# Does infinitesimals exists in $p$-adics?

Infinitesimals can't exist in $$\mathbb{R},$$ since it satisfy the Archimedean Property. That is, given any positive real number $$\varepsilon \gt 0$$ and any positive real number $$M\gt 0,$$ there exists a natural number $$n$$ such that $$n\varepsilon \gt M.$$
But intuitively an "infinitesimal" is supposed to be so small that no matter how many times we add it to itself, it never gets to $$1.$$

On the other hand, $$p$$-adic numbers is non-Archimedean. Does this mean that, we can somehow formulate "infinitesimals" in $$p$$-adics?

I am aware that the notion of infinitesimals may not be useful in $$p$$-adic reals as they used in traditional differential calculus. But still it is interesting to know how these things fit with each others.

• The main question is : how would you define the notion of infinitesimal in a reasonable way. If we mimic the definition in non standard analysis (using hyperreals), this would be something like that : let $(K,\vert \cdot \vert)$ be a valued field, where the absolute value takes values in a totally ordered abelian group. Then $x\in K$ is an infinitesimal if $n\vert x\vert <1$ for all $n\geq 1$. Clearly, this can't happen if the ordered group is a subgroup or real numbers with the standard order, since the reals do not have infinitesimal. Unless you have another definition in mind... – GreginGre Jun 27 at 8:47
• What exactly does it mean for the p-adic numbers to be “non-Archimedean”? The definition of “Archimedean” I know depends on having an ordered field, which AFAIK the p-adic numbers aren't. – celtschk Jun 27 at 8:47
• This may be a model theory question, I don't know. But it's not a set theory question. :-) – Asaf Karagila Jun 27 at 9:04
• @celtschk: An absolute value on a field is called non-archimedean if its restriction to the image of Z in F is bounded. This is equivalent to satisfying ultra metric inequality. Perhaps this post can tell you more. – Bumblebee Jun 27 at 9:06

As you said, intuitively, an "infinitesimal" is supposed to be so small that no matter how many times we add it to itself, it never gets to 1. Or in mathematical symbols, an infinitesimal $$x$$ satisfies $$nx<1$$ for any $$n\in\mathbb{N}$$.

The constraint with this notion is that it requires an order $$<$$ in the field. In the case of the p-adic numbers field $$\mathbb{Q_p}$$, it cannot have an order that behaves well will the field operations. In other words, there is no total order $$\leq$$ in $$\mathbb{Q_p}$$ that satisfies $$0\leq x^2$$ for all $$x\in\mathbb{Q_p}$$.

So we need a different way to compare elements in $$\mathbb{Q_p}$$: the distance of the elements of $$\mathbb{Q_p}$$ to $$0$$, in order words, we need a valuation. The p-adic valuation in $$\mathbb{Q_p}$$ is non-archimedean in the sense of valuations: it satisfies the strong triangular inequality $$|x+y|\leq\max\{|x|,|y|\}$$ for all $$x,y\in\mathbb{Q_p}$$, which is equivalent to the fact that $$|n|\leq 1$$ for all $$n\in\mathbb{N}$$.

Now if you choose $$x\in\mathbb{Q_p},\ x\neq0$$ such that $$0<|x|<|1|$$, then $$0<|nx|<|1|$$ for all $$n\in\mathbb{N}$$. Now if you think of $$|x|$$ as the distance of $$x$$ to $$0$$, then $$0<|x|<|1|$$ is a good candidate to consider as infinitesimal in $$\mathbb{Q_p}$$.

Using this valuation is enough to create a nice calculus theory over $$\mathbb{Q_p}$$. You can check the book:

Schikhof, Wilhelmus Hendricus. Ultrametric Calculus: an introduction to p-adic analysis. Vol. 4. Cambridge University Press, 2007.

• I don't know. Sure you can call the punctured unit disk of an ultrametric field "the infinitesimals", but I think you should better call it the punctured unit disk. Or the maximal ideal of the integer ring (without $0$). These are well-studied objects with well-established names, and throwing names from other theories at them is without merit unless that somehow teaches you something through analogy, which I would not see happening here. – Torsten Schoeneberg Jun 29 at 21:59

I suppose you could take an ultrapower of $$\Bbb Q_p$$ and see what kind of things lurk within. I suppose then the additive valuation on non-zero elements would take values in an ultrapower of $$\Bbb Z$$ rather than $$\Bbb Z$$. So you'd get elements $$x$$ with $$v(x)$$ a positive or negative infinite hyperinteger. I suggest the former would be "infinitesimal" and the latter "infinite".

Something else one could do is take the ultraproduct of $$\Bbb Q_p$$ over all primes $$p$$. That sounds an interesting thing to study, and I presume someone out there has studied it.

• I am not familiar enough with ultrafilter language to fully appreciate your answer. But thanks for the directions. – Bumblebee Jun 28 at 2:57

The $$p$$-adic numbers are non-Archimedean according to their absolute-value function, which is a different definition of “non-Archimedean” than the definition in ordered fields. That is, that absolute-value function is integral to the $$p$$-adic numbers being non-Archimedean. This absolute-value function defines a metric, and that metric is all we have to determine infinitesimalness, as algebrically without order, the concept of infinitesimal makes no sense (e.g. algebraically, there's no difference between $$\mathbb Q(\epsilon)$$, where $$\epsilon$$ is an infinitesimal, and $$\mathbb Q(\tau)$$ where $$\tau$$ is a transcendental real number).

Note that for reasons that will become apparent below, I will refer to the absolute-value function as defined here as “metric absolute-value function”.

Now a meaningful definition for an infinitesimal number is a number that is closer to zero than any rational number. In terms of the norm, this would mean: $$x\text{ is infinitesimal}\iff\forall q\in\mathbb Q: \lvert x\rvert < \lvert q\rvert$$ However the metric absolute-value function is real-valued (in particular, there are no infintesimal absolute values) and multiplicative, thus we have for any $$n\in\mathbb Z$$ that $$\lvert q^n\rvert=\lvert q\rvert^n$$. Therefore if $$\lvert q\rvert\ne 1$$ then the absolute value goes arbitrary close to $$0$$, and therefore there is no way one could have an infinitesimal value (since the real numbers are Archimedean).

Or in other words, a field with a metric absolute-value function can only have infinitesimals if the metric absolute-value function maps all rational values to $$1$$.

However the standard absolute-value function on the $$p$$-adic numbers does not have the property (and couldn't have it, or else it would be useless in defining the $$p$$-adic numbers in the first place). Therefore according to the definition above, there are no infinitesimal numbers in the $$p$$-adic numbers.

Now one might argue that according to this argument the hyperreal numbers should not have infinitesimals either, as they have an absolute-value function (by extension from the absolute-value function of $$\mathbb R$$) which maps every positive rational number to itself (and thus, generally not to $$1$$). But that hyperreal absolute-value function is not a metric absolute-value function, as it can give infinitesimal values (or more generally non-real hyperreal values).

Note that it may be possible to define a metric absolute-value function on the hyperreals with the property that $$\lvert a\lvert < \lvert b\rvert$$ iff $$a$$ is infinitesimal relative to $$b$$ (I don't know enough about hyperreals to say whether it is possible). In that case, the restriction to $$\mathbb Q$$ (and indeed to $$\mathbb R$$) would be constant $$1$$, and therefore the hyperreal infinitesimals would be infinitesimal also according to the definition above.

Similarly, one may be able to find a non-standard metric absolute-value function on the $$p$$-adic numbers which is $$1$$ for all rational numbers, but $$<1$$ for some non-rational $$p$$-adic numbers. In that case, one might call those numbers infinitesimal.

• Do you assume the infinitesimal given in your first sentence is invertible? – Bumblebee Jun 28 at 1:40
• Well, in my first sentence I don't mention infinitesimals (indeed, I don't mention them anywhere in the first two paragraphs). I assume you mean the definition in the third paragraph. Anyway, in a field, every non-zero element is invertible, this of course includes infinitesimals. The inverse of an infinitesimal is an infinite number. Of course, with absolute-value functions, we have $\lvert a^{-1}\rvert = \lvert a\rvert^{-1}$. Thus when infinitesimal numbers have $\lvert \epsilon\rvert < 1$, then infinite numbers have $\lvert x\rvert > 1$. – celtschk Jun 28 at 6:56