I have to show that two forms of the Archimedean Property are equivalent. The first form is "for every positive real number there is a smaller positive rational number". The second form is that "for every positive real number there is a bigger positive integer".
I tried to show they are equivalent by showing the first implies the second and the second implies the first. However, I am having trouble doing this because one statement involves smaller and the other involves bigger. I would appreciate any hint or suggestion in the right direction.
I apologize for the lack of clarity. The first form is "For any positive real number there exists a smaller positive rational number". The second form is ""for any real x > 0; there exists a positive integer z such that z > x."
Pretty easy to show: the number is real and so must be either a positive rational or irrational number. If the number is rational divide by two to get a smaller positive rational number (for example if the number is 1/64 then a smaller rational number is 1/128). If the number is irrational simply tuncate it at the first non-zero digit (for example if the number is 1/e^5 = 0.006737947... , truncate after the 6 to get 0.006, which is a smaller rational number than 1/e^5).
To the OP: your two statemenst are not equivalent, as the first one says there is a smaller positive rational number whereas the second says a larger integer. If the second statement was "for every positive real number there is a larger rational number" then it would be the inverse of the first statement.
Thanks for the feedback. I have reproduced the statements as I given, but it is always possible that there was a mistake from the original source. Although, I don't think the second statement would be the inverse of the first if you replace integer with rational because you would also have to interchange the quantifers when negating the statement.
All of this depends upon the basic fact of the density of the rationals: between any two numbers there is a rational number.
Apply that to .
To prove density we do need completeness and well ordering.
But I cannot how either of the posted questions relate.
"Reverse Mathematics" in Wikipedia). For example, the Heine–Borel theorem (every covering of a closed real interval by a sequence of open intervals has a finite subcovering) is equivalent to the fact that every continuous real function on the closed interval is uniformly continuous. This equivalence holds over a suitable theory that is weaker than the axioms of real numbers.
Here similarly we need a weaker theory. It seems that the theory of ordered fields is appropriate because there is an embedding of rational numbers into such field and we can formulate the Archimedean property since order is present. Compared with the axioms of real numbers, the theory of ordered fields lacks the completeness axiom, so an ordered field is not necessarily Archimedean. In other words, the problem is to prove the equivalence without using the completeness axiom. In fact, it would make more sense to state the problem in terms of ordered fields to begin with, without referring to real numbers.
The fact that the second form implies the first one is trivial: given an x > 0, find an n such that 1/x < n and take 1/n. For the converse, one has to prove that every positive rational number is exceeded by some integer.
As emakarov points out either of those two implies the other. (Although I disagree that the reals necessarily have the Archimedean Property).
Are you asked to show that those two together are equivalent to the usual statement of the Archimedean Property?
The usual statement of the Archimedean Property:
If each of is a positive real number then
Is that the question?
Opened this to confirm that I really got a "Thanks" from emakarov. Then spotted a typo in second line. It should be b>r, not r>b. I'm sure emakarov spotted it. Noble of him not to point it out. Don't know that I would have been that generous.