Undefined vs Indeterminate in Mathematics

In mathematics, term undefined may characterize an object in two circumstances: a pedestrian situation in which an object has not been defined - perhaps, as yet; and a situation in which an object can't in principle be defined meaningfully. Impossibility of "division by zero" is a manifest example of the latter case. For, say, a real or complex $a\ne 0,$ there is no fruitful way to define $a/0.$ In most number and algebraic systems, zero does not and cannot have a multiplicative inverse, such that $a/0$ is left undefined. (Occasionally, when there is a single "infinite" element $\infty$ as it happens in real or complex projective geometry, one defines $a/0=\infty.)$

For $a=0,$ $a/0$ falls into a different category; $0/0$ is referred to as indeterminate, which means that, depending on the circumstances, the expression $0/0$ may be defined, or may be left undefined as a matter of experience.

An aside: quite often the limit $\displaystyle\lim_{x\rightarrow 0}\frac{f(x)}{g(x)},$ where $\displaystyle\lim_{x\rightarrow 0}f(x)=\lim_{x\rightarrow 0}g(x)=0$ appears in this context. I find this confusing and only accept $0/0$ as a tag that helps distinguish this case from $\displaystyle\lim_{x\rightarrow 0}\frac{f(x)}{g(x)},$ where $\displaystyle\lim_{x\rightarrow 0}f(x)=\lim_{x\rightarrow 0}g(x)=\infty$ which is then described as $\infty /\infty.$ Limits of functions are in the purview of calculus, where limits like the above are calculated by special techniques, e.g., factoring or L'Hôlpital's rule, never by direct substitution.

A classical case of an indeterminate expression is "zero to the power of zero," $0^0.$ If looked at as a limit value $\displaystyle\lim_{x\rightarrow 0}x^{0}$ it ought to be declared $1.$ On the other hand, as the limit value $\displaystyle\lim_{x\rightarrow 0}0^{x}$ $0$ appears more suitable. Function of two variables $f(x,y)=x^y$ is identically $1$ on the positive $x$-axis and is identically $0$ on the positive $y$-axis. So, in calculus, the expression $0^0$ is mostly left undefined (but may be used to tag the class of limits $\displaystyle\lim_{x\rightarrow 0}f(x)^{g(x)},$ where $\displaystyle\lim_{x\rightarrow 0}f(x)=\lim_{x\rightarrow 0}g(x)=0.$

In combinatorics, on the other hand, it was found convenient to set $0^{0}=1$ because, for example, this allows expressions like $\displaystyle (x+y)^{n}=\sum_{k=0}^{n}x^{k}y^{n-k}{n \choose k}$ without making exceptions for $k=0$ or $k=n.$ Also, $m^n$ counts the number of ways to map a set of $n$ elements into a set of $m$ elements. If $n=0,$ so that the first set is empty, there is only one way to map it anywhere, even to "another" empty set. (As a reasoning goes, there is only one way to do nothing.) Thus, it is commonly accepted that "an empty product" is equal to $1.$

By analogy, an empty sum should be defined as $0$ but, in principle, is an indeterminate object. If so, what could be said about the average of the elements of an empty set: is it undefined or indeterminate?

Grégoire Nicollier (University of Applied Sciences of Western Switzerland) has kindly communicated the following observation:

You may want to figure out by yourself what pulled Grégoire's attention. To be on sure side I'll refresh below a few properties of the average.

Given a set $X=\{x_{1},x_{2},\ldots,x_{n}\}$ of $n$ elements (a better term would be multiset because $X$ may have repeated terms), the average of the set $X$ is defined as $M(X)=\displaystyle\frac{1}{n}\sum_{i=1}^{n}x_{i}.$

Let $Y=\{y_{1},y_{2},\ldots,y_{m}\}$ be a set with $m$-elements. Then, taking the union of two multisets,

$M(X\cup Y)=\displaystyle\frac{nM(X)+mM(Y)}{n+m}.$

Indeed, $nM(X)=\displaystyle\sum_{i=1}^{n}x_{i}$ and $nM(Y)=\displaystyle\sum_{i=1}^{m}y_{i},$ implying that $nM(X)+mM(Y)$ is the sum of all the elements in $X\cup Y.$ The number of such elements is $n+m,$ and the formula follows.

If $n=m$ we obtain an apparent simplification:

$M(X\cup Y)=\displaystyle\frac{nM(X)+nM(Y)}{n+n}=\frac{M(X)+M(Y)}{2}.$

The notable point here is that, for $X$ and $Y$ of equal size,

(1)

$M(X\cup Y)=\displaystyle\frac{M(X)+M(Y)}{2},$

independent of actual size of the two sets.

Another worthy property of the average is that, if every element of a set has shifted by the same amount, the average is shifted by the same amount:

(2)

$\displaystyle\frac{1}{n}\sum_{i=1}^{n}(x_{i}+a)=M(X+a)=M(X)+a.$

Now, I'll proceed with Grégoire Nicollier's observation. As never actually happens in real life or meant in statistical reports, when a population is said to have doubled, there bound to be a margin of error, which suggests that the quote should not be taken literally, for the face value. No doubt, however, Grégoire's intention was for an exact, abstract identity.

Let $P=\{p_{1},p_{2},\ldots,p_{n}\}$ be the set of ages of all the inhabitants of the city in question. Let $S=\{s_{1},s_{2},\ldots,s_{n}\}$ be the set of all ages of the student population at the local university. The quote from Håkan Nesser together with (1) show that

$M(P\cup S)=\displaystyle\frac{M(P)+M(S)}{2}=\frac{1}{2}M(P),$

from which $M(S)=0.$ With a further assumption that we deal with ages expressed with whole numbers, $M(S)=0$ means that the set of students is empty, its size $n=0$ and, so, $P$ is also empty - the city is a ghost city with no inhabitants, and the local university is an abandoned institution, with no students, let alone tenured staff.

So the two sets $P$ and $S$ are empty, their size is $0,$ and the sum of the ages of their elements is also $0.$ We also found that $M(P)=M(S)=0.$ Thus, arguably,

$0=M(P)=\displaystyle\frac{\displaystyle\sum_{p\in P}p}{0}=\frac{0}{0}.$

Which appears to assert that the average of the elements of an empty set is $0.$ Grégoire Nicollier came up with an objections based on the identity (2). Since all empty sets are he same, $\emptyset +a=\emptyset,$ which, if the average of its elements were to be defined, would imply $M(\emptyset)=M(\emptyset+a)=M(\emptyset)+a,$ for any number $a,$ which is absurd, with the conclusion that the average of the empty set should be declared undefined not just indeterminate. I tend to agree, especially because, I see no useful purpose in declaring otherwise. Still, I am amazed that under the assumptions (concerning the ages and accuracy of population estimates), the quote has more than just one unintended consequences.

So, is the derivation above of $\displaystyle 0=\frac{0}{0}$ meaningful or not. It is questionable in the very least because (1) has been obtained under an implicit assumption that $n\ne 0.$ Employing (1) with the empty sets is akin to getting $1=2$ from $1-1=2-2$ which is a known fallacy, usually better disguised

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