# What is the relationship between the Archimedean Property and Calculus?

My textbook begin with a chapter dedicaded to the Real Numbers. It introduces firstly integers and rational numbers, then it introduces the irrational numbers, which appeared in the first place as a necessity of expressing the length of incommesurable line segments. To my understanding, all of this helps to introduce the notion of continuity of a variable, which is put into work when treating the continuity of a variable dependent of another variable (if we restrict ourselves to single-valued functions).

The Archimedean Property is stated as follows (or as it is written in my textbook):

Given any number $c > 0$, there exists a natural number $n$, such that $n > c$.

Given any positive number $\epsilon$, there always exists a natural number $n$ such that the inequality $1/n < \epsilon$ is fulfilled.

The last statement is the first one with $c = 1/\epsilon$. Although I understand what is meant by it (I think it can readily be proved by contradiction), what I'm not getting is how the Archimedean Property fits in the preparatory material preceding that dealing with supremum, infimum and then with continuity and limits.

Does it have something to do with the concept of limit, perhaps?

Yes, the proof of this argument uses the fact that a non-empty subset bounded above in $\mathbb R$ has a supremum.

As an example using this notion try to show that $$\lim_{n \to \infty} \frac{n}{n^2 + 1} = 0$$

And also that $$0 =\inf \Big\{\frac{1}{n} ; n \in \mathbb N\Big\}$$

The fact that $\mathbb R$ is a complete ordered field, makes possible the idea of finding a number such as the limit or the infimum from the examples above.

• Feel free to ask any questions. – Aaron Maroja Feb 16 '15 at 14:33
• So if $\epsilon$ is very small, $1/\epsilon$ is very big, but we always can pick a $n \in \mathbb{N}$ such that the A.P holds, right? This enables to deal with the notion of the limit without violating the A.P itself (since $(-\infty, \infty) = \mathbb{R}$). Am I understanding this? – Jazz Feb 16 '15 at 14:55
• Yes, you're on the right track. The idea of supremum and infimum are very important to the theory of calculus, and A.P. is a tool that enables us to find certain limits. – Aaron Maroja Feb 16 '15 at 14:58
• @AaronMaroja, The Archimedean property is not valid for any ordered field: see this for example. True, that such fields are 'bigger' than $\Bbb R$, but they exist. – Silent Oct 7 '17 at 10:27
• @Silent You're right, I edited. One may consider the subset of polynomials $f/g$ suh that $g$ is nonzero and the leading coefficients of $f$ and $g$ have the same sign. It is an ordered field but any rational function $f/1$ is a bound for the constant rational functions of the form $n/1$, with $n \in \mathbb Q$. – Aaron Maroja Oct 9 '17 at 13:51

The Archimedean property states that the real numbers do not posses infinitely large number, nor does it have infinitesimals. It's a fundamental property of the reals setting it apart from what are known as nonstandard models of the reals. So the relationship between it and calculus is that classical calculus states no infinitesimals exist, and thus takes the reals of be an Archimedean field. In nonstandard analysis the reals do have infinitesimals, and thus are not an Archimedean field.

• I've been looking for more information about it and found this: Non-standard analysis. Does it mean I have to consider the A.P. as a kind of constraint that prohibit us to apply the properties used with numbers in an Archimedean Ordered Field (i.e. $\mathbb{R}$)? The axioms of ordered field applies to $\frac{\Delta y}{\Delta x}$ but not to $\frac{dy}{dx}$, for example? – Jazz Feb 16 '15 at 14:41

(I have written this before)

Proving the Archimedean principle first for ${\mathbb R}$, using the $\sup$, is in a way cheating. This principle is already present in ${\mathbb N}$ and should be proven from the Peano axioms.

Archimedean principle for ${\mathbb N}$: Given natural numbers $x\geq 1$ and $y\geq0$ there is an $n\geq1$ such that $n\>x>y$.

Proof. This is true for $y=0$ and all $x\geq1$. Assume that it is true for some $y\geq0$ and all $x\geq1$. Consider now $y+1$ and an arbitrary $x\geq1$. By the induction assumption there is an $n\geq1$ with $n\>x>y$, and putting $n':=n+1$ we have $$n'\>x=n x+x>y+x\geq y+1\ .$$

It is not difficult to extend this from ${\mathbb N}$ to ${\mathbb Q}_{>0}$ and then to ${\mathbb R}$.

The classical calculus as developed by the pioneers of mathematical analysis like Leibniz and Euler exploited numbers called infinitesimals (the term was possibly introduced by Leibniz himself). These are numbers that violate the Archimedean property when compared to $1$. Thus, adding an infinitesimal to itself finitely many times produces a number that is again an infinitesimal. Thus the number systems used by the pioneers of analysis were non-Archimedean.

At the end of the 19th century a different approach to analysis was developed by Cantor, Dedekind, Weierstrass, and others. In this approach one only uses the Archimedean complete ordered field. Some of the arguments and procedures of the pioneers of the calculus have to be reformulated in this approach using long-winded paraphrases that make the subject harder to learn.

An approach to calculus using infinitesimals is developed in Keisler's textbook.