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Comment by wcarey | Hacker News Reader

The author's take on V.18 is interesting. Clarke has the proposition here: https://mathcs.clarku.edu/~djoyce/java/elements/bookV/propV1...

It's not three pages long in Clarke (nor in Heath), and either way, the argument is only 10 statements long.

> But the reader who takes the trouble to decode the proposition will see that it is trivial primary school arithmetic.

This is also not super obvious from Clarke or Heath. I'd love to know how the author would render this proposition trivial to a fourth grader.

To be sure, the mathematics of Euclid is a very foreign country from what's commonly taught in schools today. But Newton's thought is firmly rooted in that foreign country.

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> I'd love to know how the author would render this proposition trivial to a fourth grader.

Looking at the diagram in your link, and labelling AE x, EB y, CF a and FD b, "Let AE, EB, CF, and FD be magnitudes proportional taken separately, so that AE is to EB as CF is to FD. I say that they are also proportional taken jointly, that is, AB is to BE as CD is to FD.", means :

If x/y = a/b, then also (x+y) / y = (a+b) / b.

To prove this is true, operate on the second equation to make it the same as the first:

1. Expand: x/y + y/y = a/b + b/b

2. Subtract 1 from both sides: x/y = a/b. QED.

Notation really makes all the difference!

I came to the comments to say this, and you beat me to it. Its a better proof than the ones in that link, IMHO. I'm not sure why the author felt the need for all those inequalities!

Right - that argument, I'm suggesting, is way beyond most fourth graders even in its modern algebraic form. It relies on lots of algebraic techniques that aren't generally taught until 7th or 8th grade.

It's not, like, rocket science, but it's not trivial either.

Well, it does seem to me fair to say that "it is trivial primary school arithmetic". But I guess the truth of whether it is or it isn't trivial, isn't really something that can be scientifically tested – it rests on how exactly you define "trivial", what you count as the involved operations etc. Well, we could go further into it if we wanted (e.g. What are these "lots of algebraic techniques" involved in these 2 little steps?!) but having asserted our rivals claims without much evidence feels like a good place to stop – this time. :-)

Edit: Or I suppose "tabooing"[0] the word trivial would have helped – it was the trouble-maker.

[0] https://www.lesswrong.com/posts/WBdvyyHLdxZSAMmoz/taboo-your...

Euclid, Book 5, Proposition 18. If magnitudes, taken separately, be proportionals, they shall also be proportionals when taken jointly; that is, if the first be to the second as the third to the fourth, then the first and second together shall be to the second as the third and fourth together to the fourth.

This is from the Todhunter edition of Euclid's elements. the proof occupies precisely three pages. in post-Euclidean notation, if a/b = c/d then (a+b)/b = (c+d)/d. For young children replace the letters by small positive integers.

To convert Euclid's formulation to the symbolic formulation requires turning a magnitude into a real number, and defining the division of two real numbers. Euclid's proof avoids these difficulties. His magnitudes are the lengths straight line sections. He can decode a/b = m/n where m and n are positive integers as meaning na = mb. And similarly he can define a/b > m/n and m/n > a/b. So now the statement a/b > c/d can be decoded as the existence of positive integers m and n such that a/b > m/n > c/d. Finally the statement a/b = c/d is decoded as asserting that both a/b > c/d and c/d > a/b are false.

In modern jargon, Euclid is using the Dedekind cut definition of a real number. Using this definition, and working from first principles, the proof unsurprisingly requires three pages.

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