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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.

Simon Fraser
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kjo
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    Firstly, it seems like the result being right or wrong is rather unimportant if the means are wrong to begin with. Secondly, "shockingly wrong" and "seemingly valid" are totally subjective. What is seemingly valid to one person (but wrong) is probably shockingly invalid to another. Mathematical fallicies are probably the *only* interesting thing to discuss, here. Both of your examples illustrate fallacies that one could hold for fraction addition and "power addition". – rschwieb Dec 17 '12 at 14:00
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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
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    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 Is Fiddling By Firelight Dec 17 '12 at 16:52
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    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
  • @TMM I just meant that I think the OP is making a meaningless distinction between "wrong ends by invalid methods" and "right ends by invalid methods." There is no reason to set aside fallacies: fallacies are the most interesting thing to look at! Also, "seemingly" and "shockingly" are mostly subjective, so it adds little to the question. – rschwieb Dec 17 '12 at 19:17
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    Highly related: http://math.stackexchange.com/questions/123633 – BlueRaja - Danny Pflughoeft Dec 17 '12 at 20:21
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    Why is this *on topic*? – Anko Dec 17 '12 at 21:11
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    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
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    Also somewhat related: [Successful absurd formalities](http://math.stackexchange.com/questions/168051/successful-absurd-formalities) – MJD Dec 17 '12 at 21:42
  • @Dan: It also reminds me of another anecdote about a physicist from the early 20th century. He had a horseshoe on his wall. Someone asked, "Surely you, a world-renowned physicist, don't believe that horseshoes bring good luck?!" He replied: "Of course not, but they say it works even if you don't believe in it." (The somewhat stretched analogy being that he got the right result (good luck) despite applying the wrong method (not believing)) – joriki Dec 18 '12 at 14:04
  • You might find that "Howler" is a good search term for this, which I've seen used before. – Benedict Dec 18 '12 at 14:26
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    @DanNeely: There was something like this in the Journal of Irreproducible Results. – Cheerful Parsnip Dec 18 '12 at 17:25
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    @Anko: (a) Some ‘successful’ errors of this type lead to at least mildly interesting mathematics, e.g., when does this particular error yield the right answer? (b) Good examples are potentially useful for teachers of mathematics. – Brian M. Scott Dec 23 '12 at 22:47
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    I see no good reason to have closed this question and have therefore voted to reopen it. – Brian M. Scott Dec 23 '12 at 22:48
  • Try searching for "Lucky Larry" sometime... – J. M. ain't a mathematician Apr 03 '13 at 15:47
  • @JM: searching for "Lucky Larry" was exceptionally irrelevant to this question, AFAICT. – kjo Apr 04 '13 at 18:10
  • @kjo, I had [this](http://www.amatyc.org/publications/AMATYC-Review/Fall2008/) in mind. Search that site for other issues, all featuring Lucky Larry. (For some reason that message did not ping me; I only came upon this thread again by accident.) – J. M. ain't a mathematician May 18 '13 at 18:49
  • i guess a *lesson* here is that if one is *consistent* in the *inconsistencies* one uses, then indeed it can be just as valid (remember Poincare and conventionalism??) :) – Nikos M. Jun 16 '14 at 13:00
  • He means when young studens who don't get at all the basic operations have mistakes and do something wrong but the answer is right, I saw this all the time. – L F May 10 '15 at 13:26

42 Answers42

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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 $x+y$.

Pedro
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Old John
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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
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    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
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    I’ve always been fascinated by students’ ability to come up with the *wrong* generalization. – Brian M. Scott Dec 17 '12 at 12:15
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    @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
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    @DilipSarwate: I say give the generalizing instinct free rein, with 202/101, 616/112 = 66/12, 217/124 = 7/4, 136/34 = 16/4, 435/145 = 3/1, 728/273 = 8/3, 784/448 = 7/4... – kjo Dec 17 '12 at 16:14
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    @OldJohn: this one looks like a shoo-in in the Poetic-License-to-Kill category... (Don't know whether to feel sorry for the cluelessness and bewilderment, or to be impressed by the creativity and spunk.) – kjo Dec 17 '12 at 16:18
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    In my answer to this [question](http://math.stackexchange.com/questions/49657/bad-fraction-reduction-that-actually-works/49731#49731) I wrote about the equation $\frac{ap + b}{bp + c} = \frac{a}{c}$. – Jay Dec 17 '12 at 17:31
  • I'm no mathematician (by any means), but I don't understand what is wrong with stating that the addition of `x` and `y` in this example will always lead to the same result as the division performed in the first example (e.g., `(831^2 - 139^2) / (831 - 139)` == `831 + 139`)... or are you all talking about something different? – orokusaki Dec 17 '12 at 19:35
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    @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
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    I wonder what kind of numerical system we could build out of this... – romeovs Dec 18 '12 at 14:14
  • If there was a competition to come up with such an (invalid) idea to get the correct answer, I would rate this as the most original and creative. – dineshdileep Dec 18 '12 at 17:17
  • Oh I never thought math stack exchange could be so funny. – TheBluegrassMathematician Apr 17 '14 at 23:49
  • And where did the division symbol go? – Ribz Jun 13 '14 at 19:29
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    on internet i saw this: $\dfrac{\sqrt{\not{x}}}{{\not{x}}}=\dfrac{\sqrt{\;}}{}$ – L F May 10 '15 at 13:31
  • I think this is a joke, not a real student product. – user21820 May 19 '21 at 17:25
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enter image description here

Here's a pretty funny one from xkcd.

draks ...
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B H
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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
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    It is ridiculous how a student knows about radicals and division before learning their multiplication tables. ;) – P.K. Dec 19 '12 at 09:51
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    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 00:13
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    Algebraically, $x \cdot x^2 = x^4/x$. – Antonio Vargas Feb 27 '14 at 14:45
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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 $\sqrt{2}+\sqrt{3}=\pi$.

Ry-
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the L
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107

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.

Hans Engler
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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
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    I guess this is an example of using a correct method for incorrect reasons. – Cheerful Parsnip Dec 18 '12 at 17:32
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    @ 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
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    You could also "prove" the product rule like this. – Baby Dragon Oct 12 '13 at 06:05
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    This "rule" also is useful for deriving an integral in which the variable is both inside the integral and in the limits http://en.wikipedia.org/wiki/Leibniz_integral_rule#Formal_statement – leonbloy Nov 02 '13 at 21:46
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    @GrumpyParsnip that counts as luck in my book – nicolas Nov 10 '14 at 18:40
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    @Squirtle I just did – nicolas Nov 10 '14 at 18:41
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    @joriki if only mathematics was using "expressions", binders, higher order functions, and other rigorous CS notations (de brujin), the work would be a better place.. how many bad conceptual math have been produced by wrong notations... – nicolas Nov 10 '14 at 18:46
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    @joriki Does this theorem have a name that I could google? – Ovi May 12 '17 at 14:52
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    @Ovi: If you like, this is an instance of the "multivariable chain rule", or thinking about the "total derivative". – Sridhar Ramesh Nov 06 '17 at 19:35
  • @joriki What you said was very interesting, where can I learn more about it? – Some Guy Mar 30 '21 at 08:02
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From MathWorld / Printer's Errors:

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

TMM
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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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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:

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

NeverBeenHere
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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
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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
  • @Joriki Yeah, perhaps I shouldn't have added the first $dx$ as well! – NeverBeenHere Dec 18 '12 at 19:08
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    That was a test by the student to see if teacher reads the proof – nicolas Nov 10 '14 at 18:47
  • Furthermore, log(a)+log(b) is not equal to log(a+b) – BadAtAlgebra Mar 31 '19 at 00:04
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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".

Jonas Meyer
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Hans Giebenrath
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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. –  Dec 17 '12 at 18:16
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    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
  • After having studied linear algebra, I think this one is my favourite one. – Joe Z. Dec 17 '12 at 18:55
  • Unfortunately, this is almost literally how Cayley-Hamilton is proved in Stoll's syllabus on linear algebra (see Theorem 2.1): http://www.math.leidenuniv.nl/~desmit/edu/la2_2012/LinAlg2-index.pdf – akkkk Dec 17 '12 at 21:35
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    @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
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    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
  • This argument does _not_ lead to a valid conclusion (as is mentioned on [Wikipedia](http://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem#A_bogus_.22proof.22:_p.28A.29_.3D_det.28AIn.C2.A0.E2.88.92.C2.A0A.29_.3D_det.28A.C2.A0.E2.88.92.C2.A0A.29_.3D_0)): $\chi(A)$ is the null *matrix* but the determinant of anything, however null, can never be a matrix. (And by the way the determinant of the (certainly null) $0\times0$ matrix is actually $1$.) – Marc van Leeuwen Dec 18 '12 at 12:53
  • @MarcvanLeeuwen: What do you want to say? Of course this "proof" is wrong, but "implying" a correct result. – Hans Giebenrath Dec 18 '12 at 13:03
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    @Marc: Compared to the big interpretative leap you're (correctly) making in interpreting $\chi(A)$ on the left as the result of evaluating the polynomial $\chi$ at $A$ with addition and multiplication referring to matrix addition and matrix multiplication, interpreting the $0$ on the right as the null matrix, as one usually would, is a small step. Under these standard interpretations, the conclusion is valid. You could argue that the student probably doesn't understand these interpretations, but it's inconsistent to apply that only to the right and not to the left. – joriki Dec 18 '12 at 13:49
  • @HansGiebenrath: What I mean is that not only the argument is not correct, it does _not_ arrive at the correct conclusion. Just to be overly explicit, "plugging in" can only produce the _number_ $0$, and that is not what it should be. To me this is not quite the same as other examples here where the two ends of the reasoning are equivalent regardless of the junk in the middle. Of course one can claim that there is one more fallicious and invisible step that replaces the $0$ found from "plugging in" by a null matrix of the proper size, and then the final result _is_ correct. – Marc van Leeuwen Dec 18 '12 at 14:57
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    @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 $K[A]$. –  Dec 18 '12 at 16:37
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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:

$$456+159-156=459$$

Amusing, but fairly unhelpful!

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

Artem
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  • What's wrong with this one? – superAnnoyingUser Feb 16 '13 at 23:46
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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
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    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 02 '13 at 21:57
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    @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 01:43
  • Alternatively, I believe you can L'Hopital and prove by induction? – Mateen Ulhaq Oct 20 '15 at 08:39
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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.

Per Manne
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    Reference, please? – Willie Wong Feb 08 '13 at 12:51
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    @WillieWong See C. H. Edwards, Jr: [*The Historical Development of the Calculus*](http://www.amazon.com/Historical-Development-Calculus-Springer-Edition/dp/0387943137/ref=sr_1_1?ie=UTF8&qid=1360328286&sr=8-1&keywords=historical+development+of+the+calculus), p. 100. An entertaining version can also be found in Arthur Koestler: [*The Sleepwalkers*](http://www.amazon.com/Sleepwalkers-History-Changing-Universe-Compass/dp/0140192468/ref=sr_1_1?s=books&ie=UTF8&qid=1360328341&sr=1-1&keywords=koestler+sleepwalkers), p. 331. – Per Manne Feb 08 '13 at 13:00
  • Thank you. I will check those out when I get a chance. – Willie Wong Feb 08 '13 at 14:22
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    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 08 '13 at 19:14
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    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
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    @kjo, Kepler came up with his laws by staring at a giant pile of data, collected by Tycho Brahe. I'm not surprised that he was able to come up with a flawed proof of an accurate law, because he presumably observed the law directly from Brahe's measurements. Notice that (1) isn't a mathematical error: the account in Dreyer's *A History of Astronomy from Thales to Kepler* suggests to me that Kepler treated it as a physical assumption, which turns out to be incompatible with both Brahe's data and Kepler's (accurate) model. – Vectornaut Oct 31 '15 at 17:14
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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 (http://math.stackexchange.com/questions/260656/cant-argue-with-success-looking-for-bad-math-that-gets-away-with-it#comment3068816_298012). – Vectornaut Oct 31 '15 at 17:16
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$\sin(x) = 0$

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

genepeer
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Jack M
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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!

Did
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dtldarek
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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
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    @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
  • @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
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    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
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    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. – JavaMan Feb 07 '13 at 06:07
  • The freshman's dream here is correct because of Fermat's little Theorem – Maximilian Janisch Jul 29 '20 at 13:16
27

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

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

Ruslan
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P.K.
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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.

Jemmy
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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...

Paul Siegel
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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 07:21
  • @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. – Paul Siegel Dec 28 '12 at 16:52
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    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
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    Really? This is going to lead to an afternoon spent with a complex analysis book... – Paul Siegel Dec 28 '12 at 17:02
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    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: http://www.ams.org/bookstore/pspdf/gsm-97-prev.pdf – Jonas Meyer Dec 28 '12 at 17:04
  • You must have miscounted powers of $\pi$, as this method will give no other value of $\zeta(2)$. Since $\sin(kx)=kx\prod_{n\ge1}(1-k^2x^2/(\pi^2n^2))$ has $x^3$ coefficient $-k^3/6=-k\cdot k^2\sum_n1/(\pi^2n^2)$, any $k\ne0$, including $k=\pi$, implies $\sum_n1/n^2=\pi^2/6$. – J.G. Apr 08 '21 at 22:08
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My "favorite" error with complex numbers is $$ \frac{i}{i}=\frac{\sqrt[2\,\,]{-1}}{\sqrt[2]{-1}}=\sqrt[2\,\,]{\frac{-1}{-1}}=\sqrt[2\,\,]{1}=1. $$

Elias Costa
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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 01:11
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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: http://math.stackexchange.com/questions/84436/. – Jonas Meyer Dec 20 '12 at 03:53
  • @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? – ADP Dec 20 '12 at 04:02
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Slightly contrived:

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

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

awwalker
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irrelephant
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One example from me:

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

Jack L.
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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
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    There's no such thing as implicit `+`. That's multiplication, and because of that, wrong. – Karoly Horvath Dec 18 '12 at 01:28
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    @yi_H I would argue that a mixed number has an implicit `+`. – Code-Guru Dec 18 '12 at 03:57
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    an implicit +? maybe in elementary school...? – Jonathan Dec 18 '12 at 05:38
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    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 01:13
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I would like to add a nice "method" for differentiating $1/x$:

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

Aiden Chow
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Jan
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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.

Integrand
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    I am not familiar with this sine identity, but if it is true, this false proof is truly astonishing. – kjo May 10 '20 at 21:13
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    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." – Integrand May 10 '20 at 21:17
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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.

Norbert
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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:

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

Oria Gruber
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By linearity

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

(just don't try to generalize this)

flawr
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    $\log\left(1+1+2+4\right)=\log\left(1\right)+\log\left(1\right)+\log\left(2\right)+\log\left(4\right)$ and $\log\left(1+1+1+2+5\right)=\log\left(1\right)+\log\left(1\right)+\log\left(1\right)+\log\left(2\right)+\log\left(5\right)$ – Joshua Tilley Apr 27 '18 at 11:29
6

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.

ADP
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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
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    I hate to be that guy (no pun intended) but that whole post is a work of fiction, see e.g. [this](http://www.snopes.com/college/exam/hell.asp). It's also, of course, off topic :) – guy Dec 18 '12 at 16:43
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    And this whole thread is off topic from this entire site. A whole thread of incorrect proofs, please... – ADP Dec 19 '12 at 02:42
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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 $K$.

Elias Costa
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    @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
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    @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
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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.

infinitesimal to circle area

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.

Integral
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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
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    @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
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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!

Eleven-Eleven
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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
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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
4

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

Felix Marin
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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.....

Eleven-Eleven
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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

$\sin(x+y)\sin(x-y)=\sin^2(x)-\sin^2(y)$

Correct method

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

zwim
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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."

André Nicolas
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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
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    What's wrong with that? They taught us this in secondary! – The Chaz 2.0 Dec 17 '12 at 22:40
  • http://en.wikipedia.org/wiki/Logarithmic_differentiation – NeverBeenHere Dec 17 '12 at 22:57
  • 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
  • @André Nicolas Cancelled out the down vote. I think you do have a point. However, I feel there's a justification somewhere as to why it's valid to do so. (In the same way that when doing integration sometimes we make a substitution that restricts the domain but in the end we get the correct answer). – NeverBeenHere Dec 18 '12 at 00:04
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    @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
  • @N.S. Makes sense. Thank you. – NeverBeenHere Dec 22 '12 at 00:03
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    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 i`s. – Charles Roddie Oct 07 '18 at 16:08
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I'm not sure where it originates from, but I first came across this wonderful example in the "Why I should not teach first-year analysis" section at the end of Samir Siksek's Introduction to Abstract Algebra lecture notes (having ignored the warning about irrepairable damage to my soul):

Let $\int = \int_0^x$. We wish to solve the integral equation $f + \int f = 1$. Obviously, we begin by factoring out the $f$, to obtain $f(1 + \int) = 1$. Dividing both sides by $1 + \int$ then gives $f = (1 + \int)^{-1}$, and the binomial theorem allows us to rewrite that as \begin{align*}f(x) &=1 - \int 1 + \iint 1 - \iiint 1 + \cdots\\&= 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \cdots \\ &= \sum\limits_{n=0}^\infty \dfrac{(-x)^n}{n!}\\&= e^{-x}.\end{align*} Substituting that back into the original equation we note that $$e^x + \int_0^xe^{-y}dy = e^x - e^x + 1 = 1,$$ as expected.

user3482749
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  • I wonder if this can be made valid by placing the operator $\int$ into an appropriate Banach Space of formal power series. When a geometric series pops up, this is sometimes the case. – Jason DeVito Aug 23 '21 at 16:00
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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 $c^2 \mid ab$.

dtldarek
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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
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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

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

Daniel McLaury
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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...

imranfat
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    This is a general principle since it works also for $$\int_0^{\infty}\frac x{x^3+1}dx=\int_0^{\infty}\frac 1{x^3+1}dx=\frac{2\pi\sqrt{3}}9$$ – zwim Sep 27 '17 at 01:18
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$$\lim_{x\to\infty}x\times e^{-x}=\infty ×0=0$$ $$\lim_{x\to\infty}\log(x+1)-\log(x)=\infty -\infty=0$$

ultralegend5385
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Nitin Uniyal
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This classic one:

Seven people went in a bar, and drank beer. Then, the waiter came for the bill which was 28 dollars. Then one of the seven people calculated the bill per person, and found it to be 13 dollars. Check out how:

28÷7=13

They paid the bill. When the manager came back, he asked the waiter to which the waiter showed him how 13 added 7 times is 28. Check out how:

13+13+13+13+13+13+13=28

ultralegend5385
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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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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 :)

Neil
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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.

Ivan Hieno
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I know there are many answers already but my TEACHER did this: \begin{align} \int_{-\infty}^{+\infty}xdx&=\int_{-\infty}^0xdx+\int_0^{+\infty}xdx&&(\text{split integrals})\\ &=\int_{+\infty}^0tdt+\int_0^{+\infty}xdx&&(t=-x)\\ &=-\int^{+\infty}_0tdt+\int_0^{+\infty}xdx&&(\text{change integral limits})\\ &=0&&\mathbf{QED}. \end{align}

moqui
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  • This does not qualify for this thread. Here we should arrive to the good result by a wrong way, but $\int_{-\infty}^{+\infty}x\,dx$ is divergent so not the proper result. What you got is in fact Cauchy principal value. – zwim Apr 04 '22 at 19:41
  • No. Read again the question. – moqui Apr 04 '22 at 20:40
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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

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

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

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

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.

Isomorphic
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    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. – Jonas Meyer Nov 30 '13 at 10:13
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    There are multiple definitions of $\mathrm dy\ldots$ – Mark S. Dec 10 '13 at 00:11
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    -1, what about non-standard analysis? – JMCF125 Apr 08 '14 at 10:20
  • Well you could actually treat dy/dx as the ratio of two quantities, so dy/dx would be a secant slope, and for the derivative, you then take the limit as dx tends to 0. – Divide1918 Dec 04 '19 at 11:19