# Can't argue with success? Looking for "bad math" that "gets away with it"

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 6's in

$$\frac{64}{16}.$$

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 all cases where one arrives at valid ends by (shockingly) wrong means.

Edit: fixed typo in last example.

• @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.
– TMM
Dec 17, 2012 at 14:30
• 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. Dec 17, 2012 at 16:52
• The College Mathematics Journal used to have a section entitled "Fallacies, Flaws, and Flimflam." It regularly featured exactly these kinds of things. Dec 17, 2012 at 18:54
• Why is this on topic?
– Anko
Dec 17, 2012 at 21:11
• 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! Dec 17, 2012 at 21:17

I was quite amused when a student produced the following when cancelling a fraction:

$$\frac{x^2-y^2}{x-y}$$

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

$$\frac{x-y}{-}$$

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

• That’s downright spectacular; I don’t think that I ever had a student do something quite that ... impressive. Dec 17, 2012 at 12:09
• When I showed another student that you can cancel 6s in $16/64$ and $26/65$ to get the right answer, his response was: "so ... you can cancel digits only if they are 6?" Dec 17, 2012 at 12:12
• I’ve always been fascinated by students’ ability to come up with the wrong generalization. Dec 17, 2012 at 12:15
• @OldJohn Maybe you could have shown him $\frac{95}{19}=\frac{5}{1} = 5$ after "cancelling the $9$'s" so he could have generalized further to "cancel digits only if they are $6$, right-side-up or upside down" Dec 17, 2012 at 12:31
• @orokusaki What we are really talking about is that the correct answer IS obtained, but by a hopelessly invalid method. Dec 17, 2012 at 19:37

Here's a pretty funny one from xkcd.

• 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. Dec 18, 2012 at 13:50
• It is ridiculous how a student knows about radicals and division before learning their multiplication tables. ;)
– P.K.
Dec 19, 2012 at 9:51
• In case anyone doesn't get it, it reinterprets $3\sqrt{81}$ as $3)\overline{81}$ which is American long division notation for 81/3. He covers for failing to change the symbol by actually writing out the schoolbook division steps. Dec 22, 2012 at 0:13
• Algebraically, $x \cdot x^2 = x^4/x$. Feb 27, 2014 at 14:45

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

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

$x = 1.41$, and $y = 1.73$, which implies that $x+y = 3.14$.

The student concluded that $\sqrt{2}+\sqrt{3}=\pi$.

• Well he was approximately right! Dec 17, 2012 at 22:39
• Hmm but OP is looking for true conclusions based on a flawed process. Dec 18, 2012 at 0:30
• You are right, Alex. But it's so funny that I couldn't resist... Dec 18, 2012 at 9:10
• @alex.jordan: Well the conclusion that $\sqrt{2}+\sqrt{3}$ is irrational is indeed true. Dec 18, 2012 at 17:33
• math.stackexchange.com/questions/701822/… Sep 18, 2014 at 11:11

Here's another classical freshman calculus example:

Find $\frac{d}{dx}x^x$.

Alice says "this is like $\frac{d}{dx}x^n = nx^{n-1}$, so the answer is $x x^{x-1} = x^x$." Bob says "no, this is like $\frac{d}{dx}a^x = \log a \cdot a^x$, so the answer is $\log x \cdot x^x$." Charlie says "if you're not sure, just add the two terms, so you'll get partial credit".

The answer $\frac{d}{dx}x^x = (1 + \log x)x^x$ turns out to be correct.

• That's not wrong; that's a perfectly valid method. You get the derivative of any expression with respect to $x$ as the sums of all the derivatives with respect to the individual instances of $x$ while holding all other instances constant. Dec 18, 2012 at 13:34
• I guess this is an example of using a correct method for incorrect reasons. Dec 18, 2012 at 17:32
• @ Joriki: You just gave me a "eureka" moment with that comment (generalizing multiple instances of the same variable to multivariable calculus). Feb 12, 2013 at 16:25
• You could also "prove" the product rule like this. Oct 12, 2013 at 6:05
• This "rule" also is useful for deriving an integral in which the variable is both inside the integral and in the limits en.wikipedia.org/wiki/Leibniz_integral_rule#Formal_statement Nov 2, 2013 at 21:46

Typesetting "errors" in which exponents or multiplication signs are omitted but the resulting expression is equivalent to the original one. Examples include \begin{align} 2^5 9^2 &= 2592, \\ 3^4 425 &= 34425, \\ 31^2 325 &= 312325,\end{align} and $$2^5 \cdot \frac{25}{31} = 25 \ \frac{25}{31},$$ where a whole number followed by a fraction is interpreted as a mixed fraction (e.g., $1 \frac{1}{2} = 1 + \frac{1}{2} = \frac{3}{2}$).

That page also contains a link to your first example of "cancelling" 6s, denoted "Anomalous Cancellation", and containing three other examples with both numerator and denominator less than $100$: $$\frac{98}{49} = \frac{8}{4} = 2, \qquad \qquad \frac{95}{19} = \frac{5}{1} = 5, \qquad \qquad \frac{65}{26} = \frac{5}{2}.$$

• 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, 2012 at 15:27

This one is from Mathematical Fallacies, Flaws, and Flimflam - Edward J. Barbeau.

A student on a quiz was asked to integrate $\displaystyle \int \frac{1}{1+x}\;{dx}$. His/her answer was as follows:

\displaystyle \begin{aligned} \int \frac{1}{1+x}\;{dx} &= \int \bigg(\frac{1}{x}+\frac{1}{1}\bigg)\;{dx} \\& = \int \frac{1}{x}\;{dx}+\int\frac{1}{1}\;{dx} \\&= \log(x)+\log(1) \\&= \log(x+1)+C. \end{aligned}

• 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!). Dec 17, 2012 at 22:27
• Strange that the $C$ appears in the last line but not in the one before. As you say, it must be how the student wrote it. Dec 18, 2012 at 13:55
• @Joriki Yeah, perhaps I shouldn't have added the first $dx$ as well! Dec 18, 2012 at 19:08
• That was a test by the student to see if teacher reads the proof Nov 10, 2014 at 18:47
• Furthermore, log(a)+log(b) is not equal to log(a+b) Mar 31, 2019 at 0:04

I dont't know if this counts. But I really like it. Let $A$ be a square matrix over a field $K$ and $$\chi = \det(X \cdot \operatorname{Id} - A) \in K[X]$$ the characteristic polynomial of $A$. Then $\chi(A) = 0$, because "it's just plugging in".

• I bet you can set things up in terms of matrices over the ring $K[A]$ so that you really can just plug things in to prove the theorem.
– user14972
Dec 17, 2012 at 18:16
• Yes I think "we" can, but not the students. Most often they have difficulties to see what they are doing wrong by just plugging in. Dec 17, 2012 at 18:27
• @akkkk: What do you mean? He mentions the above "proof", but calls it wrong and encourages students to find the error. Hence I don't understand the "Unfortunately". Dec 17, 2012 at 22:15
• This is a great example which cannot be made rigorous, at least not in any simple way. For example such an argument has to take into account that $A$ does not satisfy the equation $\mathrm{tr}(A-X\cdot I)=0$. Dec 17, 2012 at 22:27
• @Marc: You can define the determinant for matrices over any commutative ring. The ring $K[A]$ is commutative, and so matrices over $K[A]$ have determinants in $K[A]$.
– user14972
Dec 18, 2012 at 16:37

Kepler's second law famously states that the radius vector from the sun to a planet will sweep out equal areas under equal times. His proof of this law included the following errors:

(1) He assumed that the velocity of the planet, as the planet traversed its orbit, was inversely proportional to the distance from the sun.

(2) Let $P_1P_2\dots P_{n+1}$ be points on an arc of the orbit of the planet, and such that the distances $|P_{i+1}-P_i|$ for $i=1,\dots,n$ are all equal to some small $\Delta s$. Let $S$ be the position of the sun, and let $r_i = |P_i -S|$ be the radial distance between the sun and the planet's position at $P_i$. Kepler then assumed that the area swept out by the radius vector from $S$, as the planet moved from $P_1$ to $P_n$, was proportional to the sum $(r_1+r_2\dots+r_n)\Delta s$.

Both these assumptions are wrong, but fortunately the effects of these errors cancel each other, and so Kepler was able to state his correct second law of planetary motion.

• Reference, please? Feb 8, 2013 at 12:51
• @WillieWong See C. H. Edwards, Jr: The Historical Development of the Calculus, p. 100. An entertaining version can also be found in Arthur Koestler: The Sleepwalkers, p. 331. Feb 8, 2013 at 13:00
• My jaw dropped to the floor when I read this one. I think it's a bit outrageous that these historical facts are not as well known as the laws themselves. After all, to the layperson, these laws are important not for their (correct) content, but as icons of the scientific method. If their derivation was as flawed as this, then this iconic status is one big lie...
– kjo
Feb 8, 2013 at 19:14
• It's possible, though, that Kepler had derived the result by some correct but circuitous method, then attempted to clean up and simplify the derivation for publication and failed. Sep 18, 2014 at 11:22
• @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 (math.stackexchange.com/questions/260656/…). Oct 31, 2015 at 17:16

I was once writing something where for stylistic reasons it made sense to change the way I wrote a vector of non-negative integers by writing $(a_1,\dotsc,a_n)$ as $1^{a_1}\dotsm n^{a_n}$, like a product, and omitting $i$ from the string if $a_i=0$, so for example $(1,0,0,3,0,0,0,1)$ would be $14^38$ (all the vectors in the actual problem had a large number of zeros, one of the reasons to change to this more concise notation). Most of the time all of the non-zero numbers were $1$s, so they all ended up looking like integers.

Naturally the first time I actually did a calculation in this notation, I wanted to remind anybody reading that the numbers had to be read back as these vectors, not as integers. Unfortunately, the first line was:

$$(0,0,0,1,1,1,0,0,0)+(1,0,0,0,1,0,0,0,1)-(1,0,0,0,1,1,0,0,0)=(0,0,0,1,1,0,0,0,1)$$

or in my notation:

$$456+159-156=459$$

• Maybe you accidentally wrote one of the sixes or nines upside-down? Dec 17, 2012 at 23:44

Here is my example: $$\lim_{n\to\infty}\frac{1+2^2+3^3+\ldots+n^n}{n^n}=\lim_{n\to\infty}\left(\frac{1}{n^n}+\frac{2^2}{n^n}+\ldots+\frac{n^{n}}{n^n}\right)=0+0+\ldots+ 1=1.$$

• What's wrong with this one? Feb 16, 2013 at 23:46
• @George Similar thing that is wrong with $\lim_{n\to\infty} \frac{n}{n} = \lim_{n\to\infty} \left( \frac{1}{n} + \frac{1}{n} + \cdots + \frac{1}{n} \right) = 0+0+ \cdots +0=0.$ Feb 17, 2013 at 14:13
• 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. Nov 2, 2013 at 21:57
• @JonasMeyer No, the correct limit is 1. you can see it by squeezing the limit between 1 and $(n^1+n^2+\ldots n^n)/n^n$. Apr 25, 2015 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. Apr 26, 2015 at 1:43

$\sin(x) = 0$

Thus we have either $x = 0$ or $\sin = 0$. A function cannot be equal to a number, therefore we must have x = 0.

I knew someone who once got as far as the first step, although in their defense I think it was just a temporary brain fart. The conclusion is correct if you're working with a restriction of the sine function to, say, $(-\pi, \pi)$.

Old John's example is gorgeous, but consider famous freshman's dream $$(a+b)^p = a^p + b^p \pmod p .$$

Various things which are true because of complex numbers, could be derived incorrect way in reals, e.g. using symbols like $\sqrt{-2}$, etc.

In probability theory there are lot of issues with dependent random variables, which can still yield correct results.

Also, check out this: \begin{align} S(a,b) &= \sum_{k=a}^{b} 2^k\\ T(a) &= \sum_{k=a}^{\infty} 2^k = 2^a + 2\sum_{k={a+1}}^{\infty} 2^{k-1} = 2^a + 2T(a)\\ T(a) &= \frac{2^a}{1-2} = -2^a \quad(\text{sum of positive elements is negative!})\\ S(a,b) &= T(a) - T(b+1) = -2^a - (-2^{b+1}) = 2^{b+1}-2^a \end{align} and this:

\begin{align} \sum_{k=0}^{n} 2^k &= \frac{2^{n+1}-1}{1-2} \\ \frac{d}{d2}\sum_{k=0}^{n} 2^k &= \frac{d}{d2}\frac{2^{n+1}-1}{1-2}\quad(\text{differentiate over two!})\\ \sum_{k=0}^{n} k2^{k-1} &= \frac{1-(n+1)2^n+n2^{n+1}}{(1-2)^2} \end{align}

Cheers!

• For the last, you could first replace $2$ by $x$, differentiate, and replace $x$ by $2$ again, so how is that "wrong"? And for the "sum of positive elements is negative", this also happens in e.g. the $2$-adic numbers, where $1 + 2 + 4 + \ldots = -1$.
– TMM
Dec 17, 2012 at 13:42
• @TMM This is how those equations got created ;-) Still, for most people it is not only strange, but wrong - we work with "normal" numbers, not 2-adic, and then the presented series does not converge at all, why should it yield a correct result? Dec 17, 2012 at 13:54
• @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, 2012 at 15:47
• These are not very good examples, because they actually can be (and are being) used reliably. (As opposed to cancelling digits, say.) Dec 17, 2012 at 22:22
• It should also be pointed out the Freshman's dream example could be false if $n$ is not prime. For example, $$(1 + 3)^4 = 256 \equiv 0 \pmod{4}$$ but $$1^4 + 3^4 = 82 \equiv 2 \pmod{4}.$$ The freshman's dream (including the url to which you linked) deals only with the case of prime characteristic. Feb 7, 2013 at 6:07

Earlier, I asked my friend to simplify $$\dfrac{\cos^2 (73°) + \cos^2(17°)}{\cos^2(63°) + \cos^2(27°)}$$. Here is his work:

$$\frac{\cos^2(73°) + \cos^2(17°)}{\cos^2(63°) + \cos^2(27°)} = \frac{\cos^2(73° + 17°)}{\cos^2(63° + 27°)} = \frac{\cos^2{(90°)}}{\cos^2(90°)} = \frac{\cos^2}{\cos^2} = 1$$

If $G$ is a group and $K,N$ are normal in $G$ with $K \subseteq N$ then $$G/N \cong (G/K)\large/\normalsize(N/K)$$ which is obviously true by just cancelling the terms on the rhs.

• Can this be made rigorous somehow? Oct 31, 2015 at 17:44
• @YoTengoUnLCD It's the 3rd isomorphism theorem. There are lots of proofs around. Nov 2, 2015 at 23:57

A classical example due to Euler, I believe:

Notice that the roots of $\sin(x)$ are precisely the numbers $k \pi$ where $k$ is any integer. But the same is true of the product

$$x \left(1 - \frac{x^2}{\pi^2 1^2}\right) \left(1 - \frac{x^2}{\pi^2 2^2}\right)\ldots$$

so the two must be equal. The coefficient of $x^3$ in the product is $-\frac{1}{\pi^2} \sum_{n=1}^\infty \frac{1}{n^2}$, and the coefficient of $x^3$ in the Taylor series of $\sin(x)$ is simply $-\frac{1}{6}$. Therefore,

$$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$

If part of your brain is tempting you to think that this argument might be right after all, note that if you apply exactly the same reasoning to the function $\sin(\pi x)$ then you get the value $\frac{\pi^3}{6}$. Nevertheless, this is so eerie that I can't help but wonder if there's something to it...

• This is a good one, although it is a case of lack of proper rigor and justification (which hardly existed at the time) rather than simply invalid nonsense as in many of the other answers. There is definitely something to it. I do not understand your point at the end about $\sin(\pi x)$. Dec 28, 2012 at 7:21
• @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. Dec 28, 2012 at 16:52
• That is not true. The product is valid on the entire complex plane, converging to $\sin$ uniformly on compact subsets. I now see your point about $\sin(\pi x)$; I misread your exponent, which is off because the correct product for $\sin(\pi x)$ has $\pi x$ as the first term rather than just $x$. You are correct that the method of argument for why the product is valid is incorrect (basically there isn't any here), and therefore easily leads to other incorrect products like your implied one for $\sin(\pi x)$. Alternatively, note that $e^x\sin(x)$ also has the same zeros. Dec 28, 2012 at 17:00
• Really? This is going to lead to an afternoon spent with a complex analysis book... Dec 28, 2012 at 17:02
• 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: ams.org/bookstore/pspdf/gsm-97-prev.pdf Dec 28, 2012 at 17:04

My "favorite" error with complex numbers is $$\frac{i}{i}=\frac{\sqrt[2\,\,]{-1}}{\sqrt[2]{-1}}=\sqrt[2\,\,]{\frac{-1}{-1}}=\sqrt[2\,\,]{1}=1.$$

• Whats wrong with that? The answer could have been achieved directly, apart from that, I don't actually see anything wrong?
Dec 20, 2012 at 1:11
• @ADP: $\dfrac{\sqrt{a}}{\sqrt{b}}=\sqrt{\dfrac{a}{b}}$ and $\sqrt{a}\sqrt{b}=\sqrt{ab}$ are identities that hold when $a$ and $b$ are positive and $\sqrt{x}$ is defined to be the positive square root of a positive number $x$. When square roots are extended to negative (or even complex) numbers, first one has to be careful about which square root, and secondly one has to give up hope of such identities always holding. E.g., consider the classic nonsense line $-1=\sqrt{-1}\sqrt{-1}=\sqrt{(-1)(-1)}=\sqrt{1}=1$. Related: math.stackexchange.com/questions/84436. Dec 20, 2012 at 3:53
• @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?
Dec 20, 2012 at 4:02

Slightly contrived:

Given $n = \frac{2}{15}$ and $x=\arccos(\frac{3}{5})$, find $\frac{\sin(x)}{n}$.

$$\frac{\sin(x)}{n} = \mathrm{si}(x) = \mathrm{si}x = \boxed{6}$$

Given that $$\sum_{k=1}^{n} k = \frac{\frac{n}{2}\cdot\frac{n+1}{2}}{\frac{1}{2}},$$ we have

$$\sum_{k=1}^{n} \sin(k)=\frac{\sin(\frac{n}{2})\cdot\sin(\frac{n+1}{2})}{\sin(\frac{1}{2})}.$$

Proof: take the sine of everything.

• I am not familiar with this sine identity, but if it is true, this false proof is truly astonishing.
– kjo
May 10, 2020 at 21:13
• It is true. One (proper) proof involves summing the imaginary part of $e^{ik}$, $1\leq k\leq n$, where the former identity is in fact used. Or you can use induction and angle-addition. But yeah, I prefer the brazen, "take the sine of everything, that's how math works." May 10, 2020 at 21:17

One example from me:

$$\sqrt{5 \frac{5}{24}} = 5 \sqrt{\frac{5}{24}}$$ $$\sqrt{12 \frac{12}{143}} = 12 \sqrt{\frac{12}{143}}$$

• My first reaction was that this should have been generalized to, say, any positive $x \neq 1$, but then I realized that in the general form, the $+$ on the LHS is no longer implicit, $$\sqrt{x + \frac{x}{x^2 - 1}} = x\sqrt{\frac{x}{x^2 - 1}}\;,$$ which makes the maneuver less amusing somehow.
– kjo
Dec 17, 2012 at 18:05
• There's no such thing as implicit +. That's multiplication, and because of that, wrong. Dec 18, 2012 at 1:28
• @yi_H I would argue that a mixed number has an implicit +. Dec 18, 2012 at 3:57
• an implicit +? maybe in elementary school...? Dec 18, 2012 at 5:38
• probably... that would make it ambigous.. that's why you never use it to denote addition. No operator means multiplication. Dec 19, 2012 at 1:13

I would like to add a nice "method" for differentiating $$1/x$$:

$$\require{cancel}\frac{\mathrm d}{\mathrm d{x}} \frac{1}{x} = \frac{\mathrm d}{\mathrm d}\frac{1}{\mathrm {x^2}} = \frac{\cancel{d}}{\cancel{d}}\frac{1}{\mathrm {x^2}} = - \frac{1}{x^2}.$$

• The way the minus signed is arrived at shows attention to detail. Classy.
– kjo
May 1, 2020 at 22:37

When I asked my student to get rid of irrationality in the denominator of fraction $$\frac{1}{\sqrt[3]{3}+\sqrt[3]{5}}$$ He gave an immediate solution $$\frac{1}{3^{1/3}+5^{1/3}}$$ What can I say, no roots no irrationalities :-)

I must confess, this example doesn't fit in the original citeria.

• The denominator is still irrational. Nov 2, 2013 at 22:06
• I'm not sure this really qualifies for this question. The question is about bad mathematical steps that lead to the correct answer. May 10, 2015 at 13:45

A classical example.

Cayley Hamilton Theorem states that for any matrix $A \in Mat(\mathbb R)_{n \times n}$, $A$ is a root of its characteristic polynomial $P_A(t)=det(A-tI)$.

Proof:

$P_A(A)=det(A-AI)=det(A-A)=det(0)=0$

By linearity

$$\log(1+2+3) = \log(1)+\log(2)+\log(3).$$

(just don't try to generalize this)

• $\log\left(1+1+2+4\right)=\log\left(1\right)+\log\left(1\right)+\log\left(2\right)+\log\left(4\right)$ and $\log\left(1+1+1+2+5\right)=\log\left(1\right)+\log\left(1\right)+\log\left(1\right)+\log\left(2\right)+\log\left(5\right)$ Apr 27, 2018 at 11:29

You all probably, no doubt, have seen the proof to the question: Is Hell Endo or Exothermic.

This one always makes me laugh...

Dr. Schambaugh, of the University of Oklahoma School of Chemical Engineering, Final Exam question for May of 1997. Dr. Schambaugh is known for asking questions such as, "why do airplanes fly?" on his final exams. His one and only final exam question in May 1997 for his Momentum, Heat and Mass Transfer II class was: "Is hell exothermic or endothermic? Support your answer with proof."

Most of the students wrote proofs of their beliefs using Boyle's Law or some variant. One student, however, wrote the following:

"First, We postulate that if souls exist, then they must have some mass. If they do, then a mole of souls can also have a mass. So, at what rate are souls moving into hell and at what rate are souls leaving? I think we can safely assume that once a soul gets to hell, it will not leave.

Therefore, no souls are leaving. As for souls entering hell, let's look at the different religions that exist in the world today. Some of these religions state that if you are not a member of their religion, then you will go to hell. Since there are more than one of these religions and people do not belong to more than one religion, we can project that all people and souls go to hell. With birth and death rates as they are, we can expect the number of souls in hell to increase exponentially.

Now, we look at the rate of change in volume in hell. Boyle's Law states that in order for the temperature and pressure in hell to stay the same, the ratio of the mass of souls and volume needs to stay constant. Two options exist:

If hell is expanding at a slower rate than the rate at which souls enter hell, then the temperature and pressure in hell will increase until all hell breaks loose. If hell is expanding at a rate faster than the increase of souls in hell, then the temperature and pressure will drop until hell freezes over. So which is it? If we accept the quote given to me by Theresa Manyan during Freshman year, "that it will be a cold night in hell before I sleep with you" and take into account the fact that I still have NOT succeeded in having sexual relations with her, then Option 2 cannot be true...Thus, hell is exothermic."

The student, Tim Graham, got the only A.

• Although this is a classic, I do not think this is a good answer for this question (wrong method, correct answer), or this site in general (focused on mathematics).
– TMM
Dec 18, 2012 at 13:21
• I hate to be that guy (no pun intended) but that whole post is a work of fiction, see e.g. this. It's also, of course, off topic :)
– guy
Dec 18, 2012 at 16:43
• And this whole thread is off topic from this entire site. A whole thread of incorrect proofs, please...
Dec 19, 2012 at 2:42

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 $K$.

• @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. Feb 17, 2013 at 14:11
• @RagibZaman "connect up the u and v via a path through the Vx balls" What if K is not convex? Oct 18, 2013 at 23:44

There is a lot of treatment with little rigor around the first infinitesimal calculations. For example consider a circle of radius $r$, we can approximate it's area by filling it with equal triangles.

The area of ​​each triangle is given(approximately) by $lr/2$, where $l$ is the base of the triangle. If we have $n$ triangles, the sum of their areas is given by $nlr/2$. By increasing the number of triangles to infinity, the sum of the bases of the triangles approximates the length of the circumference, i.e, if $n\rightarrow\infty$ then $nl\rightarrow2\pi r$. Thus, the sum of the areas of the triangles tends to $2\pi r^2/2 = \pi r^2$, which is the area of the circle.

The result is correct but we calculates $\infty\cdot 0$ with almost no rigor at all.

• 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". Oct 18, 2013 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. Apr 19, 2015 at 20:46

This happened to me last week with a quiz I gave to my Algebra 2 class doing radical equations.

Solve $\sqrt{-5x+35}+7=x$

WRONG $$\sqrt{-5x+35}=x-7$$ $$-5x+35=x^2-49$$ $$0=x^2+5x-84=0$$ $$0=(x+12)(x-7)$$ Thus $x=-12, 7$. Checking for extraneous solutions yields the only solution as $x=7$

RIGHT $$\sqrt{-5x+35}=x-7$$ $$-5x+35=x^2-14x+49$$ $$0=x^2-9x+14$$ $$0=(x-2)(x-7)$$ Thus $x=2,7$. Checking for extraneous solutions yields the only solution as $x=7$

I assumed most of them were cheating.... I was wrong!

• 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? Feb 19, 2014 at 11:19
• A couple of early quizzes I graded had the first two lines, or variations of them, but didn't finish the step by step method and just circled $x=7$ as the answer. It is a basic Algebra 2 class, so I definitely have students who do this often. But after a while, I noticed that most students made that error above. The students I asked afterward all said the same thing; I thought when you squared (x-7) you got $x^2-49$. Feb 19, 2014 at 12:01

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

Another one I came across was a function composition problem.

Let $g(x)=x^2.$ Find $(g\circ g)(x)$.

Well it should be $$(g\circ g)(x)=g(g(x))=g(x^2)=(x^2)^2=x^{2\cdot{2}}=x^4$$ But of course I should have caught that my students would do the "natural thing" and say $$(g\circ g)(x)=g(x)\cdot g(x)=x^2\cdot x^2=x^{2+2}=x^4$$

I blame myself for not catching this one, but.....

• A rare instance where the ambiguity of the notation $g^2(x)$ wouldn't matter.
– Dan
Aug 19, 2022 at 22:57

Does this count? It can be shown that following the steps will give the correct answer, but the steps themselves are sometimes questionable. Let $y=(x-1)^3(x-2)^5(x-3)^7$. Find $\dfrac{dy}{dx}$.

Take the log of both sides. We get $$\log y=3\log(x-1)+5\log(x-2)+7\log(x-3).$$ Thus $$\frac{1}{y}\frac{dy}{dx}=\frac{3}{x-1}+\frac{5}{x-2}+\frac{7}{x-3},$$ and therefore $$\frac{dy}{dx}=3(x-1)^2(x-2)^5(x-3)^7 + 5(x-1)^3(x-2)^4(x-3)^7+7(x-1)^3(x-2)^5(x-3)^6.$$ Simple, generalizes, true for all $x$, including many $x$ at which log is not defined.

Remark: One can find many examples in Euler: formal manipulations that lead to the correct answer through in principle unjustified steps. About this, Euler wrote something like "Sometimes my pencil is more clever than I am."

• That's a standard method to prove polynomial identities by proving them for enough special values. Dec 17, 2012 at 22:40
• What's wrong with that? They taught us this in secondary! Dec 17, 2012 at 22:40
• en.wikipedia.org/wiki/Logarithmic_differentiation Dec 17, 2012 at 22:57
• @Nur For negatives you can multiply both sides by $-$ and do exactly the same computations, the minuses cancel. Why the result holds at $x=1,2,3$ is a little more subtle in general, in this case it is easy to argue because everything is polynomial...Formally, when we apply logarithmic differentiation to $y=f(x)$ we should actually apply it to $sy=s f(x)$ where $s(x)$ is the sign of $f(x)$... Dec 20, 2012 at 17:19
• The argument can be completed by defining logs of negative numbers. You can can take the log as set valued (defined up to addition of multiples of $2i\pi$), define a derivative, and the argument will work for all complex numbers other than -1, -2, or -3. Or you can take a branch cut away from a point you are looking at to choose a particular complex number as the log, and the equation will hold up to a difference of a constant number of 2\pi is. Oct 7, 2018 at 16:08