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One example would be "cancelling" the 6s in


Another one would be something like


Yet another one would be



Edit: fixed typo in last example.

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32 Answers 32






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enter image description here

Here's a pretty funny one from xkcd.

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Maybe this could be explained? I had to Google for what it meant because I have never learned to write down a division like that. – sam hocevar Dec 18 '12 at 13:50
Algebraically, $x \cdot x^2 = x^4/x$. – Antonio Vargas Feb 27 '14 at 14:45

A student in a test was asked to give an example of two irrational numbers whose sum is irrational.

He choose $x = \sqrt{2}$, and $y=\sqrt{3}$, and computed the sum $x+y$ using a calculator. Unfortunately, he only took two digits, which led to the following:



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@Jim Ah, I see! – alex.jordan Dec 18 '12 at 18:20

Here's another classical freshman calculus example:




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From MathWorld / Printer's Errors:



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Thanks for the pointer. That page also lists all the "all proper solutions up to 3-digit denominators" (where "solutions" are fractions amenable to "anomalous cancellation"), but for some reason it does not include fractions like 101/202. It also does not mention that the four examples with 2-digit denominators are particularly productive: 2666666/6666665 = 2/5, 499999/999998 = 1/2, etc. – kjo Dec 17 '12 at 15:27




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Actually the first $dx$ is missing in the book. I'm not sure whether the solution of the student was as such or whether it's a typo in the book. It's probably how the student wrote it. Anyway, that book contains many interesting examples of this, as well as non-trivial faults with solutions that arrive preposterous conclusions! (The one above this one for example arrives at the conclusion that every derivative is continuous!). – NeverBeenHere Dec 17 '12 at 22:27
@Joriki Yeah, perhaps I shouldn't have added the first $dx$ as well! – NeverBeenHere Dec 18 '12 at 19:08
That was a test by the student to see if teacher reads the proof – nicolas Nov 10 '14 at 18:47




or in my notation:


Amusing, but fairly unhelpful!

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$\sin(x) = 0$



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What's wrong with this one? – Student Feb 16 '13 at 23:46
To spell out the problem, you can't interchange a limit with a sum when the number of terms in the sum involves the same variable that you're taking the limit with respect to. – Keshav Srinivasan Nov 2 '13 at 21:57
@JonasMeyer No, the correct limit is 1. you can see it by squeezing the limit between 1 and $(n^1+n^2+\ldots n^n)/n^n$. – Artem Apr 25 '15 at 23:03
Oh thanks @Artem, sorry for not checking better; I was misreading because of the other thread and taking all the exponents as $n$, so $1^n + 2^n +\cdots +n^n$ in the numerator instead of $1^1+2^2+\cdots+n^n$. Thanks for pointing out a good way to find this limit, too. I'll delete my wrong comments. – Jonas Meyer Apr 26 '15 at 1:43







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@dtldarek, I think the first equation involving $T(a)$ should be $$T(a) = \sum_{k=a}^{\infty}2^k = 2^a + 2 \sum_{k=a+1}^{\infty}2^{k-1} = 2^a + 2T(a).$$ – kjo Dec 17 '12 at 15:47
@kjo Yeah, thanks! – dtldarek Dec 17 '12 at 16:12


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Can this be made rigorous somehow? – YoTengoUnLCD Oct 31 '15 at 17:44
@YoTengoUnLCD It's the 3rd isomorphism theorem. There are lots of proofs around. – Jeremy Nov 2 '15 at 23:57





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@DanielMcLaury, I think it's more accurate to say that Kepler observed the result, rather than deriving it, as I commented to kjo a minute ago (…). – Vectornaut Oct 31 '15 at 17:16



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Whats wrong with that? The answer could have been achieved directly, apart from that, I don't actually see anything wrong? – ADP Dec 20 '12 at 1:11
@Jonas Meyer, thanks. I didn't realize the power laws were constrained like that, to me they were just identities. Doesn't that then undermine the proof for Euler's Formula, where $e^{-ix}$ is divided by $e^{-ix}$ (once the derivative is proved to be constant), yielding the desired identity? – ADP Dec 20 '12 at 4:02

A classical example due to Euler, I believe:






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@Jonas Meyer: The "product formula" at the beginning is complete baloney; I know of nothing even close to it which is correct. If you apply the same argument to $\sin(\pi x)$, then the product formula involves terms of the form $(1 - \frac{x^2}{n^2})$ and so the coefficient of $x^3$ is $-\sum \frac{1}{n^2}$. On the other hand the coefficient of $x^3$ in the Taylor series of $\sin(\pi x)$ is $-\pi^3/6$. Actually, it seems that if you apply the argument to $\sin(ax)$ for an appropriately chosen value of $a$, you get get any value for the sum that you want. – Paul Siegel Dec 28 '12 at 16:52
I know I've seen this in Conway's Functions of one complex variable if you happen to have that handy. Here's something just found by Googling: – Jonas Meyer Dec 28 '12 at 17:04

Slightly contrived:



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One example from me:


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I'm not sure this really qualifies for this question. The question is about bad mathematical steps that lead to the correct answer. – Thomas Andrews May 10 '15 at 13:45


This one always makes me laugh...







The student, Tim Graham, got the only A.

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You assumed they were cheating because they all used the same wrong method, so you thought they copied? How do you know you were wrong? – Daniel Fischer Feb 19 '14 at 11:19

Another one I came across was a function composition problem.




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A very common mistake in analysis.

Exercise: Let's $K\subset\mathbb{R}^{n}$compact and a function $f:K\to \mathbb{R}$ locally Lipschitz, i.e. for all $x$ in the compact $K$ there is an open set $V_x$ containing $x$ and a constant $L_x$ such that $$ \left|\,f(u)-f(v)\,\right|<L_x \|u-v\|, \quad \forall u,v\in V_x $$ Proof that $f$ is too Lipschitz in all $K$.

"Proof:" Let $\{V_x\}_{x\in K}$ the open cover of $K$ where each $V_x$ is as in the hipotesis of exercise. As $K$ is compact there is a finite pen cover $\left\{V_{x_{_1}},\dots, V_{x_{_N}}\right\}$. Then for all $u,v\in K$ we have $$ \left| f(u)-f(v) \right|<\max\left\{L_{x_{_1}},\dots,L_{x_{_N}}\right\}\cdot \|u-v\|, \quad \forall u,v\in K. $$

So just $L=\max\left\{L_{x_{_1}},\dots,L_{x_{_N}}\right\}$ to a constant that is valid for all $K$. Then $f$ is too Lipschitz in all $K$.

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What is the problem with this? – Student Feb 17 '13 at 0:02
@George At least two things wrong I can see are that this proof seems to assume that $K$ is connected, and also the $u,v$ went from being in each $V_x$ to magically being able to be picked from $K$. To fix the second issue, we need to connect up the $u$ and $v$ via a path through the $V_x$ balls. For the first one, prove the theorem assuming $K$ is connected, and then show being Lipschitz on all connected components of $K$ implies the result. – Ragib Zaman Feb 17 '13 at 14:11
@RagibZaman "connect up the u and v via a path through the Vx balls" What if K is not convex? – Jisang Yoo Oct 18 '13 at 23:44


infinitesimal to circle area



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There is something about this kind of arguments that I always found awkward. I mean some of us have seen that picture that goes like "here's a right triangle with sides of length 1, 1, and square root of 2. we draw a staircase that approximates the hypotenuse. therefore square root of 2 is 2". – Jisang Yoo Oct 18 '13 at 22:57
@JisangYoo There is something fundamentally different between lengths and areas. If shape A contains shape B, we know that area(A)>area(B), but we do not always have that perimeter(A)>perimeter(B). I think that's the basic idea behind why nonrigorous arguments about area, like the OP's, work, while the same sort of reasoning, like the one you mentioned, doesn't. – Akiva Weinberger Apr 19 '15 at 20:46

A classical example.




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Certainly it is familiar. However, it is (for functions defined on the reals) formally incorrect, since for example at $x=0.5$ we are using the logarithm of three negative numbers. – André Nicolas Dec 17 '12 at 23:01

Recently seen:


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If you think about the meaning, $\cdot$ does really "distribute over" |. – Utku Alhan Mar 18 '14 at 18:19
@UtkuAlhan This is not distribution, the left side contains $c$ while the right has $c^2$. I'm not saying you cannot make it formal, after all, the result is true. I'm just saying that this particular derivation (without any additional comment) is wrong. – dtldarek Mar 18 '14 at 18:24
Yes, distribution was the wrong word to use. You're right. Thanks for the correction. This particular derivation is wrong of course: for starters, what does (statement 1) = (statement 2) mean? – Utku Alhan Mar 18 '14 at 18:32

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $$\large \lim_{x \to 0^{+}}{8 \over x} = \infty\quad\imp\quad\lim_{x \to 0^{+}}{3 \over x} =\omega $$

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this really did happened to my friends in 9th grade :D – ADG Apr 25 '15 at 14:28



because you'll get back a bunch of


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My answer is : Divide and rule!!


Now using L'Hospital rule!


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protected by Eric Naslund Dec 18 '12 at 17:55


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