Theorem: If $x,y \in \mathbb{R}$ and $x > 0$, $\exists$ a positive integer $n$ such that $nx > y$

I read the proof by Rudin and understood it. I think it is very elegant and uses the LUB property of $\mathbb{R}$. But the way I tried to prove it is the following:

If $x > y$, then $n = 1$. If $x < y$, then $y > 0$ and it makes sense to define $m = \frac{y}{x}$. $m \in \mathbb{R}$ using one of the field axioms. Since the set of natural numbers is unbounded $\exists n \in \mathbb{N}$ such that $n > m$. This implies $nx > y$. This proves the Archimedean property.

Is there something wrong with the proof or am I assuming something that is not obvious?

This is something I frequently wonder while I am trying to learn analysis. I really like the way Rudin proves the theorems in his book. But I find it really difficult to adapt that kind of thinking and rigor. I think a big reason why I find this difficult is because I have an engineering background and I am never used to being so rigorous with proofs. Can anyone suggest some tips (or your experience) that deal with improving the ability to write rigorous proofs (and the ability to find logical gaps in proofs).


But you don't know that the set of natural numbers is unbounded in the reals. In fact that statement is equivalent to the statement that $\mathbb{R}$ has the Archimedean property.

I presume Rudin shows it in pretty much the same way as suggested in Exercise 3 in another answer of mine (man, I've gotten a lot of mileage out of that!)

  • $\begingroup$ You are right!! I am embarrassed I did not realize it. Guess I didn't completely understand Rudin's proof then. Thanks. $\endgroup$ – Srikanth Feb 26 '11 at 16:50
  • $\begingroup$ That "answer of mine" you refer to is amazing! I had never thought about uniqueness of $\mathbb{R}$ in that context - thanks for linking to it! $\endgroup$ – Jason DeVito Feb 26 '11 at 17:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.