# 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 6s 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 are cases where one arrives at valid ends by (shockingly) wrong means.

Edit: fixed typo in last example.

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@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 '12 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. – Dan Neely Dec 17 '12 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. – Mike Spivey Dec 17 '12 at 18:54
Why is this on topic? – Anko Dec 17 '12 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! – Fixed Point Dec 17 '12 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 $x+y$.

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That’s downright spectacular; I don’t think that I ever had a student do something quite that ... impressive. – Brian M. Scott Dec 17 '12 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?" – Old John Dec 17 '12 at 12:12
I’ve always been fascinated by students’ ability to come up with the wrong generalization. – Brian M. Scott Dec 17 '12 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" – Dilip Sarwate Dec 17 '12 at 12:31
@orokusaki What we are really talking about is that the correct answer IS obtained, but by a hopelessly invalid method. – Old John Dec 17 '12 at 19:37

Here's a pretty funny one from xkcd.

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legendary example. – narayanpatra Dec 18 '12 at 12:34
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
It is ridiculous how a student knows about radicals and division before learning their multiplication tables. ;) – Parth Kohli Dec 19 '12 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. – Random832 Dec 22 '12 at 0:13
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:

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

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

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Well he was approximately right! – fluffy Dec 17 '12 at 22:39
Hmm but OP is looking for true conclusions based on a flawed process. – alex.jordan Dec 18 '12 at 0:30
You are right, Alex. But it's so funny that I couldn't resist... – the L Dec 18 '12 at 9:10
@alex.jordan: Well the conclusion that $\sqrt{2}+\sqrt{3}$ is irrational is indeed true. – Grumpy Parsnip Dec 18 '12 at 17:33

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.

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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. – joriki Dec 18 '12 at 13:34
I guess this is an example of using a correct method for incorrect reasons. – Grumpy Parsnip Dec 18 '12 at 17:32
@ Joriki: You just gave me a "eureka" moment with that comment (generalizing multiple instances of the same variable to multivariable calculus). – Joe Z. Feb 12 '13 at 16:25
Jokes on everyone that up-voted this. – Squirtle Mar 19 '13 at 22:30
You could also "prove" the product rule like this. – Baby Dragon Oct 12 '13 at 6:05

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 $100$: $$\frac{98}{49} = \frac{8}{4} = 2, \qquad \qquad \frac{95}{19} = \frac{5}{1} = 5, \qquad \qquad \frac{65}{26} = \frac{5}{2}.$$

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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

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}

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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. – joriki Dec 18 '12 at 13:55
This is beautiful. – Alexander Gruber Dec 18 '12 at 18:37
That was a test by the student to see if teacher reads the proof – nicolas Nov 10 '14 at 18:47

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:

$$456+159-156=459$$

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Maybe you accidentally wrote one of the sixes or nines upside-down? – Joe Z. Dec 17 '12 at 23:44

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".

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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. – Hurkyl Dec 17 '12 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. – Hans Giebenrath Dec 17 '12 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". – InvisiblePanda Dec 17 '12 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 $\mathrm{tr}(A-X\cdot I)=0$. – Morgan Sherman Dec 17 '12 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 $K[A]$. – Hurkyl Dec 18 '12 at 16:37

$\sin(x) = 0$

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, $(-\pi, \pi)$.

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

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@George Similar thing that is wrong with $\lim_{n\to\infty} \frac{n}{n} = \lim_{n\to\infty} \left( \frac{1}{n} + \frac{1}{n} + \cdots + \frac{1}{n} \right) = 0+0+ \cdots +0=0.$ – Ragib Zaman Feb 17 '13 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. – Keshav Srinivasan Nov 2 '13 at 21:57

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!

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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 '12 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? – dtldarek Dec 17 '12 at 13:54
These are not very good examples, because they actually can be (and are being) used reliably. (As opposed to cancelling digits, say.) – Phira Dec 17 '12 at 22:22

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.

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I always used that trick to remember this theorem. – Integral May 24 '13 at 22:08

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.

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Reference, please? – Willie Wong Feb 8 '13 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. – Per Manne Feb 8 '13 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 '13 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. – Daniel McLaury Sep 18 '14 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/…). – Vectornaut Oct 31 '15 at 17:16

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

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

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

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@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. – Jonas Meyer Dec 20 '12 at 3:53

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...

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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 $\sin(\pi x)$. – Jonas Meyer Dec 28 '12 at 7:21
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. – Jonas Meyer Dec 28 '12 at 17:00
Really? This is going to lead to an afternoon spent with a complex analysis book... – Paul Siegel Dec 28 '12 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 – Jonas Meyer Dec 28 '12 at 17:04

Slightly contrived:

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

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

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One example from me:

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

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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 '12 at 18:05
There's no such thing as implicit +. That's multiplication, and because of that, wrong. – Karoly Horvath Dec 18 '12 at 1:28
@yi_H I would argue that a mixed number has an implicit +. – Code-Guru Dec 18 '12 at 3:57
an implicit +? maybe in elementary school...? – Jonathan Dec 18 '12 at 5:38
probably... that would make it ambigous.. that's why you never use it to denote addition. No operator means multiplication. – Karoly Horvath Dec 19 '12 at 1:13

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.

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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 '12 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 '12 at 16:43
And this whole thread is off topic from this entire site. A whole thread of incorrect proofs, please... – ADP Dec 19 '12 at 2:42

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.

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The denominator is still irrational. – Brusko651 Nov 2 '13 at 22:06

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.....

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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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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.

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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

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!

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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 $x^2-49$. – Eleven-Eleven Feb 19 '14 at 12:01

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:

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

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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."

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That's a standard method to prove polynomial identities by proving them for enough special values. – Phira Dec 17 '12 at 22:40
What's wrong with that? They taught us this in secondary! – The Chaz 2.0 Dec 17 '12 at 22:40
@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 $f(x)$... – N. S. Dec 20 '12 at 17:19

Recently seen:

Given $\gcd(a, b) = c$ we have $c\mid a$ and $c \mid b$, so $$(c\mid a) \cdot (c \mid b) = (c \cdot c) \mid (a \cdot b) = c^2 \mid ab,$$ hence $c^2 \mid ab$.

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

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I'm learning right now why you don't ask first-year calculus students to determine

$$\lim_{x \to -\infty} \frac{x}{\sqrt{x^2+1}}:$$

because you'll get back a bunch of

$$\frac{x}{\sqrt{x^2+1}} \to \frac{-\infty}{\infty} = \frac{-\not\infty^1}{\not \infty^1} = -1.$$

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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!

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

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## protected by Eric NaslundDec 18 '12 at 17:55

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