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: $(x + y)^{2} = x^{2} + y^{2}$.
False beliefs in mathematics (conceptual errors made despite, or because of, mathematical education)
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19$\begingroup$ 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. $\endgroup$– BBischofCommented Oct 26, 2010 at 4:34
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$\begingroup$ math.stackexchange.com/questions/6848/… this is another instance. $\endgroup$– anonymousCommented Oct 26, 2010 at 4:36
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5$\begingroup$ Voted to close. This is a dup, as per the question itself. $\endgroup$– Mariano Suárez-ÁlvarezCommented Oct 26, 2010 at 5:05
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4$\begingroup$ 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. $\endgroup$– T..Commented Oct 26, 2010 at 15:22
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1$\begingroup$ @T.. Title has been edited as per your suggestion. $\endgroup$– AmiCommented Oct 26, 2010 at 15:34
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15 Answers
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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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3$\begingroup$ It works if e.g. $D$ and $C$ commute. $\endgroup$– PlopCommented Oct 27, 2010 at 12:55
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Every torsion-free Abelian group is free.
(This only holds for finitely generated Abelian groups.)
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3$\begingroup$ The counterexample which is obvious in hindsight: Q. $\endgroup$ Commented Oct 26, 2010 at 15:49
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$\begingroup$ @Qiaochu Yuan: Yes, my problem when I had this false belief was, that I only knew Q, the ring---not Q, the Abelian group. ;) $\endgroup$– RasmusCommented Oct 26, 2010 at 17:53
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1$\begingroup$ What about Q the omnipotent being? $\endgroup$ Commented Jul 26, 2011 at 21:38
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4$\begingroup$ @OghmaOsiris: please try to only include relevant comments... $\endgroup$ Commented Jul 28, 2011 at 22:35
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7$\begingroup$ @Mariano: please try to lighten up. $\endgroup$ Commented Jul 30, 2011 at 21:33
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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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3$\begingroup$ It seems they mistook your question for "What is *a* square root of $a^2$?" $\endgroup$– user856Commented Oct 27, 2010 at 1:12
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11$\begingroup$ er... in $\mathbb{R}$ I suppose, but surely not in $\mathbb{C}$. $\endgroup$ Commented Oct 27, 2010 at 1:14
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$\begingroup$ I've recently seen it in two answers to a question here. :( $\endgroup$– egregCommented Dec 25, 2014 at 21:40
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$\begingroup$ That's a bad question. $a^2$ has two square roots, $a$ and $-a$ $\endgroup$ Commented Oct 30, 2019 at 11:06
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I have seen this one time too many $$\frac{a}{b}+\frac{c}{d}=\frac{a+c}{b+d}$$
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6$\begingroup$ 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. $\endgroup$ Commented Oct 27, 2010 at 11:36
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$\begingroup$ @John D. Cook: Agree, I did not know this. I found the following en.wikipedia.org/wiki/Mediant_%28mathematics%29 $\endgroup$ Commented Oct 27, 2010 at 12:33
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1$\begingroup$ Somewhat related with Simpson's paradox en.wikipedia.org/wiki/Simpson's_paradox $\endgroup$– leonbloyCommented Sep 3, 2011 at 17:13
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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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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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$\begingroup$ I had posted this on MO, and i am posting this here as well as this appears elementary. $\endgroup$– anonymousCommented Oct 26, 2010 at 4:35
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2$\begingroup$ 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". $\endgroup$– user2469Commented Apr 5, 2011 at 5:28
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3$\begingroup$ 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. $\endgroup$– gfesCommented Jun 6, 2011 at 2:37
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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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$\begingroup$ So $y=\exp(-x)\cos(x)$ is asymptotic to the horizontal axis, yes? $\endgroup$ Commented Oct 28, 2010 at 12:25
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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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1$\begingroup$ how can one think of this as true? it's obviously false. apart from that it is certainly not a common error $\endgroup$ Commented Dec 22, 2013 at 18:57
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3$\begingroup$ @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". $\endgroup$ Commented Dec 22, 2013 at 19:46
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$\begingroup$ maybe those not aware of Liouville's theorem.... $\endgroup$ Commented Dec 22, 2013 at 21:51
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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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I have seen this many times:
$$a^2 + a^3 = a^5$$
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12$\begingroup$ it depends on what the meaning of + is.... $\endgroup$ Commented Oct 28, 2010 at 12:16
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2$\begingroup$ The meaning of
a little don's comment depends on the meaning of all the words he typed. $\endgroup$ Commented Dec 19, 2013 at 23:13
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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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$\begingroup$ 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. $\endgroup$– StenCommented Jun 18, 2014 at 13:43
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$\begingroup$ @Sten, yes, it is the same mistake. $\endgroup$– XoffCommented Jun 18, 2014 at 19:07
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1$\begingroup$ 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! $\endgroup$ Commented Jun 18, 2014 at 23:00
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$\begingroup$ @FireGarden, as many little mistakes, it's not a real problem if done only once. $\endgroup$– XoffCommented Jun 19, 2014 at 15:25
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- 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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Many students struggle to understand why the dream of Freshmen is true in some cases. More precisely, they just cannot accept in a commutative ring of characteristic $p$, the fomula $$ (x+y)^p=x^p+y^p $$ is true. Probably many people believe such an equlity is false for their whole life, because it is false in $\mathbb{R}$.
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A question on this site asks for examples of normed vector spaces, an answer says something to the effect:
The field of rationals , seen as a vector space over itself, together with the norm inherited from $\mathbb{R}$, forms an incomplete normed space.
This is wrong because the definition of a normed vector space requires a vector space over $\mathbb{R}$ or $\mathbb{C}$ from the outset. Besides, the vector space of $\mathbb{Q}$ over $\mathbb{Q}$ is one-dimensional, but finite-dimensional normed spaces are complete.