It was established in Baby Rudin that

If x is a positive real and y is a real, then there exists a natural number n such that nx>y.

It is then widely referenced in subsequent theorems without further explanation. I would like to verify whether this is correct or not.

In proving the existence of decimals in positive real, it was said that

Let x be a positive real and n be the greatest integer such that n is less than or equal to x.

The statement is immediately followed by "The existence of such n depends on the Archimedean property", However, this is less than or equal to instead of the greater than, which means it is not a direct application. So I wonder, is the following proof right, or can it be simplified?

  1. Denote $A=\{n\in\mathbb{Z^+}:n\geq x\}$
  2. By the Archimedean property, there exists $n_0\in\mathbb{Z^+}$ such that $n_0>x$
  3. The set is not empty
  4. Now consider the set $B=\{n-x:n\in\mathbb{Z^+}\land n\leq n_0\land n-x\geq 0\}$
  5. We know $n_0$ is finite so the set has finite members. As a result, the infimum equals minimum
  6. $x\in[\inf B-1+x,\inf B+x]$
  7. $n_0=\inf B-1+x\leq x$

Here I show there is an integer. Am I correct? The problem is not about the proof. The matter is about the degree which was ignored by Rudin. The proof applies the concept of infimum and order set. Archimedean property is just part of it.

So, is there better proof that uses the archimedean property?

  • 1
    $\begingroup$ The archimedean property insures the existence of some $n\ge 1$ with $x\in(0,n]$. Consider now the finite set $\{1,2,\dots,n\}\cup\{x\}$, ordered by the usual order of the real numbers. Pick the subset of all elements $\le x$ of it, and take the maximum. $\endgroup$
    – dan_fulea
    Jul 26, 2021 at 11:00
  • $\begingroup$ @dan_fulea Thanks, this makes a lot of sense. I believe the "finite" can hit home earlier had i written in your way in my post. $\endgroup$
    – Andes Lam
    Jul 26, 2021 at 11:02
  • $\begingroup$ @dan_fulea Excuse the naive question, but isn't proving the existence of decimals numbers in the reals extremely trivial; just take $1$, divide it by $10$ - which you're allowed to do by construction of the reals, and now you have $0.1$... $\endgroup$
    – FShrike
    Jul 26, 2021 at 11:02
  • $\begingroup$ @AndesLam Maybe you could explain? Sorry for distracting from your question, but I do not see why all the set theory is necessary $\endgroup$
    – FShrike
    Jul 26, 2021 at 11:06
  • $\begingroup$ @IdioticShrike I think you catch the drift of intuition but not rigor. Division, insofar as baby rudin is concerned at this point, which is chapter 1, is merely "defined" as the inverse of a non-zero real. It's not even defined tbh. What I am doing here is mimicking the behavior of division, which is yet defined, in a most rigorous, or primitive, fashion. $\endgroup$
    – Andes Lam
    Jul 26, 2021 at 11:06

1 Answer 1


Since an answer is still missing, I will try to fill this up:


Let $x \in \mathbf{R}^+$. By the Archimedean Property, there exists $n \in \mathbf{Z}^+$ big enough such that $n > x$. Define the nonempty (why?) set $$ A := \{ m \in \mathbf{Z}: 0 \leq m \leq n \}. $$ Note that $A$ is finite as $n$ is. Now consider the nonempty (why?) set $$ B := \{ m \in \mathbf{Z}: 0 \leq m \leq x \}. $$ Then $B \subseteq A$ since $n > x$, so $B$ is finite and bounded above. Thus we can take $n_0 = \sup(B) \equiv \max(B)$ as desired. We leave why $\sup(B) \equiv \max(B)$ to the exercise below.


For any finite set $A \subseteq F$ on an ordered field $F$ (with the least upper bound principle, if we wish), we have $$ \sup(A) = \max(A). $$


Taking supremum and the exercise is not necessary if one knows any finite set on a totally ordered field has a maximum and a minimum. See for example, here: A finite set always has a maximum and a minimum.


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