I'm looking for cases of invalid math operations producing (in spite of it all) correct results (aka "every math teacher's nightmare").

One example would be "cancelling" the 6s in


Another one would be something like

$$\frac{9}{2} - \frac{25}{10} = \frac{9 - 25}{2 - 10} = \frac{-16}{-8} = 2 \;\;.$$

Yet another one would be

$$x^1 - 1^0 = (x - 1)^{(1 - 0)} = x - 1\;\;.$$

Note that I am specifically not interested in mathematical fallacies (aka spurious proofs). Such fallacies produce shockingly wrong ends by (seemingly) valid means, whereas what I am looking for are cases where one arrives at valid ends by (shockingly) wrong means.

Edit: fixed typo in last example.

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    $\begingroup$ @rschwieb: I don't understand your comment at all. The OP is asking for examples where, by wrong means, you end up with a correct result, and not where by "seemingly" valid means, you end up with a wrong result. It does not matter how you interpret "seemingly"; any "proof" that ends up with a wrong conclusion should not be posted as an answer here. That's the OP's choice. $\endgroup$ – TMM Dec 17 '12 at 14:30
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    $\begingroup$ This is reminding me of an anecdote about a physicist from the early 20th century with a reputation for making arithmetic errors who as a joke intentionally made a huge order of magnitude error (10^10???) in a published paper; and then published a correction the next month noting that the error didn't affect the results of the computation. Unfortunately I'm failing to Google it so I can't see if what he did would be relevant to this question or not. $\endgroup$ – Dan Neely Dec 17 '12 at 16:52
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    $\begingroup$ The College Mathematics Journal used to have a section entitled "Fallacies, Flaws, and Flimflam." It regularly featured exactly these kinds of things. $\endgroup$ – Mike Spivey Dec 17 '12 at 18:54
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    $\begingroup$ Why is this on topic? $\endgroup$ – Anko Dec 17 '12 at 21:11
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    $\begingroup$ There is a whole book about this "Mathematical Fallacies, Flaws, and Flimflam". This book is exactly what you are looking for. It has scores of such examples in a whole bunch of categories like algebra, calculus, multi-variable calculus, and so on. Love this book! $\endgroup$ – Fixed Point Dec 17 '12 at 21:17

37 Answers 37


Question is : Find $\lim_{x \to 1}\left(\frac{x}{x-1}-\frac{1}{\ln(x)}\right)$

My answer is : Divide and rule!!

$\lim_{x \to 1}\left(\frac{x}{x-1}-\frac{1}{ln(x)}\right) = \lim_{x \to 1}\left(\frac{x}{x-1}\right)-\lim_{x \to 1}\left(\frac{1}{ln(x)}\right)$

Now using L'Hospital rule!

$\lim_{x \to 1}\left(\frac{x}{x-1}-\frac{1}{ln(x)}\right) = \lim_{x \to 1}\left(\frac{x}{x-1}\right)-\lim_{x \to 1}\left(\frac{1}{ln(x)}\right)=\frac{1}{1}-0=1$


When I was in kinder-garden I was sick one day so I missed class and the next day the teacher gave us a quiz on basic multiplication. But I didn't know what multiplication was so when I got the test paper I thought that she wrote the + signs really badly. I didn't want to make her feel bad by pointing it out (not that my writing was any good either).

So I took the test. She graded it with me and explained that multiplication was something different so I said that I must have missed every question. Then she went through and checked every problem. Except it turned out that I got one problem right. 2x2 which happens to equal 2+2. If 0x0 was on there I would have gotten 2 problems right :)

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    $\begingroup$ … yawn. But, – you did multiplication in kindergarten? $\endgroup$ – k.stm Jun 30 '16 at 15:56

Consider: $$\int_{-\infty}^{+\infty}\dfrac{e^{2x}-e^x}{x(1+e^{2x})(1+e^x)}dx =ln2$$ This integral has been worked out here: Finding value of integration with infinite limit But it is rather difficult because the linear $x$ term in the denominator. So I am deciding to omit $x$ and instead do $$\int_{-\infty}^{+\infty}\dfrac{e^{2x}-e^x}{(1+e^{2x})(1+e^x)}dx ?$$ This one is LOTS easier and the answer is still $ln2$. Works for me...

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    $\begingroup$ This is a general principle since it works also for $$\int_0^{\infty}\frac x{x^3+1}dx=\int_0^{\infty}\frac 1{x^3+1}dx=\frac{2\pi\sqrt{3}}9$$ $\endgroup$ – zwim Sep 27 '17 at 1:18
  • $Lim_{x\rightarrow\infty}x.e^{-x}=\infty ×0=0$
  • $Lim_{x\rightarrow\infty}log(x+1)-log(x)=\infty -\infty=0$

Well, using $\frac{dy}{dx}$ as a ratio is also one such example. It carries no meaning at all to use $\frac{dy}{dx}$ as ratio but it works each time to get you to the right result instead of appreciating the limit process behind it. I think it is as much abuse of notation as is eliminating 6's from a fraction, it is just that in this case the notation is designed such that each time you abuse it, you will get the right answer but you won't know actually why if you just rely on the notation. For example: $\frac{dy}{dx}=\frac{dy}{dt} \frac{dt}{dx}$ due to chain rule and not due to treating $\frac{dy}{dx}$ as a fraction just like $\frac{x^2-y^2}{x-y}=x+y$ by algebra and not by


"cancelling" the $x$ and the $y$ on top and bottom, to get:


and then conclude that "two negatives make a positive", so the final answer is be $x+y$.

Now if you see a student prove $\frac{5^2-3^2}{5-3}=8$ by the wrong method, will he be correct as this method of proof will hold for all $x,y \in R$ until $x \neq y$.

Also in three dimensions you have stuff like $ \frac{\partial x}{\partial y}\cdot\frac{\partial y}{\partial z}\cdot\frac{\partial z}{\partial x}=-1. $

Also, $dy$ isn't even defined , let alone be manipulated like a real number.

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    $\begingroup$ Depending on your perspective, often such use is not simply (shockingly) wrong, just lacking in proper justification and rigor. If it works each time, there is something to it. $\endgroup$ – Jonas Meyer Nov 30 '13 at 10:13
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    $\begingroup$ There are multiple definitions of $\mathrm dy\ldots$ $\endgroup$ – Mark S. Dec 10 '13 at 0:11
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    $\begingroup$ -1, what about non-standard analysis? $\endgroup$ – JMCF125 Apr 8 '14 at 10:20

I notice that most of the answers are cute and humorous, so I may incur the wrath of the forum by posting an answer that is neither cute nor humorous.

I would have to say that most informal proofs are not valid, but there are formal proofs that come to the same conclusion, so the conclusion is correct, but informal proofs tend to be incomplete and lack rigor.

Cantor's diagonal argument is a good example. How long will it take to get a seventh grader to understand something like this, http://us.metamath.org/mpegif/ruc.html ? I don't even know if that is right.

Yet, most seventh graders will understand vsauce's 2 minute explanation of the theory. But it is incomplete. https://www.youtube.com/watch?v=s86-Z-CbaHA&feature=youtu.be&t=4m35s

Vsauce is correct when he says (paraphrased) "If there is a 1 to 1 correspondence between the two, then we can match one whole number to each real number, then the sets are the same size".

The hidden assumption is that we must use the whole numbers in order of magnitude. Which is not true.

So he is wrong when he says "We have used up every single whole number, the entire infinity of them, and yet we can still come up with more real numbers." The truth is that he did not need to "use up every single whole number".

To avoid this, all he had to do is match a subset of the whole numbers with his purported list of all real numbers. Like this

$4 \to r_1$

$44 \to r_2$

$444 \to r_3$

$4444 \to r_4$

and so on.

Then vsauce can generate another infinite list of real numbers that were not on his first list and again we can match them to a subset of the whole numbers like this

$14 \to rr_1$

$144 \to rr_2$

$1444 \to rr_3$

$14444 \to rr_4$

and so on.

And we can repeat this a countable infinite number of times, while vsauce can repeat his operation an uncountable infinite number of times.

But that is what vsauce needs to prove, and he can only do it with the formal proof, and that will take a very long and involved video.


The problem : solve the equation $$e^{(x-2)}+e^{(x+8)}=e^{(4-x)}+e^{(3x+2)}$$

Incorrect method

As everybody knows $$e^a+e^b=e^{ab}$$

So let's apply it to solve the equation :

$\begin{array}{lll} e^{(x-2)}+e^{(x+8)}=e^{(4-x)}+e^{(3x+2)} & \iff & e^{(x-2)(x+8)}=e^{(4-x)(3x+2)} \\ & \iff & (x-2)(x+8)=(4-x)(3x+2) \\ & \iff & x^2+6x-16=10x-3x^2+8 \\ & \iff & 4x^2-4x-24=0 \\ & \iff & x^2-x-6=0 \\ & \iff & (x+2)(x-3)=0 \\ \end{array}$

Thus $x=-2$ or $x=3$.

Correct method:

Let's set $X=e^x$ and $a=e^2$ and substitute

$\frac Xa+a^4X=\frac {a^2}X+aX^3\iff X^2+a^5X^2=a^3+a^2X^4\iff(a^2X^2-1)(X^2-a^3)=0$

Leading also to $x=-2$ and $x=3$.

The problem : simplify the expression $\sin(x+y)\sin(x-y)$

Incorrect method

As everybody knows $$f(a+b)f(a-b)=f(a)^2-f(b)^2$$

Applying the formula gives immediately


Correct method

$\begin{array}{lll} \sin(x+y)\sin(x−y) &=& \big[\sin(x)\cos(y)+\cos(x)\sin(y)\big]\ \big[\sin(x)\cos(y)−\cos(x)\sin(y)\big]\\ &=&\sin^2(x)\cos^2(y)−\cos^2(x)\sin^2(y)\\ &=&\sin^2(x)(1−\sin^2(y))−(1−\sin^2(x))\sin^2(y)\\ \require{cancel}&=&\sin^2(x)−\cancel{\sin^2(x)\sin^2(y)}−\sin^2(y)+\cancel{\sin^2(x)\sin^2(y)}\\ &=&\sin^2(x)−\sin^2(y)\\ \end{array}$


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

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