Archimedean property

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.

So why exactly is the natural numbers greater than the real numbers?

>>1
"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 ?"

This is true, but it is not the opposite of the archimedean property. The opposite of the archimedean property would be "there exists some x such that no natural number n is greater than x."

"It uses the Axiom of Completeness on the naturals but I thought it could only be used for the reals. It makes no sense."

It uses the axiom of completeness on the naturals as a subset of the reals. Also, it does not state that the reals are bounded above.

>>2
Okay. I guess it makes sense because you can use the Axiom of Completeness on the naturals.

>>2
Thanks. I re-read the proof and it makes a bit more sense to me. It's just not so revealing as the 'lol square root of 2 is irrational' proof.

>>4
err...what I meant by revealing was that I understood it and liked it I guess... Ignore the last sentence there lol.