Archimedean property (2)

1 Name: Anonymous Scientist : 2007-01-28 03:00 ID:oenHq1li

Let x be any real number. Then there exists a natural number n such that n > x.

I don't get this or the proof of this. Can't you just as well say the opposite: Let n be a natural number, then there exists a real number x such that x > n ?

The second part of the Archimedean property makes sense though: Given any real number y > 0, there exists a natural number n satisfying 1/n < y. And you can get the first part from this. But I don't get the weird proof the book gives and why the natural numbers have to be higher than the real numbers. It says it should be obvious but I don't see it.

The book also say 'Assume N (the natural numbers) is bounded above, then by the Axiom of Completeness it should have a least upper bound and you reach a contradiction because you can do n+1 [or something] and so N is not bounded above but somehow the reals are.' It uses the Axiom of Completeness on the naturals but I thought it could only be used for the reals. It makes no sense.

2 Name: Anonymous Scientist : 2007-01-28 10:27 ID:oenHq1li


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