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In the question Why do students have problems with showing that something is well-defined? How can this be improved?, it was suggested that perhaps students have never seen something that is not well-defined.

I agree with this and would like to hear what you think is a good starting example for a not-well-defined object. I think a good starting example should require as little background knowledge as possible while avoiding the pitfalls of the following examples:

  • Too fixable: You could claim that "the antiderivative of a function" is not well-defined, since there are lots of them. But students will easily protest that the antiderivative can just be something with a +C on it, which seems to clear up the issue. This doesn't help anyone who doesn't already understand the idea.

  • Too obvious: You could say that "the denominator of a rational function" is not well-defined (or similar things in the link above), but the students see it immediately. This gives the wrong impression; in fact checking well-definedness carefully is critical and you can almost never see it immediately.

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    $\begingroup$ Maybe instead of antiderivative of a function, you could talk about "anti-curl" of a vector field, which will only be well defined up to a gradient of a function, a much larger class than simply the constants. $\endgroup$ – Steven Gubkin Aug 21 '14 at 2:04
  • $\begingroup$ Also the thread you link too does provide a great example: exponentiation of numbers in a cyclic group. $\endgroup$ – Steven Gubkin Aug 21 '14 at 3:35
  • $\begingroup$ I think ordinal arithmetic is one place to find such examples, but I feel this might take a bit of explanation (possibly in a direction that you don't want to bring the students). The fundamental idea of making precise "$\infty + 1$" involves (at some point) taking the size of the "smallest infinity," $|\mathbb{Z}|:=\omega$, and then beginning to extend the arithmetic operations. But ensuring this is well-defined is nontrivial, since, e.g., addition is not commutative: $1+\omega \neq \omega +1$, etc. Not sure if this is the sort of thing in which you are interested... $\endgroup$ – Benjamin Dickman Aug 21 '14 at 17:48
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    $\begingroup$ @StevenGubkin Can you move your anti-curl example to an answer, and I'll copy-paste the cyclic exponentiation as a separate answer? So far I kind of think all three of the comments here are actually answers. $\endgroup$ – Chris Cunningham Aug 21 '14 at 20:08
  • $\begingroup$ complex analysis is awash in multiply-valued "functions" for which the formula appears innocent from our real number experience. $\endgroup$ – James S. Cook Aug 22 '14 at 14:13
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The naive "addition" on rational numbers, denote it $\oplus$, where you simply add numerator with numerator and denominator with denominator.

Example: $\tfrac 1 3 \oplus \tfrac 3 4 = \tfrac 4 7$.

I think this is less obviously non-well-defined than the other examples involving fractions.

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  • $\begingroup$ This is known as the mediant fraction. It has some interesting applications in number theory. $\endgroup$ – Bill Dubuque May 15 '18 at 0:13
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Supposedly defining a ring homomorphism $f:\mathbb Z/m \to \mathbb Z/n$ by $f(k)=k$ is a popular, traditional attempt to define a map ... initially $\mathbb Z\to \mathbb Z/n$, that may be imagined to "factor through" the quotient $\mathbb Z/m$, but which does not without assumptions on $m,n$. Yet, in my experience, students look at the "formula" for it, and cannot see the difficulty, since the formula has such a straightforward appearance. :)

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The linked question has a decent example, from mweiss:

"Well-defined" only becomes a meaningful concept if you have experience with cases in which something is not well-defined. Here is a simple case in which something seems entirely reasonable: Let $m,n$ be two integers and $[m],[n]$ their equivalence classes mod $p$. Define $[m]^{[n]}=[m^n]$. Seems reasonable, especially because of the way we define the other arithmetic operations mod $p$. But as soon as you try to calculate particular examples you realize the definition is broken; different representatives of $[n]$ yield different results.

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    $\begingroup$ Since the example came from my own comment on the linked post, I've taken the liberty of editing the formatting here to make it clearer. Yes, the example was about exponentiation, not multiplication. $\endgroup$ – mweiss Aug 21 '14 at 21:08
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Here is a troubling one that requires very little background:

Given $x \in \mathbf{R}$ and $q \in \mathbf{Q}$, define $x^q$ by:

  • Write q as a fraction $\frac{a}{b}$.
  • Evaluate $\sqrt[b]{x^a}$.

Then, for example, we compute something unpleasant like

$i = \sqrt{-1} = (-1)^{1/2} = (-1)^{2/4} = \sqrt[4]{(-1)^2} = 1$

or even

$-1 = \sqrt[3]{-1} = (-1)^{1/3} = (-1)^{2/6} = \sqrt[6]{(-1)^2} = 1$

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    $\begingroup$ ah yes, $-1 = \sqrt{-1}\sqrt{-1}=\sqrt{(-1)(-1)} = \sqrt{1}=1$. Wait. What! $\endgroup$ – James S. Cook Aug 23 '14 at 3:43
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You can't get much more prototypical than Euclid. Book I, proposition 1 assumes that the two circles intersect, but there is no postulate that guarantees that they intersect. From a modern point of view, the hidden assumption is that you're working in a space like $\mathbb{R}^2$, with completeness, rather than, say, $\mathbb{Q}^2$.

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The simple example I use is an operation on rational numbers, call it $\star$ where

$$ \frac{a}{b}\star\frac{c}{d} := \frac{\max(a,c)}{\max(b,d)} $$

This then motivates the need usual proof that one needs to check that $+$ and $\cdot$ are still well defined modulo an ideal.

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