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

Question

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.

Answer 1

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 $x+y$.

Answer 2

Here's a pretty funny one from xkcd.

Answer 3

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

Answer 4

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.

Answer 5

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

Answer 6

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:

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

Answer 7

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".

Answer 8

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!

Answer 9

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

Answer 10

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.

Answer 11

$\sin(x) = 0$

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

Answer 12

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

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

Answer 13

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!

Answer 14

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.

Answer 15

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...

Answer 16

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

Answer 17

Slightly contrived:

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

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

Answer 18

One example from me:

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

Answer 19

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

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

Answer 20

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.

Answer 21

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.

Answer 22

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:

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

Answer 23

By linearity

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

(just don't try to generalize this)

Answer 24

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.

Answer 25

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 $K$.

Answer 26

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.

Answer 27

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!

Answer 28

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

Answer 29

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.....

Answer 30

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."