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Many believe (I think rightly so) that the presentation of counterexamples should play an important role in the teaching upper level mathematics courses such as real analysis and topology. Counterexamples show why the hypotheses of various theorems are important. Further, counterexamples very often add to students' intuition and ability to quickly recognize false propositions.

Some examples of counterexamples that could be provided in a first year calculus course include:

  • Removing or varying hypotheses in L'Hopital's Rule,
  • Removing or varying hypotheses in Intermediate Value Theorem
  • Converse of Differentiability implies continuity.

Should counterexamples play an important role in a first year calculus course taken largely by non-majors?

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I think it depends on your goals with your calculus class. If your goal is primarily to prepare students to apply calculus in other subjects, then perhaps counterexamples are not so important. And if your goal is to "filter", then perhaps you want to hit them with difficult stuff for no particular reason, including counterexamples.

But there are other goals that you might have. For instance, at my university, among the goals of calculus is to teach logical thinking and the meaning of things like definitions and theorems in mathematics. For this purpose, counterexamples (within reason) are essential. I do not discuss things that I consider "pathologies", such as bounded non-integrable functions or nowhere differentiable functions, but simple counterexamples showing that the converse of a theorem may not hold are necessary. This isn't the same as "rigor", e.g. we don't use the $\epsilon\delta$ definition of limits — rather we're trying to convey an idea of the logical structure of mathematics, which must be understood prior to any attempt at rigor.

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    $\begingroup$ When I first saw "And if your goal is to "filter", then perhaps you want to hit them with difficult stuff for no particular reason, including counterexamples." I was like, well that's a bit unfair there are many reasons ... then proceed to read second paragraph ... then proceed to upvote $\endgroup$
    – WetlabStudent
    May 6, 2014 at 5:24
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I tend to think that the plusses and wonderful power of calculus to address natural, simple questions about geometry and mechanics should be communicated before students are taught to worry. After all, not counting the classical Greek protocalculus, from Fermat and Descarte, and Newton and Leibniz, and Bernoullis and Euler, up to the 19th century, mathematicians thought calculus was wonderful because of what it could do, not because or despite pathologies or worries or counter-examples might come up. (Cf. Bishop Berkeley.)

As it is, many universities' use of basic calculus is as a filter, with no intellectual purpose whatsoever. This is already undignified enough, and subliminally understood by many students. Bringing up pathologies or dangers, historically inaccurate anyway, will most often just seem like adding insult to injury. More crazy nonsense.

Being positive about what calculus can do I think is much better than using it as testbed for "rigor"... which no one would care about, anyway, if calculus could not do amazing things.

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    $\begingroup$ I disagree with the premise that counter-examples are for later, but your answer is so well written I'm doubting myself. $\endgroup$
    – Kevin O'Bryant
    Mar 14, 2014 at 0:29
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    $\begingroup$ I agree with the answer with respect to pathological counter-examples, but not basic counterexamples like "Is every continuous function differentiable?" $\endgroup$
    – Steven Gubkin
    Dec 10, 2015 at 22:01
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    $\begingroup$ @StevenGubkin, it still would seem to me that for calculus students it might be too negative to emphasize that continuity does not imply differentiability... which, after all, was a difficult thing for many people in the 19th century to get straight. And, from the other side, in practice, most of those concerns are misplaced, since, if necessary, continuous functions on reasonable spaces are locally $L^1$ so give distributions, so can be differentiated distributionally. No problem. What starts to matter more is just the interpretation of the outcome, not whether it's possible... $\endgroup$
    – paul garrett
    Dec 10, 2015 at 22:21
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    $\begingroup$ @paulgarrett I guess we can agree to disagree, but if we are even going to teach them words like "continuity" and "differentiable", I think they should be able to understand the implication in one direction, and have counterexamples in the other. It is just part of the low level logical exploration needed to learn the language properly. $\endgroup$
    – Steven Gubkin
    Dec 10, 2015 at 22:34
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This question should be "community wiki", if there is such a thing.

I think counterexamples need to be used to make the "obvious" statements of calculus meaningful. For example, the Intermediate Value Theorem says that $f(x)=x^2-2$ has a root between 0 and 2, but in fact this is only true if you believe that irrational numbers exist. This example motivates proving that $\sqrt{2}$ is irrational, and it also affords an opportunity to tell the students that their instinct (the IVT is obvious) is good, but that the IVT is saying something subtle about real numbers.

Students also find Rolle's theorem obvious, but examples like $g(x)=|x|^{|x|}$ (with the removable discontinuity at $x=0$ removed) demonstrate that maximums don't always go along with $g'(x)=0$.

Another favorite type of counterexample are the polynomials that are always positive but don't achieve their minimum. I'll leave that one to the reader.

Another example that can be used early, in my opinion, so that it doesn't seem so difficulty later, is to remove the singularity of $h(x)=e^{-1/x^2}$. Then show that every derivative of $h$ is 0 at $x=0$. This requires the limit definition of derivative, and (later) is an example of a function that has a Taylor Series that converges for all $x$, but doesn't converge to the function it comes from.

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    $\begingroup$ Huh? If a polynomial is always positive on $\mathbb{R}$, it is even; and in that case it has a minimum somewhere. $\endgroup$
    – vonbrand
    Mar 14, 2014 at 13:00
  • $\begingroup$ @vonbrand: A positive polynomial on $\mathbb{R}^2$, on the other hand, need not achieve it's minimum. $\endgroup$
    – Jason DeVito - on hiatus
    Mar 15, 2014 at 3:28
  • $\begingroup$ An example, please? Slicing along an avid gives an one-variable polynomial that is positive, so attains is minimum. $\endgroup$
    – vonbrand
    Mar 16, 2014 at 1:47
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    $\begingroup$ As a sum of two squares, $p(x,y) = x^2 y^2 + (1-xy^2)^2$ can never be negative, and thinking about how to make both summands 0 at the same time shows it can't be 0. But $p(\epsilon^2, 1/\epsilon) = \epsilon^2 \to 0$. $\endgroup$
    – Kevin O'Bryant
    Mar 17, 2014 at 4:42
  • $\begingroup$ Replace "polynomial" with rational function, then it works for $\mathbb{R}$. $\endgroup$
    – kjetil b halvorsen
    Feb 1, 2016 at 16:14

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