# Plane/composite numbers as lines?

In the Elements of Euclid a plane number (i.e. a composite number), was represented by a line AB. But, being a plane number a multiplication of two numbers (i.e. two lines, in the mind of a pythagorean) it should have been represented by an area. In fact it is believed that the pythagoreans represented plane numbers by means of rectangles. When did begin to be conceived that a surface could be also represented by a line, to divide it in fractions?

In the Babylonian tablet Plimpton 322 (1800 B.C.) there is a list of very big pythagorean triples ($a^2 + b^2 = c^2$). They must have been generated by means of some formula. But if the formula is the classic $a=p^2-q^2$, $b=2pq$, $c=p^2+q^2$, then they would have added/subctracted two squares to obtain a side of the triangle. Did the babylonians have this capacity for abstraction?

• Your interpretation of Euclid doesn't seem right to me. It seems clear to me from Elements VII 15-19 that a particular integer may have as many as three different representations: as a line, as a plane, or as a solid. – Ben Crowell Aug 14 '15 at 13:20
• I didn't interpreted any Euclid assertion. In fact, as you say, Euclid uses different representations for integers. My question concerns when and how ->before Euclid<- these different representations arose. – Benzio Aug 14 '15 at 14:08
• If I understand your question correctly you will find the answer by searching for "the area and the side i added". – user2255 Oct 15 '15 at 18:49

As for Babylonians their system of numerals was positional with base $60$ and excellent for doing calculations, which they were very good at, even their astronomical models used numerical sequences rather than geometry. So they are much more likely to have found small Pythagorean triples by numerical experimentation and then figured out the pattern without squares or triangles. A suggestion of Robson based on refined translations is that for pedagogical purposes they were interested in reciprocal pairs of numbers $x=p/q$ and $y=q/p$ with only $2,3,5$ (divisors of $60$) as prime divisors of $p,q$. Pythagorean triples then come as a side effect: if $xy=1$ then $1+\big(\frac{x-y}{2}\big)^2=\big(\frac{x+y}{2}\big)^2$, and in terms of $p,q$ we get the formula for the triples.