similar to
ABC. The similarities lead to the two ratios:
These can be written as
Summing these two equalities, we obtain
In other words, the Pythagorean theorem:
Proof using similar trianglesLike most of the proofs of the Pythagorean theorem, this one is based on the proportionality of the sides of two similar triangles.Let ABC represent a right triangle, with the right angle located at C, as shown on the figure. We draw the altitude from point C, and call H its intersection with the side AB. The new triangle ACH is similar to our triangle ABC, because they both have a right angle (by definition of the altitude), and they share the angle at A, meaning that the third angle will be the same in both triangles as well. By a similar reasoning, the triangle CBH is also similar to ABC. The similarities lead to the two ratios: These can be written as Summing these two equalities, we obtain In other words, the Pythagorean theorem: Euclid's proofIn Euclid's
Elements, Proposition 47 of Book 1, the Pythagorean theorem is proved by an argument along the following lines. Let
A,
B,
C be the vertices of a right triangle, with a right angle at
A. Drop a perpendicular from
A to the side opposite the hypotenuse in the square on the hypotenuse. That line divides the square on the hypotenuse into two rectangles, each having the same area as one of the two squares on the legs.
For the formal proof, we require four elementary lemmata:
- If two triangles have two sides of the one equal to two sides of the other, each to each, and the angles included by those sides equal, then the triangles are congruent. (Side - Angle - Side Theorem)
- The area of a triangle is half the area of any parallelogram on the same base and having the same altitude.
- The area of any square is equal to the product of two of its sides.
- The area of any rectangle is equal to the product of two adjacent sides (follows from Lemma 3).
The intuitive idea behind this proof, which can make it easier to follow, is that the top squares are morphed into parallelograms with the same size, then turned and morphed into the left and right rectangles in the lower square, again at constant area.
The proof is as follows:
- Let ACB be a right-angled triangle with right angle CAB.
- On each of the sides BC, AB, and CA, squares are drawn, CBDE, BAGF, and ACIH, in that order.
- From A, draw a line parallel to BD and CE. It will perpendicularly intersect BC and DE at K and L, respectively.
- Join CF and AD, to form the triangles BCF and BDA.
- Angles CAB and BAG are both right angles; therefore C, A, and G are collinear. Similarly for B, A, and H.
- Angles CBD and FBA are both right angles; therefore angle ABD equals angle FBC, since both are the sum of a right angle and angle ABC.
- Since AB and BD are equal to FB and BC, respectively, triangle ABD must be equal to triangle FBC.
- Since A is collinear with K and L, rectangle BDLK must be twice in area to triangle ABD.
- Since C is collinear with A and G, square BAGF must be twice in area to triangle FBC.
- Therefore rectangle BDLK must have the same area as square BAGF = AB2.
- Similarly, it can be shown that rectangle CKLE must have the same area as square ACIH = AC2.
- Adding these two results, AB2 + AC2 = BD × BK + KL × KC
- Since BD = KL, BD* BK + KL × KC = BD(BK + KC) = BD × BC
- Therefore AB2 + AC2 = BC2, since CBDE is a square.
Garfield's proofJames A. Garfield (later President of the United States) is credited with a novel algebraic proof.
The whole trapezoid is half of an (
a +
b) by (
a +
b) square, so its area = (
a +
b)
2/2 =
a2/2 +
b2/2 +
ab.
Triangle 1 and triangle 2 each have area
ab/2.
Triangle 3 has area
c2/2, and it is half of the square on the hypotenuse.
But the area of triangle 3 also = (area of trapezoid) − (sum of areas of triangles 1 and 2)
=
a2/2 +
b2/2 +
ab −
ab/2 −
ab/2
=
a2/2 +
b2/2
= half the sum of the squares on the other two sides.
Therefore the square on the hypotenuse = the sum of the squares on the other two sides.
Proof by subtractionIn this proof, the square on the hypotenuse plus four copies of the triangle can be asssembled into the same shape as the squares on the other two sides plus four copies of the triangle. This proof is recorded from China.
[b]Similarity proofFrom the same diagram as that in Euclid's proof above, we can see three similar figures, each being "a square with a triangle on top". Since the large triangle is made of the two smaller triangles, its area is the sum of areas of the two smaller ones. By similarity, the three squares are in the same proportions relative to each other as the three triangles, and so likewise the area of the larger square is the sum of the areas of the two smaller squares.
Proof by rearrangementA proof by rearrangement is given by the illustration and the animation. In the illustration, the area of each large square is (
a +
b)
2. In both, the area of four identical triangles is removed. The remaining areas,
a2 +
b2 and
c2, are equal. Q.E.D.
This proof is indeed very simple, but it is not
elementary, in the sense that it does not depend solely upon the most basic axioms and theorems of Euclidean geometry. In particular, while it is quite easy to give a formula for area of triangles and squares, it is not as easy to prove that the area of a square is the sum of areas of its pieces. In fact, proving the necessary properties is harder than proving the Pythagorean theorem itself (see Lebesgue measure and Banach-Tarski paradox). Actually, this difficulty affects all simple Euclidean proofs involving area; for instance, deriving the area of a right triangle involves the assumption that it is half the area of a rectangle with the same height and base. For this reason, axiomatic introductions to geometry usually employ another proof based on the similarity of triangles (see above).
A third graphic illustration of the Pythagorean theorem (in yellow and blue to the right) fits parts of the sides' squares into the hypotenuse's square. A related proof would show that the repositioned parts are identical with the originals and, since the sum of equals are equal, that the corresponding areas are equal. To show that a square is the result one must show that the length of the new sides equals
c. Note that for this proof to work, one must provide a way to handle cutting the small square in more and more slices as the corresponding side gets smaller and smaller.