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I'm currently preparing for a talk to be delivered to a general audience, consisting primarily of undergraduate students from diverse majors. My proposed topic would be Examples of fallacies in arithmetic and/or algebra.

So my question would be:

What are some examples of arithmetic/algebraic fallacies that you know of?

One example per answer please.

Let me give my own example, which is one of my personal favorites:

Let $$a = b.$$ Multiplying both sides by $a$, we get $$a^2 = ab.$$ Subtracting $b^2$ from both sides, we obtain $$a^2 - b^2 = ab - b^2.$$ Factoring both sides, we have $$(a + b)(a - b) = b(a - b).$$ Dividing both sides by $(a - b)$, $$a + b = b.$$ Substituting $a = b$ and simplifying, $$b + b = b,$$ and $$2b = b.$$ Dividing both sides by $b$, $$2 = 1.$$

Of course, this fallacious argument breaks down because we divided by $a - b = 0$, since $a = b$ by assumption, and division by zero is not allowed.

Jose Arnaldo Bebita Dris
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8 Answers8

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I like the following one. It's kinda silly, but still interesting.

We know $1\$=100c$. But then:

$$\begin{align}1\$&=100c\\ &=10c\times 10c\\ &=0.1\$\times 0.1\$\\ &=0.01\$\\ &=1c\end{align}$$

So a dollar is worth just a penny!

Wojowu
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  • Thank you Wojowu! Of course, this argument is fallacious because it violates the rules of dimensional analysis. =) – Jose Arnaldo Bebita Dris Jan 14 '16 at 10:51
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    To be precise ${$}^2 \neq {$}$. –  Jan 14 '16 at 10:56
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    @mistermarko I think the essential flaw here is $c\neq c^2$ and $$^2\neq$$. – Wojowu Jan 14 '16 at 10:56
  • Needed 2 min to check working! –  Jan 14 '16 at 10:59
  • From $100c=10c\times10c$ we could simplify to $1c=1$ !? –  Jan 14 '16 at 11:08
  • @YvesDaoust Indeed we could, if only the former equality were true :) – Wojowu Jan 14 '16 at 11:10
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    In the US, at least, we write $$1$ and $$0.01$ rather than $1$$ and $0.01$$. (I know some other countries also call their currency the "dollar" and I'm not sure if they do it differently.) The cents sign comes after the number, as you have written, though. – Akiva Weinberger Jan 14 '16 at 11:32
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    @AkivaWeinberger Thanks, good to know. I have originaly seen this one with Polish currency, and I just changed currency to dollars. I didn't realize that it is written in a different way. – Wojowu Jan 14 '16 at 11:34
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$$1 = \sqrt{1} = \sqrt{(-1)(-1)} = \sqrt{-1}\sqrt{-1} = i\cdot i = -1$$

fosho
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You might want to look at Edward J. Barbeau's books "Mathematical Fallacies, Flaws, and Flimflam" (MAA, 2000) and "More Fallacies, Flaws and Flimflam" (MAA, 2013).

vonbrand
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Without resorting to $i$:

$$1-3=4-6\\ 1-3+\frac94=4-6+\frac94\\ 1-2\cdot1\cdot\frac32+\left(\frac32\right)^2=4-2\cdot2\cdot\frac32+\left(\frac32\right)^2\\ \left(1-\frac32\right)^2=\left(2-\frac32\right)^2\\ 1-\frac32=2-\frac32\\ 1=2$$

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I have an answer to this question!

Proposition: Every positive number can be described in less than $15$ English words.

Note: I don't necessarily mean the spelling expansion of its form like $256$ as "two hundred and fifty-six" but also as the more economical "sixteen squared".

Base Step: $1$ can be written in less than fifteen English words.

Hypothesis: Let us assume that till a given positive number $k$, we are able to express all the numbers in less than fifteen English words.

Induction Step: Let us consider $k+1$. Either it can be written in less than fifteen English words or it can't. The first case solves our problem. If it can't be expressed in less than $15$ English words, then it is "the smallest positive integer that cannot be expressed in less than fifteen English words", which is a fourteen-word description!

Therefore, $P(k+1)$ is true wherever $P(k)$ is true.

This implies that all positive integers can be expressed in less than fifteen English words!

(However, it is false. The set of positive integers is infinite. The set of English words is not. Therefore a bijective mapping cannot be made. I'll leave it to you to ponder the fallacy.)

Jose Arnaldo Bebita Dris
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Saikat
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  • Thank you for your contribution, @user230452! – Jose Arnaldo Bebita Dris Jan 14 '16 at 13:44
  • I also posted this question on here a while. Shows a common fallacy made while using strong induction. http://math.stackexchange.com/q/1400331/230452 – Saikat Jan 14 '16 at 13:47
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    Did you figure out the fallacy ? :P – Saikat Jan 14 '16 at 13:47
  • Care to elaborate? =) – Jose Arnaldo Bebita Dris Jan 14 '16 at 13:52
  • We can modify the argument from induction to an indirect proof: If the set of the numbers which cannot be expressed by less than 15 words is nonempty, then there must be a smallest number. This number is "the smallest positive integer that cannot be expressed in less than fifteen English words", which is obviously not an element of this set. Thus, the set has to be empty. The fallacy here is, that the 'set' is not a 'set' at all. – Roland Jan 14 '16 at 13:52
  • Why not ? It is a set. An empty set is still a set. The fallacy is a logical mistake called self-reference – Saikat Jan 14 '16 at 13:59
  • Indeed, I found an explanation for this fallacy in Weitzenkamp's A Logical Zoo: Interesting Fallacious Mathematical Arguments. It is described from pages 4 to 5. – Jose Arnaldo Bebita Dris Jan 14 '16 at 14:04
  • You could use an inductive argument to prove that it is impossible to form a crowd, since one person is not a crowd and adding one won't make it a crowd and that everybody is bald, because someone with one hair is bald, and adding one hair can't change that. Nobody is rich because having a penny doesn't make someone rich and adding one penny more does not alleviate ones financial condition. – Saikat Jan 14 '16 at 14:10
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    I don't understand Weitzenkamp's explanation of the paradox/fallacy. To me it sounds like their resolution is "saying this leads to a contradiction, hence we can't do that", which is just like avoiding the paradox by not talking about it. – Wojowu Jan 14 '16 at 14:33
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    @Wojowu The simple explanation in this case is the ambiguity of what it means to "describe a number". Here's a finite version: a prisoner is told he is to be executed next week, but he'll only know the exact date on the morning of execution day. Executions are performed at midday. The prisoner thinks: they can't execute me on Sunday because then I'd already know on Saturday afternoon. But that means they can't execute me on Saturday either because I'd already know on Friday. And so on, until he concludes that he won't be executed. So he's very surprised when he's executed on Wednesday. – biziclop Jan 14 '16 at 16:26
  • A teacher tells his class that there will be a surprise test some time in the next week. The students conclude that the test cannot happen on Friday because if a test was not conducted till Thursday, it would have to be on Friday and no longer be a surprise. This same logic crosses out all the days other than Monday. If it has to be on Monday, it's not a surprise. Hence, the students concluded that there would be no test. Yet, to their dismay, they had to answer a test on Thursday. – Saikat Jan 14 '16 at 17:01
  • @user230452 Yes, this is slightly less brutal than my version. :) – biziclop Jan 15 '16 at 17:21
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$$S=1+2+4+8+\cdots$$ $$2S=2+4+8+16+\cdots$$ Subtracting like this: $$S-2S=(1+2+4+8+\cdots)-(0+2+4+8+\cdots)=(1-0)+(2-2)+(4-4)+\cdots=1$$ $$S=-1$$

Aditya Dev
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    Thank you Aditya! Of course, this argument is fallacious because pairwise addition / subtraction of terms breaks down for infinite series. – Jose Arnaldo Bebita Dris Jan 14 '16 at 11:08
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    @ArnieDris: in this case I'd rather say that the problem is with a diverging series. $S$ is not defined. –  Jan 14 '16 at 11:11
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    Termwise addition and subtraction is perfectly valid. The problem is, as Yves points out, that $S$ is undefined. We simply have indeterminate $\infty-\infty$. – Wojowu Jan 14 '16 at 11:12
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    Indeed, thanks Yves! =) – Jose Arnaldo Bebita Dris Jan 14 '16 at 11:12
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    Now, it is possible to define a function $S:{\subseteq\Bbb R^{\Bbb N}}\to\Bbb R$ on sequences such that $S(1,2,4,8,\dots)=-1$ and such that $S$ behaves like addition (i.e. $S$ has some properties that infinite addition also has). For example, a very famous thing that crops up on this site every now and then is the false statement $1+2+3+\dotsb=-\frac1{12}$, which really just means that $S(1,2,3,\dots)=-\frac1{12}$ when $S$ is Ramanujan summation or zeta regularization. (One can theoretically say that the normal way to sum infinite series isn't any better than these other methods, though…) – Akiva Weinberger Jan 14 '16 at 11:28
  • That sounds interesting, @AkivaWeinberger! Care to share some introductory material on Ramanujan summation and zeta regularization that you can recommend for me? Thanks! – Jose Arnaldo Bebita Dris Jan 14 '16 at 11:32
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    @ArnieDris https://en.m.wikipedia.org/wiki/Divergent_series – Akiva Weinberger Jan 14 '16 at 11:46
  • Thanks again, @AkivaWeinberger! =) – Jose Arnaldo Bebita Dris Jan 14 '16 at 11:50
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    Don't tell quantum physicists that this isn't valid. :) – biziclop Jan 14 '16 at 16:17
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I am fond of fallacies where a property of members of set of things, and the properties of the limit of that set, are assumed to be equal. But the limit need not be a member of the set, and therefore need have nothing in common with members of the set.

For example, imagine a collection of line segments that goes straight up one unit and straight right one unit. The total length is two.

Now imagine it goes up a half, right a half, up a half, right a half. Again, the length is two. And now we have something that looks like a staircase.

Now up a third, right a third, up a third, right a third, up a third, right a third. Another staircase. Length is still two.

Obviously as we continue this sequence the line segments more and more closely approximate a line of length root-two going diagonally. The conclusion we fallaciously reach is that two and its square root are equal.

Eric Lippert
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  • This is a good example of a fallacy, but I wouldn't really call it a fallacy "in arithmetic and/or algebra". – Wojowu Jan 14 '16 at 17:03
  • Very well done. What is the fallacy here ? I don't know it's a fallacy but a very non intuitive result is that the number of points on any line are always the same and the number of points in a surface is equal to the number of points in a line. – Saikat Jan 14 '16 at 17:07
  • @Wojowu I think it would look algebraic if he wrote it represented the lines as vectors in the i and j directions and calculated Euclidean distance. It's just that an intuitive geometric explanation was given here. – Saikat Jan 14 '16 at 17:09
  • @user230452 I think that if this was written in terms of vectors and what not, then the fallacy would become something I would rather not call a fallacy, but plain wrong proof, since we would have a limit without any kind of justification, apart from a geometrical one. – Wojowu Jan 14 '16 at 17:20
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    @Wojowu: Right, I gave the "skeleton" of the fallacy here. You'd want to dress it up in appropriately misleading algebra and throw in some Pythagorean Theorem and whatnot to hide where the incorrect step goes. Also, students not having a clear definition of "limit" in their head helps immensely. – Eric Lippert Jan 14 '16 at 17:23
  • Thank you for this contribution, @EricLippert! =) – Jose Arnaldo Bebita Dris Jan 14 '16 at 23:05
  • But, I have still not understood the fallacy here. What is it . – Saikat Jan 16 '16 at 06:17
  • @user230452: It's the fallacy of hasty generalization. In this fallacy a large number of examples are given that a particular claim is true -- an infinite number of the shapes I described have lengths that add up to two -- and then a conclusion is drawn that generalizes the claim -- the length of the limiting line must also be equal to two. But the evidence is insufficient because properties of members of a set need not be properties of the limit of that set if the limit is not a member. – Eric Lippert Jan 16 '16 at 14:35
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1 = 0?

Given the progression $a_n=(-1)^n$, the sum $s_n=\sum_{i=1}^\infty a_i$ can be built in two ways, all terms in parentheses are zero.

  1. $$s_n=\sum_{i=1}^\infty a_i=(1-1)+(1-1)+(1-1)+...=0$$

  2. $$s_n=\sum_{i=1}^\infty a_i=1+(-1+1)+(-1+1)+...=1$$

Therefore, $s_n=1=0$.

Jose Arnaldo Bebita Dris
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s3lph
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  • This was rather brilliant. You could also take out two 1's and write 1 + 1 = 2 and then write the entire 1-1 series. This way you could always arrange 1 and -1 to get any integer desirable since all integers are a sum of many 1's and -1's. This leads to the paradox that all integers are equal to each other. How to resolve it, I'm not sure. – Saikat Jan 14 '16 at 16:01
  • Is this series divergent ? What is the fallacy here ? – Saikat Jan 14 '16 at 16:19
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    @user230452 Yes, the series is alternating, therefore divergent. Its Cesaro sum is 1/2. – biziclop Jan 14 '16 at 16:32
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    @user230452 Yes, this series is divergent. But this also illustrates an important point: you can't always skip the parentheses when summing a series. – Wojowu Jan 14 '16 at 16:32
  • Somewhat related. This is an illustration of the idea of this proof used in a serious impossibility proof. – Wojowu Jan 14 '16 at 16:33
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    It can be even worse. If I am not mistaken, in a conditionally convergent series one can rearrange terms so that the series will converge to any given real number. Actually, here you go: https://en.wikipedia.org/wiki/Riemann_series_theorem – Peter Kravchuk Jan 14 '16 at 17:19
  • Thank you for your contribution, Seppi! =) – Jose Arnaldo Bebita Dris Jan 14 '16 at 23:08