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My friend showed this to me and I instantly know that this is wrong. However, I cannot explain why this is wrong to my friend.

Question. Prove $\displaystyle \frac{100-100}{100-100} = 2.$

Answer. $$\begin{align*} \frac{100-100}{100-100} &= \frac{(10)^2 - (10)^2}{10(10-10)}\\ &= \frac{(10+10)(10-10)}{10(10-10)}\\ &= \frac{20}{10} = 2. \end{align*}$$

My argument is that in the third step, where it goes like this: $$\frac{(10+10)(10-10)}{10(10-10)}$$ you cannot just cancel out the $(10-10)$ - it doesn't seem right. However, I am at a loss of explaining why exactly you cannot do that and my friend has the argumentative power (is that even a word? I mean he is good with arguments, even if they are not facts) and he has me confused to the point that I am starting to think it can be done.

Can anyone please explain why this is wrong?

Thanks.

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    $\begingroup$ This is the same division by zero trick. Soon to be closed as a duplicated question I'd guess. $\endgroup$
    – Asaf Karagila
    Dec 6, 2011 at 23:35
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    $\begingroup$ How about $2=2\cdot 0$ and $1=1\cdot 0$, so $\frac{0}{0}=\frac{2\times 0}{1\times 0}=\frac{2}{1}=2$. It is the same nonsense with less algebra. $\endgroup$ Dec 6, 2011 at 23:39
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    $\begingroup$ Notice that this argument will just as easily "prove" that (100-100)/(100-100)=1/2. Clearly, there is something wrong; what's wrong is division by zero. The first three lines all have division by zero, and so are undefined-- the only valid line is 20/10=2. $\endgroup$ Dec 6, 2011 at 23:50
  • $\begingroup$ This is wrong to begin with. You cannot divide 0. $\endgroup$ Dec 7, 2011 at 0:22
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    $\begingroup$ I just noticed the typo in my comment (I meant $0=2\cdot 0$ and $0=1\cdot 0$). Sorry, my comment was actually more nonsense than the original algebra. $\endgroup$ Dec 7, 2011 at 15:33

6 Answers 6

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You can't divide by $0$; $\frac{100-100}{100-100}$ is not defined, precisely because of that kind of paradoxes.

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    $\begingroup$ well first you just cannot consider (100-100)/(100-100). Then dividing both sides by (10-10)=0 is another thing that is forbidden. If you do that, you get absurd results $\endgroup$
    – Albert
    Dec 6, 2011 at 23:40
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    $\begingroup$ okay @Glougkoubarbaki, thank you! $\endgroup$
    – iamserious
    Dec 6, 2011 at 23:47
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    $\begingroup$ It seems like an obvious joke to me that while you can't divide by "0" you can, fortunately, divide by "0!". $\endgroup$
    – JSchlather
    Dec 6, 2011 at 23:52
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    $\begingroup$ @iamserious: You cannot divide by $0$ because it is $0$, not because of whatever particular way it is written. "$(10-10)$" is a particular representation of the number $0$; the fact that there is a minus sign is unimportant. (E.g., it is OK to divide by $(10-11)$.) $\endgroup$ Dec 7, 2011 at 0:16
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    $\begingroup$ @iamserious: He's saying that (100-100)/(100-100) is undefined to begin with, because it's 0/0, and you cannot divide by 0. However, even if it was valid, the step where you cancel (10-10) is also invalid because (10-10) equals 0, and once again you can't divide by 0. It's the same issue that this related paradox has - every step is valid except the last one, where you are dividing by 0. $\endgroup$ Dec 7, 2011 at 4:08
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There is no such thing as a "cancel" operation. This is, rather, shorthand for factoring out $1$, and simplifying. In other words, suppose you have $$\frac{x^3+x}{x^2+1}$$ You could factor out $$\frac{x^2+1}{x^2+1}$$ to yield $$\frac{x(x^2+1)}{x^2+1}$$ However, since, everywhere $$\frac{x^2+1}{x^2+1}=1$$ You simply replace the former with the latter, to yield $$x.$$

At a fundamental level, where your proof goes wrong is that it skips over this subtlety and uses the "shortcut" of canceling without respect for the conditions under which that shortcut is valid.

In the end, what it comes down to is that $$\frac{(10-10)}{(10-10)}=\frac{0}{0}\ne1,$$ which is the condition you need in order to make that cancellation.

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  • $\begingroup$ Whoa, this is EXACTLY what I was looking for, great answer, thanks! $\endgroup$
    – iamserious
    Dec 7, 2011 at 12:41
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It's a special case of the fact that if $0$ is invertible or cancellable then all elements are equal, viz.

$$\rm x\cdot 0\ =\ y\cdot 0\ \ \Rightarrow\ \ x\ =\ y\quad by \ cancelling\:\ 0 $$

Clearing denominators in your proof, we see it is $\rm\ z\to 10\ $ in the following special case of above

$$\rm (z + z)\:(z-z)\ =\ z\ (z - z) \ \Rightarrow\ \ z+z\ =\ z\quad by \ cancelling\:\ z-z $$

Occasionally sneaky algebraists exploit similar proofs that conclude by deducing that $\:1 = 0\:$ in a ring $\rm\:R\:,\:$ or $\rm\:R = \{0\}$ (so all elements are equal!) e.g. see this question. But as you can see from the many confused comments there and here, it's better pedagogically to avoid such esoteric inferences.

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Another way to see it is to recast the argument as follows:

$\begin{align*} &100-100=100-100\\ &(10+10)(10-10)=10(10-10)\\ &20=10 \end{align*}$

You can check that the first two lines are correct, while the third is not. The reason is we divided by $10-10=0$ between them.

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  • $\begingroup$ aaah.. this is making it clearer to understand, thanks! $\endgroup$
    – iamserious
    Dec 6, 2011 at 23:44
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Notice that $3\cdot0=5\cdot0$.

So, dividing both sides by $0$, we conclude that $3=5$.

Same thing.

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    $\begingroup$ Why repeat what has already been said? $\endgroup$ Dec 7, 2011 at 0:48
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    $\begingroup$ I would take this over KKD's answer (buddy!)... $\endgroup$ Dec 7, 2011 at 13:33
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you can't cancel zero.... buddy...

$(10+10)(10-10) = 10(10-10)$

as $10-10 = 0$

you can't just cancel it...

it is like doing:

$7 \times 0 = 8 \times 0$

cancel zero on both sides and we get:

$7 = 8$

which is incorrect...

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