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

Mathematics Asked on January 7, 2022

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.

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:

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:

Answered by ultralegend5385 on January 7, 2022

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 \ &= sumlimits_{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.

Answered by user3482749 on January 7, 2022

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

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

Proof: take the sine of everything.

Answered by Integrand on January 7, 2022

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

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

Answered by Jan on January 7, 2022

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

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

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

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

Answered by zwim on January 7, 2022

By linearity

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

(just don't try to generalize this)

Answered by flawr on January 7, 2022

$$lim_{xtoinfty}xtimes e^{-x}=infty ×0=0$$ $$lim_{xtoinfty}log(x+1)-log(x)=infty -infty=0$$

Answered by Nitin Uniyal on January 7, 2022

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...

Answered by imranfat on January 7, 2022

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.

Answered by Ivan Hieno on January 7, 2022

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 :)

Answered by Neil on January 7, 2022

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:

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

Answered by Oria Gruber on January 7, 2022

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

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

Answered by Daniel McLaury on January 7, 2022

$newcommand{+}{^{dagger}} newcommand{angles}[1]{leftlangle, #1 ,rightrangle} newcommand{braces}[1]{leftlbrace, #1 ,rightrbrace} newcommand{bracks}[1]{leftlbrack, #1 ,rightrbrack} newcommand{ceil}[1]{,leftlceil, #1 ,rightrceil,} newcommand{dd}{{rm d}} newcommand{down}{downarrow} newcommand{ds}[1]{displaystyle{#1}} newcommand{expo}[1]{,{rm e}^{#1},} newcommand{fermi}{,{rm f}} newcommand{floor}[1]{,leftlfloor #1 rightrfloor,} newcommand{half}{{1 over 2}} newcommand{ic}{{rm i}} newcommand{iff}{Longleftrightarrow} newcommand{imp}{Longrightarrow} newcommand{isdiv}{,left.rightvert,} newcommand{ket}[1]{leftvert #1rightrangle} newcommand{ol}[1]{overline{#1}} newcommand{pars}[1]{left(, #1 ,right)} newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}} newcommand{pp}{{cal P}} newcommand{root}[2][]{,sqrt[#1]{vphantom{large A},#2,},} newcommand{sech}{,{rm sech}} newcommand{sgn}{,{rm sgn}} newcommand{totald}[3][]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}} newcommand{ul}[1]{underline{#1}} newcommand{verts}[1]{leftvert, #1 ,rightvert} newcommand{wt}[1]{widetilde{#1}}$ $$large lim_{x to 0^{+}}{8 over x} = inftyquadimpquadlim_{x to 0^{+}}{3 over x} =omega$$

Answered by Felix Marin on January 7, 2022

Another one I came across was a function composition problem.

Let $g(x)=x^2.$ Find $(gcirc g)(x)$.

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

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

Answered by Eleven-Eleven on January 7, 2022

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!

Answered by Eleven-Eleven on January 7, 2022

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 $frac{partial x}{partial y}cdotfrac{partial y}{partial z}cdotfrac{partial z}{partial x}=-1.$

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

Answered by Isomorphic on January 7, 2022

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!

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

Answered by user87543 on January 7, 2022

Recently seen:

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

Answered by dtldarek on January 7, 2022

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 $nrightarrowinfty$ then $nlrightarrow2pi r$. Thus, the sum of the areas of the triangles tends to $2pi r^2/2 = pi r^2$, which is the area of the circle.

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

Answered by Integral on January 7, 2022

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 $P_1P_2dots P_{n+1}$ 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 $(r_1+r_2dots+r_n)Delta s$.

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.

Answered by Per Manne on January 7, 2022

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

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

Answered by P.K. on January 7, 2022

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...

Answered by Paul Siegel on January 7, 2022

A very common mistake in analysis.

Exercise: Let's $Ksubsetmathbb{R}^{n}$compact and a function $f:Kto mathbb{R}$ 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 $$left|,f(u)-f(v),right|<L_x |u-v|, quad forall u,vin V_x$$ Proof that $f$ is too Lipschitz in all $K$.

"Proof:" Let ${V_x}_{xin K}$ 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 $u,vin K$ we have $$left| f(u)-f(v) right|<maxleft{L_{x_{_1}},dots,L_{x_{_N}}right}cdot |u-v|, quad forall u,vin K.$$

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

Answered by Elias Costa on January 7, 2022

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

Answered by Elias Costa on January 7, 2022

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.

$sin(x) = 0$

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, $(-pi, pi)$.

Answered by Jack M on January 7, 2022

Slightly contrived:

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

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

Answered by irrelephant on January 7, 2022

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

Answered by Artem on January 7, 2022

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.

Answered by Hans Engler on January 7, 2022

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.

Answered by Norbert on January 7, 2022

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}+intfrac{1}{1};{dx} \&= log(x)+log(1) \&= log(x+1)+C. end{aligned}

Answered by NeverBeenHere on January 7, 2022

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.

Answered by Jemmy on January 7, 2022

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 $$log y=3log(x-1)+5log(x-2)+7log(x-3).$$ 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."

Answered by André Nicolas on January 7, 2022

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 $sqrt{2}+sqrt{3}=pi$.

Answered by the L on January 7, 2022

Here's a pretty funny one from xkcd.

Answered by B H on January 7, 2022

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".

Answered by Hans Giebenrath on January 7, 2022

One example from me:

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

Answered by Jack L. on January 7, 2022

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:

$$456+159-156=459$$

Answered by mdp on January 7, 2022

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

Answered by TMM on January 7, 2022

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 + 2sum_{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!

Answered by dtldarek on January 7, 2022

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 $x+y$.

Answered by Old John on January 7, 2022

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