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by yesenadam 1930 days ago
> 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!
2 comments
javajosh 1930 days ago
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!
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wcarey 1930 days ago
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.

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yesenadam 1929 days ago
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...

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