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Over on mathoverflow, there is a popular CW question titled: Examples of common false beliefs in mathematics. I thought it would be nice to have a parallel question on this site to serve as a reference for false beliefs within less obscure mathematics. That said, it would be good not get bogged down with misconceptions that are generally assumed to be elementary such as: $(x + y)^{2} = x^{2} + y^{2}$.

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I dont really see a need for this question. :/ I think the one on Mathoverflow is welcoming of examples in other areas of math rather than "obscure". I think that it is a cool question, but I really dont think we need this duplicate. – BBischof Oct 26 '10 at 4:34… this is another instance. – anonymous Oct 26 '10 at 4:36
Voted to close. This is a dup, as per the question itself. – Mariano Suárez-Alvarez Oct 26 '10 at 5:05
Specifically, the MO discussion focused on mistakes that were (a) conceptual, and (b) known to be made by mathematicians (especially, mistakes the answerers had made). This restriction prevented trivial responses. I suggest revising the question to be "false beliefs YOU -- a presumably mathematically capable math.SE user -- have held" as a separate matter from "review every error that students make!". If the latter is interesting it would be better to explore it in another thread. – T.. Oct 26 '10 at 15:22
@T.. Title has been edited as per your suggestion. – Ami Oct 26 '10 at 15:34

13 Answers 13

Many well-educated people believe that a p-value is the probability that a study conclusion is wrong. For example, they believe that if you get a 0.05 p-value, there's a 95% chance that your conclusion is correct. In fact there may be less than a 50% chance that the conclusion is correct, depending on the context. Read more here.

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I recently caught myself thinking that the formula for the determinant of a 2-by-2 matrix also works for a block matrix, i.e. $\det (A B; C D) = \det(A)\det(D) - \det(B)\det(C)$.

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It works if e.g. $D$ and $C$ commute. – Plop Oct 27 '10 at 12:55

Every torsion-free Abelian group is free.

(This only holds for finitely generated Abelian groups.)

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The counterexample which is obvious in hindsight: Q. – Qiaochu Yuan Oct 26 '10 at 15:49
@Qiaochu Yuan: Yes, my problem when I had this false belief was, that I only knew Q, the ring---not Q, the Abelian group. ;) – Rasmus Oct 26 '10 at 17:53
What about Q the omnipotent being? – OghmaOsiris Jul 26 '11 at 21:38
@OghmaOsiris: please try to only include relevant comments... – Mariano Suárez-Alvarez Jul 28 '11 at 22:35
@Mariano: please try to lighten up. – OghmaOsiris Jul 30 '11 at 21:33

I have seen this one time too many $$\frac{a}{b}+\frac{c}{d}=\frac{a+c}{b+d}$$

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Interestingly, the fraction (a+c)/(b+d) is useful. Of course it's not the sum of a/b and c/d, but it is part of constructing the Farey sequence and is important in finding rational approximations. – John D. Cook Oct 27 '10 at 11:36
@John D. Cook: Agree, I did not know this. I found the following – AD. Oct 27 '10 at 12:33
Somewhat related with Simpson's paradox's_paradox – leonbloy Sep 3 '11 at 17:13

To generalize a few of the answers, for pretty much any function, someone somewhere will make the mistake of treating it as if it is linear in all of its variables. Thus we get: $e^a + e^b = e^{a+b}$, $\sqrt{a + b} = \sqrt{a} + \sqrt{b}$, $a/(b+c) = a/b + a/c$, ...

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The question I've heard on many levels (including the grad level): what is the square root of $a^2$? And everyone says: it's $a$!

In fact it is $|a|$.

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It seems they mistook your question for "What is *a* square root of $a^2$?" – Rahul Oct 27 '10 at 1:12
er... in $\mathbb{R}$ I suppose, but surely not in $\mathbb{C}$. – J. M. Oct 27 '10 at 1:14
I've recently seen it in two answers to a question here. :( – egreg Dec 25 '14 at 21:40

These are 2 instances which i have seen to happen with my friends. If $A$ and $B$ are 2 matrices, then they believe that $(A+B)^{2}=A^{2}+ 2 \cdot A \cdot B +B^{2}$.

Another mistake is if one i asked to solve this equation, $ \displaystyle\frac{\sqrt{x}}{2}=-1$, people generally square both the sides and do get $x$ as $4$.

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I had posted this on MO, and i am posting this here as well as this appears elementary. – anonymous Oct 26 '10 at 4:35
More generally, people simplify X to Y by saying "if X, then Y". They forget that they also need "if Y, then X". In other words, "Sqrt[x] = -2" implies "x=4" (vacuously), but the converse is false. I blame our telling students "you can do anything to both sides of an equation, as long as it's the same thing". That's true, but we should add "most of the time, you want to make sure you can undo it too". – barrycarter Apr 5 '11 at 5:28
Actually "you can do anything to both sides of an equation, as long as it's the same thing" is exactly right for deriving new equations from old ones. I think the problem is probably that most students don't realise that solving an equation is the exact opposite procedure. – gfes Jun 6 '11 at 2:37

Both my students and some of my colleagues (!) believe that the graph of a function cannot cross a horizontal asymptote. Obviously this implies that they misunderstand the definition of an asymptote. More worryingly (in my eyes), it also seems to imply that they don't understand why we even care about asymptotes.

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So $y=\exp(-x)\cos(x)$ is asymptotic to the horizontal axis, yes? – J. M. Oct 28 '10 at 12:25

I have seen this many times:

$$a^2 + a^3 = a^5$$

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it depends on what the meaning of + is.... – futurebird Oct 28 '10 at 12:16
The meaning of a little don's comment depends on the meaning of all the words he typed. – PyRulez Dec 19 '13 at 23:13

Recently, a friend of mine pointed out the following to me:

The open unit disc $D\subset\mathbb{C}$ is not biholomorphic to all of $\mathbb{C}$.

Indeed they are diffeomorphic, but we can easily see that they are not biholomorphic since if there was a biholomorphism $\phi:\mathbb{C}\rightarrow D$, then consider the function $f:D\rightarrow\mathbb{C}$ given by $f(z)=z$. It is obviously holomorphic and non-constant. But then $f\circ\phi$ would be holomorphic, non-constant and bounded, which is a contradiction to Liouville's theorem.

(In fact, the same argument holds to prove that there is no surjective holomorphic function from the whole of $\mathbb{C}$ to any bounded domain.)

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how can one think of this as true? it's obviously false. apart from that it is certainly not a common error – Alexander Grothendieck Dec 22 '13 at 18:57
@Wicht Maybe it is not a common belief, but I think that many people would say that $D$ and $\mathbb{C}$ are biholomorphic if asked, at least at first glance, because they are used to think to the open disk and the plane as "the same thing". – Daniel Robert-Nicoud Dec 22 '13 at 19:46
maybe those not aware of Liouville's theorem.... – Alexander Grothendieck Dec 22 '13 at 21:51

The null factor law is as follows: $$ab = 0 \Rightarrow (a = 0) \vee (b = 0).$$ This law applies for real numbers, as well as polynomials which is where the law is most commonly envoked. I have seen far too many instances of the following incorrect generalisation: $$ab = c \Rightarrow (a = c)\vee(b = c).$$

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An error that I often see with my students, and that I made myself when I was a student :

Let $f$ be a function on $\mathbb R\mapsto \mathbb R$ and $f'$ its derivative.

Then if $\lim_{x\rightarrow\infty} f'(x)=0$ then $f$ is bounded ($\lim_{x\rightarrow\infty} f(x)<\infty$).

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I've seen this one more explicitly with infinite sums - if $\lim_{n \to \infty} a_n = 0$, then obviously $\sum_{n=0}^\infty a_n$ converges. – Sten Jun 18 '14 at 13:43
@Sten, yes, it is the same mistake. – Xoff Jun 18 '14 at 19:07
I'm not sure whether to think this is a 'silly' mistake to make or not. I would think it's clear why someone would think this might be true, but on the other hand it isn't hard to find a counter example. $\ln(x)$ would do! – FireGarden Jun 18 '14 at 23:00
@FireGarden, as many little mistakes, it's not a real problem if done only once. – Xoff Jun 19 '14 at 15:25
  • If $V$ is a finite dimensional normed vector space and $A$ is convex compact, then there is a unique $y\in A$ minimizing the distance from $0$ to $A$ (uniqueness doesn't need to hold, for example, for $\mathbb{R}^n$ with the max-norm).
  • If $\exp(A) \exp(B)=\exp(A+B)$, then $A$ and $B$ commute (doesn't hold for some matrices)
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