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Is it possible to prove that every real number must between two integers without using the completeness property?

Martin Sleziak
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    It depends in how precisely you develop the reals. There are Archimedean ordered fields that are not complete. In that sense you do not need completeness. – André Nicolas Nov 05 '13 at 06:53
  • @Newb This is old, but I still think it's worth mentioning. What you have stated is not the Archimedean property. The property (in one of its equivalences), essentially says that no matter how large an $x \in \mathbb{R}$ is given, there is always an $n \in \mathbb{N}$ larger. The way I first saw it stated, in Rudin, gives it explicitly as: If $x,y \in \mathbb{R}$ and $x > 0$, then $\exists n \in \mathbb{N}$ such that $nx > y$. Now, what you have stated is that, given any $a,b \in \mathbb{R}_{>0}$, then there is always an $n \in \mathbb{N}$, such that $na > nb$. However this is only the case if –  May 08 '14 at 08:05
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    You can find this on ProofWiki: https://proofwiki.org/wiki/Real_Number_lies_between_Unique_Pair_of_Consecutive_Integers – Bart Michels Nov 06 '17 at 20:03

2 Answers2

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Yes. It's a consequence of the Archimedean Property of $\mathbb{R}$.

Fact: Archimedean Property: If $a,b$ are positive real numbers, then $\exists n \in \mathbb{N}$ s.t. $na > b$.

Proposition: if $a \in \mathbb{R}$, then $\exists N \in \mathbb{Z}$ s.t. $N-1 \leq a < N$.

Proof: Let $S = \{n \in \mathbb{Z} | n > a \}$. Then by the Archimedean property, $S \neq \emptyset$ and S is bounded below. By the Well-Ordering Principle, S has a least element. Then $N-1 \notin S$, so $N-1 \leq a < N$.

Newb
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  • I think you have a small error. Archimedian property tells us that for every $a,b\in\mathbb{R}$ there exists natural$n$ such that $na>b$. Correct? – Amihai Zivan Nov 05 '13 at 06:53
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    @AmihaiZivan No. Consider $a=-10, b=-5 \in \mathbb{R}$. We have $a<b$ and for no choice of $n \in \mathbb{N}$ will $an>b$ be true. (Assuming that $\mathbb{N}$ is the set of positive integers.) – Newb Nov 05 '13 at 07:02
  • OK - correct, but I think it should be $na>b$, because that's really the essence of it - for every two positive reals WLOG $a<<b$ we can find a large enough natural s.t. $na>b$. – Amihai Zivan Nov 05 '13 at 07:12
  • @AmihaiZivan I don't see how that highlights in a mistake in my answer. – Newb Nov 05 '13 at 08:17
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    @Newb $a$ and $b$ should be strictly positive that's why your example above doesn't work. Moreover, The archimedian property should be $na>b$ not $na>nb$. Counterexample: $a=1$ and $b=2$ there's no $n\in\mathbb{N}$ such that $na>nb$. – user5402 Aug 11 '14 at 16:10
  • There is somethin wrong here. As pointed out by others it should be $na>b$ because for natural $n$, it is either false or trivially true that $na>nb$ – Aritra Das Aug 02 '16 at 12:18
  • Indeed, this answer's statement of the Archimedean Property isn't right. It should be updated to read that there is some n such that na > b – bschneidr Sep 06 '17 at 06:08
  • Almost four years later, I'm fixing this mistake. Apologies to all. – Newb Sep 06 '17 at 07:02
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Assume there was a number that wasn't between two integers, than it must be greater than every integer or less than every integer which is a contradiction to the Archimedean property. While this holds, I think it implicitly relies on completeness.

Jeremy Upsal
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    I don't think this proves the title statement "very real number lies between 2 consecutive integers". – Yibo Yang Jun 20 '18 at 04:20