Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

closed as too broad by mrf, Ian Miller, Henning Makholm, G. Sassatelli, Théophile Jan 14 at 17:08

There are either too many possible answers, or good answers would be too long for this format. Please add details to narrow the answer set or to isolate an issue that can be answered in a few paragraphs.If this question can be reworded to fit the rules in the help center, please edit the question.

4  
    
Thank you, Bjorn! – user11235813 Jan 14 at 10:59
up vote 14 down vote accepted

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!

share|cite|improve this answer
    
Thank you Wojowu! Of course, this argument is fallacious because it violates the rules of dimensional analysis. =) – user11235813 Jan 14 at 10:51
2  
To be precise ${$}^2 \neq {$}$. – mistermarko Jan 14 at 10:56
1  
@mistermarko I think the essential flaw here is $c\neq c^2$ and $\$^2\neq\$$. – Wojowu Jan 14 at 10:56
3  
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 at 11:32
1  
@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 at 11:34

$$1 = \sqrt{1} = \sqrt{(-1)(-1)} = \sqrt{-1}\sqrt{-1} = i\cdot i = -1$$

share|cite|improve this answer
    
Thank you Daniel! Of course, this argument is fallacious because the formula $\sqrt{ab} = \sqrt{a}\sqrt{b}$ breaks down when $a$ and $b$ are both negative. =) – user11235813 Jan 14 at 9:55
2  
There are more questions on this site using this, or related, fallacies than anyone could count. – Wojowu Jan 14 at 11:13
    
@Wojowu, care to point me to one such particular question here in MSE? Thanks! – user11235813 Jan 14 at 11:15
    
@ArnieDris math.stackexchange.com/questions/1122038/… A few more are linked. – Wojowu Jan 14 at 11:19
    
Thanks again, Wojowu! =) – user11235813 Jan 14 at 11:22

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).

share|cite|improve this answer
    
Thanks for this, Von! =) – user11235813 Jan 14 at 11:34

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.)

share|cite|improve this answer
    
Thank you for your contribution, @user230452! – user11235813 Jan 14 at 13:44
    
I also posted this question on here a while. Shows a common fallacy made while using strong induction. math.stackexchange.com/q/1400331/230452 – user230452 Jan 14 at 13:47
1  
Did you figure out the fallacy ? :P – user230452 Jan 14 at 13:47
1  
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 at 14:33
1  
@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 at 16:26

Without resorting to $i$:

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

share|cite|improve this answer
2  
Thank you Yves! Of course, this argument is fallacious because the correct formula to use is $\sqrt{x^2} = |x|$. – user11235813 Jan 14 at 10:22

$$S=1+2+4+8+\cdots$$ $$2S=2+4+8+16+\cdots$$ Subtracting like this: $$S-2S=(1+2+4+8+\cdots)-(0+2+4+8+\cdots)=(1-0)+(2-2)+(4-4)+\cdots=1$$ $$S=-1$$

share|cite|improve this answer
1  
Thank you Aditya! Of course, this argument is fallacious because pairwise addition / subtraction of terms breaks down for infinite series. – user11235813 Jan 14 at 11:08
1  
@ArnieDris: in this case I'd rather say that the problem is with a diverging series. $S$ is not defined. – Yves Daoust Jan 14 at 11:11
1  
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 at 11:12
1  
Indeed, thanks Yves! =) – user11235813 Jan 14 at 11:12
2  
Don't tell quantum physicists that this isn't valid. :) – biziclop Jan 14 at 16:17

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.

share|cite|improve this answer
    
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 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. – user230452 Jan 14 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. – user230452 Jan 14 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 at 17:20
2  
@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 at 17:23

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, $s_n=1=0$.

share|cite|improve this answer
    
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. – user230452 Jan 14 at 16:01
    
Is this series divergent ? What is the fallacy here ? – user230452 Jan 14 at 16:19
1  
@user230452 Yes, the series is alternating, therefore divergent. Its Cesaro sum is 1/2. – biziclop Jan 14 at 16:32
1  
@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 at 16:32
1  
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: en.wikipedia.org/wiki/Riemann_series_theorem – Peter Kravchuk Jan 14 at 17:19

Not the answer you're looking for? Browse other questions tagged or ask your own question.