February 26

Plan: ( copied from the last class)

  • Questions
  • Work toward Euclid's characterization of Pythagorean triples

The discussion page can be found here, for observations and remarks. Edit this page to provide a single coherent narrative of what went on in class - don't just append your personal few notes.


Proposition II-5

gnomon.jpg

If a straight line is cut into equal and unequal segments, then the rectangle contained by the unequal segments of the whole together with the square on the straight line between the points of section equals the square on the half. Let a straight line AB be cut into equal segments at C and into unequal segments at D. I say that the rectangle AD by DB together with the square on CD equals the square on CB. Describe the square CEFB on CB, and join BE. Draw DG through D parallel to either CE or BF, again draw KM through H parallel to either AB or EF, and again draw AK through A parallel to either CL or BM.

This is the identity

(1)
\begin{equation} x^2 - y^2 = (x-y) (x+y), \end{equation}

proved geometrically. To see that, let $x = AC = BC= BF$ and $y = CD = MF$. Then the right side of (1) is the rectangle $ADHK$. The light purple rectangle just matches $DBFG$, so $ADHK$ plus the dark purple square is the square $BFEC$.

You can use this Proposition to solve quadratic equations geometrically … maybe I will put that on the homework. It points toward the material in Chapter 5 of our text, which we may do next.

The picture is from
http://my.opera.com/jehovajah/blog/2012/05/08/completing-the-square-by-the-gnomon

"gnomon" is a nice word.

  • Use Euclidean geometry to represent numbers.(not by point but by segment length)
              • unit length (In ancient Greek, the systems of ancient weights and measures evolved as needs changed)
              • Euclidean Arithmetic is able to do addition, multiplication, 1/n, n/m, etc.
              • Eudoxus' theory of proportions is in Euclid Book V. The wikipedia entry http://en.wikipedia.org/wiki/Eudoxus_of_Cnidus#Mathematics is informative.

How did Euclid use this result to find all the Pythagorean triples?

  • Rational arithmetic
  • Plane numbers

We have the result:

(2)
\begin{equation} (x-y)(x+y)+y^2=x^2 \end{equation}

If $(x+y)(x-y)$ is square, then we have a Pythagorean triple.

What would make $(x+y)(y-x)$ square? Euclid said: $(x+y)(y-x)$ is a square if and only if $(x+y)$ and $(x-y)$ are similar plane numbers.

Before trying to prove this statement, we need the following definitions:

A plane number "is" a rectangle. For example, 2*6 is a plane number, representing a rectangle with sides 2 and 6. In other words, a plane number is just two factors.

Two plane numbers a*b and c*d are similar if a/b=c/d, i.e. if the respective rectangles are similar. For example, 1*2 and 2*4 are similar plane numbers.

An (incomplete) attempt at a proof of Euclid's statement from above:

Suppose that a/b=c/d. We wish to show that there exists some integer r such that a*b*c*d=r^2.

Let a*b be less than or equal to c*d. Then we can find some integer k such that c/d=(a*k)/(b*k).
[For example, take 1*3 and 4*12: since 1*3<4*12, we can write 4*12=(1*4)/(3*4). In this case k=4.]
Then a*b*c*d=a*b*(a*k*b*k)=(a*b*k)^2. Then r=a*b*k.

Now suppose that (a*b)(c*d)=r^2 for integer r. We wish to show that a/b=c/d. I am not completely sure how to prove this part.


Euclid's Theorem on The Infinitude of Primes
* Book IX, Proposition 20: prime numbers are more than any assigned multitude of prime numbers
* The formulation deliberately avoids the word "infinite".

Theorem:
Prime numbers are more than any assigned multitude of prime numbers

Proof (Euclidean):
For finitely many primes $p_1, p_2, ... p_n$. Let $Q=(p_1\cdot p_2\cdot ... \cdot p_n)+1$ is either a prime or not. If Q is a prime, then it's a new prime another than $p_i$'s, if Q is not a prime, it can only be factorized by some new prime other than $p_i$'s

Proof (Modern):
Suppose, to the contrary, that there are only finitely many primes. Define P=$p_1p_2\dots p_n$, and furthermore, define $Q=P+1$. We arrive at two cases:
(I) Q is prime
Suppose Q is prime. Then this is a contradiction because we have found a prime outside of our finite list.
(II) Q is composite
Suppose Q is composite. Because P is the product of all primes, it must be the case that one of those primes $p_i$, divides Q. But it is also the case, since P is the product of these primes, that $p_i$ divides P. However, since $p_i$ divides Q and P, it must be able to divide Q-P. However, since this equals one, this is a contradiction, because by definition a prime must be greater than one and nothing greater than one evenly divides one.

Therefore, by these contradictions, it must be the case that there are an infinite number of primes. Q.E.D.
(See: http://www.cut-the-knot.org/proofs/primes.shtml)


Page Reformatted on 2/26/2013 by Rob Moray (robmoray)

Interesting…
Some Pythagorean triples are scalar multiples of other triples: (6, 8, 10) is twice (3, 4, 5). We call a triple (a, b, c) primitive when the three integers have no common factor. For any triple (a, b, c), if d is the greatest common divisor of all three terms then (a/d, b/d, c/d) is a primitive triple and the original triple is a scalar multiple of this, so finding all Pythagorean triples is basically the same as finding all primitive Pythagorean triples.
-Rachel Bay-Chaparro

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