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t0mek | 1 year ago

We can also assume that the p/q=√2 is already the simplest form of the fraction, since every fraction must have one, as in the first section of the article.

Then if we figure out that both p and q are even, it means that p/q can be simplified (by dividing p and q by 2), which contradicts the assumption about the simplest form - and we don't need to use the infinite descent.

discuss

order

yuliyp|1 year ago

That assumes that every fraction has a unique simplest form. The first section of the article makes no such claims about the existence of a simplest form of fraction. The proof uses just algebraic manipulation, the fact that a sequence of strictly decreasing positive integers is finite in length, and the definition of a rational number (there exist integers p, q (q != 0) such that the number can be expressed as p/q).

red_trumpet|1 year ago

The article explicitly claims

> Rational numbers or fractions must have a simplest form.

They make no claim about uniqueness, but that is not needed in the argument.

vessenes|1 year ago

I was going to say this. The proof, unfortunately, sidelines into some very deep mathematics that it didn’t need to. I imagine but don’t know for sure that Euclid stopped where you do.

I’m not sure when infinite descent would be considered to have been formally proven to a modern mathematician but I bet it wasn’t in Euclid’s time!

yuliyp|1 year ago

I think the idea is that any strictly decreasing sequence of positive integers starting at N is a subsequence of (N, N-1, N-2, ..., 1) which has N elements, so any such sequence must have a finite number of elements.

So if you manage to produce an infinite sequence of strictly decreasing positive integers starting at a particular positive integer, then you've reached a contradiction.

red_trumpet|1 year ago

Isn't the proof just claiming that you can't divide an integet infinitely many times by 2, and still expect it to be an integer?