# Example of not well-defined object

I am a teaching assistant for a first year linear algebra course for mathematics and physics students and I think that it is not so clear for my students what "well-defined" means. Therefore, I would like to dedicate a part of the next session to giving them an idea of what this notion is.

From my own experience as a first year student, I remember that somehow seeing examples of functions or operations that are not well-defined was much more helpful to understand what this concept really means.

So my question is: What are cool (not too complicated) example of definitions of mathematical objects that are not well-defined ? The examples can relate to both analysis or linear algebra but need to be quite simple. I would also be interested in not well-defined constructions that are not necessarily related to functions directly (along the lines of my last example).

I've come up with a few example but I was wondering if there were better ones :). Here are the example I have come up with.

• Under what condition is the "identity function" $$\varphi : \mathbb{Z}/n\mathbb{Z} \rightarrow \mathbb{Z}/m \mathbb{Z}, [x] \mapsto [x]$$ well-defined?
• Is the function $$f: \mathbb{Q} \rightarrow \mathbb{Z}, \frac{a}{b} \mapsto a + b$$ well-defined ?
• Why do we care about associativity in groups/rings/fields ?
• So I guess like you said, there are functions that depends on some choice, for instance, if the function depended on choosing a representative in an equivalence class it could be likely that the function would be evaluated differently for different choice of representatives. Another example would be complex logarithm. Oct 25 '20 at 9:32
• You might want to take a look at Timothy Gowers' blogpost gowers.wordpress.com/2009/06/08/… Oct 25 '20 at 11:29

I think your examples suffer because they depend on notational conventions. AFter all, why do we denote a single number by both $$\frac12$$ and $$\frac24$$? Because rationals are defined by equivalence classes, but that's hidden in our familiarity with the notation. The same goes for modular arithmetic.

I suggest the following.

1. Let's break the integers into three piles, Z, U, T, where $$Z$$ consists of all multiples of three, $$U$$ consists of $${\ldots, -5, -2, 1, 4, 7, \ldots}$$ (i.e., multiples of three, plus one) and $$T$$ is "multiples of three, plus two"). Define addition as follows:

$$A \oplus B$$ for any two of $$Z, U, T$$, is defined by taking an element of $$A$$, an element of $$B$$, and adding them in $$\Bbb Z$$. The result is in one of the three sets, and we defined the "sum" of $$A$$ and $$B$$ to be that third set.

[work out at least one example; show there's an identity for addition, namely $$Z$$].

1. Let's break the integers into three piles Let's break the integers into three piles, $$Z, E, I,$$ where $$𝑍$$ consists of just zero, $$E$$ contains all even numbers except $$0$$, and $$I$$ contains all odd numbers. Define addition as follows: $$A \oplus B$$ for any two of $$Z, E, I$$ is defined by taking an element of 𝐴, an element of 𝐵, and adding them in ℤ. The result is in one of the three sets, and we define the "sum" of 𝐴 and 𝐵 to be that third set.

Once again do an example (show that $$A \oplus B = A$$, for instance, or that $$Z \oplus I = I$$), show that $$Z$$ is an additive identity element, and then show...that it's not well-defined, because $$E+E$$, if you pick elements $$2$$ and $$-2$$, would add up to $$Z$$, but if you pick $$2$$ and $$4$$, it would add up to $$E$$.

This shows exactly where the fault in the definition lies ("The result is in one of the three sets" is true, but which of the three depends on which choices you made...)

It might also help to do this with multiplication rather than addition, because item-by-item multiplication (in Bbb Z) (in the first example) of $$Z$$ and $$Z$$ does not end up equal to $$Z$$, but instead ends up a subset of $$Z$$ (and similarly for the other products). I managed to misunderstand this (in a different context) for some time.