**Euclid of Alexandria (325 – 265 B.C.)**

Besides being a mathematician in his own right, Euclid is most famous for his treatise *The Elements* which catalogs and places on a firm foundation much of Greek mathematics. Not much is known about Euclid’s life. Even his birthplace and his education are in dispute. We do know that he was a teacher of mathematics in Alexandria and the founder of the Alexandrian School of Mathematics. His book *The Elements* is one of the most influential books in mathematics and, with the exception of the Bible, the most widely studied book in history. In fact, Abraham Lincoln felt very influenced by three works; the King James Bible, the works of William Shakespeare, and Euclid’s *Elements*.

It is an authoritative masterpiece which lays out carefully in a systematic way, what can be assumed (axioms and postulates) and what can be proven (propositions). He lays out the rules of arithmetic, the precise definitions of everyday geometric objects, and the basic rules of logic. He then goes on to prove 465 propositions from various areas of mathematics – geometry, number theory, and algebra.

*The Elements* is not without its critics. You will have to read an article (The Teaching of Euclid, Math. Gazz. 33, 1902) by the British philosopher Bertrand Russell who calls one of the propositions of Euclid “a tissue of nonsense”. His fillets several other parts of the *Elements*. Defenders of the Euclid claim that the only problem with Euclid is that he dd not study Russell! So, the lesson to be learned here is: read everything critically – no matter what famous name is attached to it.

Let’s us go through a few of the proofs of Euclid. We have already seen some of the result from the *Elements* earlier (one of the proofs of the Pythagorean theorem and the irrationality of the square root of 2). Here are a few more.

The first is a classic which proves that there are infinitely many prime numbers. A careless student may say “there are infinitely many numbers 1, 2, 3, 4,….. and so there must be an infinite number of primes”. But this logic does not hold water. If you think about it long enough, it is not as obvious as you may think. Here is the precise proof – considered one of the best proofs in mathematics.

**Theorem (Euclid – The Elements, Book IX, Prop. 20):**

*There are infinitely many prime numbers*.

**Proof:** This is another proof by contradiction (*reductio ad absurdum*). Suppose that there were only finitely many prime numbers. Call them a1, a2, a3, … an. This is the complete list of prime numbers according to your hypothesis. Consider the number x = a1 a2 a3…an + 1 (the product of your n prime numbers plus one). Now any integer must be divisible by some prime and so x is divisible by ai for some i (since a1, a2, a3, …., an is your *complete* list of primes). Since the product a1 a2 a3….an is divisible by aj and x is divisible by aj, then 1 = x – a1 a2 a3….an is also divisible by aj. But wait a minute! This would mean that aj = 1 which cannot be since the primes start with 2. Thus we have reached a contradiction and hence our hypothesis that there were only a finite number of primes can not be true. Hence there are an infinite number of primes. **QED**.

Note: Be careful here. The proof give you the impression that if you take the first n primes p1, p2, …, pn and form the number x = p1 p2….pn + 1 you will get another prime. This is not true. Can you find a counterexample?

Another note: The version of the proof presented is essentially Euclid’s but not exactly the way he said it. Here is a link to the original proof. Can you see how his proof works?

Every high school student knows that (a + b)^2 = a^2 + b^2 (the “foil” method). How do you actually prove this?

**Theorem (Euclid – The Elements, Book II, Proposition 4):** *For positive numbers a and b (a + b)^2 = a^2 + 2 a b + b^2.*

**Proof:** The proof becomes obvious when you consider the following drawing:

The identity follows by computing the area of the above square in two ways. First by computing the area as (a + b)^2. Then compute it by adding up the areas of the rectangles and squares which make up the big square. **QED.**

Note: Again, I have put Euclid’s argument into more modern language. Here is the original proof.

We all know the quadratic formula for solving ax^2 + b x + c = 0. *The Elements *has a proof of the* existence *of a solution to this equation (under certain circumstances). Solving it explicitly came later.

Here is a proof found off a very nice math history website.

**Theorem (Euclid)**: *For positive numbers a and b (with a > 2 b) the quadratic equation x^2 + b^2 = a x has a solution.*

**Proof: **Draw a line of length a. Then draw a line of length b perpendicular from the midpoint of a. Then draw a circle of radius a/2 and center at the end of this line.

We will now show that x is a solution to our quadratic equation x^2 + b^2 = a x. By the Pythagorean theorem, we have (a/2 – x)^2 + b^2 = (a/2)^2. Do a little algebra to get x^2 + b^2 = a x.** QED.**

*Note:* This proof shows an odd feature of mathematics which ones sees often. It shows that a solution *exists* but does not really show you how to find it. There are plenty of existence theorems throughout mathematics and some students find them annoying since they seem not to actually solve the problem. Perhaps this annoyance is justified. However, if one is thinking of mathematics in the Platonic sense, these existence theorems are actually ideal. We don’t need to physically know the solution. We just need to know it exists.

Another note: As usual, Euclid did not phrase the solution to the quadratic equation in exactly that way. He did not have a real notion of equation. The above is a modern rendition of Euclid.

Project: If one reads the proof carefully and uses a ruler (with measurements on it) and a compass, one can estimate the x which solve the quadratic. Try it out for a few values of a and b.

This visit with Euclid will be short. However, we will visit* The Elements *again when we talk about bisections and trisections and then again when we talk about the five Platonic solids.

Before leaving Euclid, we will mention that although The Elements is a respected book, it is not without its critics.

**Discussion of Russel’s critique of Euclid**

Bertrand Russell (British mathematician and philosopher 1872 – 1970) write a critique of Euclid in 1902 (*The teaching of Euclid*, Math. Gazz. Vol 2, 1902) where he claimed that there were some flaws in Euclid’s arguments. Here are some of his accusations:

- Euclid uses drawings to support his arguments. These do not give convincing mathematical arguments.
- Book I, Proposition I (To construct an equilateral triangle on a given finite straight line.)
- Euclid’s argument is as follows:
- Let AB be the given finite straight line.
- Describe the circle BCD with center A and radius AB. Again describe the circle ACE with center B and radius BA. Join the straight lines CA and CB from the point C at which the circles cut one another to the points A and B.
- Now, since the point A is the center of the circle CDB, therefore AC equals AB. Again, since the point B is the center of the circle CAE, therefore BC equals BA.
- Russell’s problem with this proof is that is assumed that there is a point C (which is not always true in curved space – and Euclid does not make this explicit) and it relies on a drawing. The proof requires an axiom: On any straight line there is at least one point whose distance from a given point on or off the line exceeds a given distance. The proof of the existence of C then follows from continuity.

- Book I, Proposition IV (If two triangles have two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, then they also have the base equal to the base, the triangle equals the triangle, and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides.)
- Euclid’s argument is as follows:
- Let ABC and DEF be two triangles having the two sides AB and AC equal to the two sides DE and DF respectively, namely AB equal to DE and AC equal to DF, and the angle BAC equal to the angle EDF.
- I say that the base BC also equals the base EF, the triangle ABC equals the triangle DEF, and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides, that is, the angle ABC equals the angle DEF, and the angle ACB equals the angle DFE.
- If the triangle ABC is superposed on the triangle DEF, and if the point A is placed on the point D and the straight line AB on DE, then the point B also coincides with E, because AB equals DE.
- Russell has a problem with superimposing the two triangles. Are these material triangles or abstract triangles? If they are material, they cannot be perfectly rigid. In addition when superimposed, they are bound to be slightly off.
- Russell proves that with the addition of another (kind of complicated) axiom, this theorem is still true.

- Book I, Proposition XII (To draw a straight line perpendicular to a given infinite straight line from a given point not on it.)
- Here is Euclid’s original argument.
- Let AB be the given infinite straight line, and C the given point which is not on it.
- It is required to draw a straight line perpendicular to the given infinite straight line AB from the given point C which is not on it.
- Take an arbitrary point D on the other side of the straight line AB, and describe the circle EFG with center C and radius CD. Bisect the straight line EG at H, and join the straight lines CG, CH, and CE.
- I say that CH has been drawn perpendicular to the given infinite straight line AB from the given point C which is not on it
- Since GH equals HE, and HC is common, therefore the two sides GH and HC equal the two sides EH and HC respectively, and the base CG equals the base CE. Therefore the angle CHG equals the angle EHC, and they are adjacent angles.
- But, when a straight line standing on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands.
- Therefore CH has been drawn perpendicular to the given infinite straight line AB from the given point C which is not on it.
- Russell claims that the existence of H required an axiom of continuity.