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There are many common mistakes in solving math ...like below $$\sqrt{a^2+b^2}=a+b\\\sqrt{x^2}=\pm x\\\vdots$$ But there are some mistakes can't compare with common mistakes. I call them advance mistakes. To make a sense ,see below $$1+\frac12+\frac14+\frac18+\cdots=?\\\text{name as x}\\1+\frac12+\frac14+\frac18+\cdots=x\\ 1+\frac12(\underbrace{1+\frac12+\frac14+\frac18+\cdots}_{x})=x\\1+\frac12(x)=x\\x=2 \space \checkmark $$It was true ,now see a wrong solution by this idea $$1+2+4+8+16+...=x\\1+2(\underbrace{1+2+4+8+..}_{x})=x\\1+2x=x \\x=-1 \text{ obviously wrong ,because } |q|>1$$ I am thankful if you can add some mistake example like this. thanks in advance.

Masacroso
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Khosrotash
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  • The first one is a variant of Freshman's dream. – Simply Beautiful Art Jul 28 '17 at 11:32
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    such manipulations only work for finite or converging sums. If the sum is divergent, many surrises are possible. For example, Riemanns's theorem says that we can choose a number $\alpha$ arbitary real and then we can arrange the positive numbers $1,\frac{1}{3},\frac{1}{5},\cdots$ and the negative numbers $-\frac{1}{2},-\frac{1}{4},-\frac{1}{6},\cdots$ in such a way that the value of the sum is $\alpha$ – Peter Jul 28 '17 at 11:32
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    have you seen the numberphile video about how 1+2+3+4+... = $1\over12$ ? –  Jul 28 '17 at 11:35
  • @RoddyMacPhee :sure , I saw that .$$1+2+3+4+...=-\frac{1}{12}$$ – Khosrotash Jul 28 '17 at 11:38
  • Too many people are fimilar with this "identity" :( – Peter Jul 28 '17 at 11:39
  • As @Peter mentions, see here for really advanced mistakes concerning the manipulation of a conditionally converging series. – Simply Beautiful Art Jul 28 '17 at 11:40

4 Answers4

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I have always liked the "proof" that $\sqrt 2 = 2$:

Start with the unit square. It is clear that the diagonal has length $\sqrt 2$. Now approximate the diagonal. first consider the path that goes first along one side then along another. That path has length $2$. Now improve the approximation by going along half the horizontal, then go up half the vertical, then half the horizontal again, then the last half of the vertical. That also has length $2$. Now go by thirds, fourths, and eventually steps of length $\frac 1n$. Each of these paths has length $2$ and pointwise the paths converge to the diagonal, hence the diagonal must have length $2$ and we are done.

lulu
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I know a pair of paradoxes/mistakes that could lie in some place between elementary and advanced.

1) The "stairs paradox": consider a sequence of stairs $10$ meters long and $10$ meters height. If there are, say, $40$ equal steps ($0.25$ meters long and height each), the outline of the ladder is $20$ meters long ($10$ meters upwards and $10$ meters sidewards). If you duplicate the number of the steps, they will be smaller, but the outline is still $20$ meters length. No matter how big is the number of steps and how small they are, the outline will be always the same length.

But when the number of steps goes to infinity, the stairs become a slope, which is, by Pythagorean theorem, $10\sqrt{2}$

Thus, $$\lim_{n\to\infty}20n\cdot\frac{10}n=10\sqrt 2$$

2) A paradeox abot derivatives. Let's differentiate $y=x^2$. We know that $$y=\underbrace{x+x+\cdots+x}_{x\text{ times}}$$

Then $$y'=\underbrace{1+1+\cdots+1}_{x\text{ times}}=x$$

EDIT: Well, I have realized just now that my first example has been already posted by @lulu.

I was also told about some "proof" that every triangle is isosceles, but I don't know the details. I'm sure you can find it googling.

A more advanced common mistake (this one could hardly be considered as a "paradox") is about functional analysis. Many students (myself, for example) have applied the closed graph theorem to a linear operator between non Banach spaces.

ajotatxe
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$$\sqrt{x^2-4x+4}=x-2 ,when \space x>0$$ It is a big mistake because $$\sqrt{x^2-4x+4}=|x-2|$$ for example x=1 $$\sqrt{x^2-4x+4}=\sqrt{1-4+4}\neq1-2$$

roshanaie
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Other example of mistake is : $$0=0+0+0+0+0+...\\0=(1-1)+(1-1)+(1-1)+...\\0=1+(-1+1)+(-1+1)+(-1+1)+(-1+1)+...\\\text{ simplify -1,1 } \to \\0=1$$ In this time I make a paradox to make sense "what's wrong here !" $$s=(1-1)+(1-1)+(1-)+...\\s=1-1+1-1+1-1+...\\s=1-\underbrace{(1-1)+(1-1)+(1-1)+...}_{s}\\s=1-s\\2s=1\\s=\frac12$$ now it seems that sum and subtract of integers may not be $\dfrac12$. it is totally paradox .So I can make a sense to show wrong solution...

Khosrotash
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