I think it is sometimes said that it is anachronistic to attribute to the ancient Greeks the discovery that the square root of two is irrational, because what they discovered was a properly geometrical fact, that the side and diagonal of a square are incommensurable, rather than a fact about real numbers.
It is correct to say that the Greeks discovered an incommensurability fact. But it is, I think, worth noting that this incommensurability fact is not really geometric fact: it is a geometric-cum-arithmetical fact. Here is why. The claim that two line segments are commensurable says that there are positive integers m and n such that m copies of the first segment have the same length as n copies of the second. This claim is essentially arithmetical in that it quantifies over positive integers.
And because pure (Tarskian) geometry is decidable, while the theory of the positive integers is not decidable, the positive integers are not definable in terms of pure geometry, so we cannot eliminate the quantification over positive integers. In fact, it is known that the rational numbers are not definable in terms of pure geometry either, so neither the incommensurability formulation nor theory irrationality formulation is a purely geometric claim.
I think. All this decidability and definability stuff confuses me often.
Well, some theories of the positive integers are not decidable, such as Peano arithmetic. Others are decidable such as Presburger arithmetic.
ReplyDeleteOtherwise if these things are not purely geometrically definable, then I guess the uderlying topologies here are also not purely geomtrical - I guess so.
I was thinking of the decidability of Th(N), where N is our "intended" model of the naturals, not of the decidability of any particular recursive axiomatization. (Th(N) is not recursively axiomatizable, of course.)
ReplyDeleteDon't you rather mean "Th(something)" to be our "intended" model of "something", such that "something" could be the naturals N, such that Th(N) is our "intended" model of naturals N?!?
ReplyDeleteWhy isn't Th(N) recursively axiomattizaböe though?!?