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We state the Archimedean property as follows. If $a>0$ and $b>0$, then $\exists n\in \mathbb{N}$ s.t. $an>b$. I know that the original proof uses the completeness of $\mathbb{R}$. However, my following attempt of the proof does not involve the completeness and hence should be rejected, hopefully finding out the precise loophole. In particular, I am aware that there are ordered fields that does not have the Archimedean property, although I do not know them in any detail.

My attempt of proof: Suppose that it is false, i.e. if $a>0$ and $b>0$, then $\forall n\in \mathbb{N}$, $an\leq b$. Then, $n\leq b/a$ and (by definition of field operations) $\mathbb{N}$ is bounded above. Hence, it is a contradiction.

Is this problematic because this is valid provided that [$\mathbb{N}$ is not bounded above] and [$n\leq b/a$ $\forall n\in\mathbb{N}$] are contradictory, and I have not shown that?

Nugi
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  • @SassatelliGiulio How about this. Suppose $n_0\in\mathbb{N}$ is an upper bound of $\mathbb{N}$. From the Peano axioms, the successor of $n_0$ exists and we denote it as $n_0+1\in\mathbb{N}$. By axioms of ordered fields (not completely ordered necessarily), $1>0$ and hence $n_0+1>n_0$ gives a contradiction. Thus, $\mathbb{N}$ is not bounded above even if it is considered as a subset of a non-complete ordered field. – Nugi Jun 06 '22 at 10:51
  • However, you can easily prove that there is no largest natural number. If this were the case, there would be a natural number without a successor. But the peano axioms guarantee that this is impossible. – Peter Jun 06 '22 at 10:52
  • @SassatelliGiulio That's true. But, I have (I think) showed that $\mathbb{N}$ is not bounded above in just ordered field. Hence, isn't it enough to contradict with $n<b/a $ $\forall n\in\mathbb{N}$? – Nugi Jun 06 '22 at 10:55
  • @SassatelliGiulio That's amazing to hear. Could you point me where I can find them? Actually, I now think I was wrong saying that I am aware of ordered fields that Archimedean property do not hold. – Nugi Jun 06 '22 at 10:57
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    https://math.stackexchange.com/questions/484689/ordering-the-field-of-real-rational-functions – Sassatelli Giulio Jun 06 '22 at 11:00
  • @SassatelliGiulio I will take some time to look into them, thank you! And, if possible, could you tell me in a concise way why $\textit{That's true. But,} \cdots$ is wrong? – Nugi Jun 06 '22 at 11:00
  • I see! And I really appreciate your answer. Thank you :) – Nugi Jun 06 '22 at 11:58

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Your attempt is incorrect for two reasons. Firstly, you wrongly stated the archimedean property without specifying $a,b$. To grasp the logical structure of mathematics properly, you must use the appropriate quantifiers in the appropriate order. Omitting them will not help you. Also read this post regarding a related issue. Because you wrongly stated the archimedean property, you were unable to negate it correctly. The negation of "$∀a,b{∈}ℝ\ ( \ a>0 ∧ b>0 ⇒ Q(a,b) \ )$" is certainly not "$∀a,b{∈}ℝ\ ( \ a>0 ∧ b>0 ⇒ ¬Q(a,b) \ )$"! So do the negation properly, and then think again what your argument gets you.

Secondly, you cannot prove that $ℕ$ has no upper bound without using something equivalent to the archimedean property. So you have not actually managed to obtain a contradiction as you claimed. It would be a good exercise for you to prove the equivalence for any ordered field $F$ containing $ℕ$:

  $∀a,b{∈}F\ ( \ a>0 ∧ b>0 ⇒ ∃k{∈}ℕ\ ( \ a·k > b \ ) \ ) ⇔ ¬∃c{∈}F\ ∀k{∈}ℕ\ ( \ k ≤ c \ )$.

user21820
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