Can $\frac {100-100}{100-100}=2$? \begin{align*}
\frac{0}{0} &= \frac{100-100}{100-100} \\
&= \frac{10^2-10^2}{10(10-10)} \\
&= \frac{(10+10)(10-10)}{10(10-10)} \\
&= \frac{10+10}{10} \\
&= \frac{20}{10} \\
&= 2
\end{align*}
I know that $0/0$ isn't equal to $2$, the what is wrong in this proof? I wasn't satisfied after searching it on Google, any help would be appreciated.
 A: I think the best way to see the folly here is to work backwards. The problem occurs when we encounter the "equality"
$$
\frac{10+10}{10}=\frac{(10+10)(10-10)}{10(10-10)}
$$
But to obtain this equality, we must multiply the rational number $\displaystyle\frac{10+10}{10}$ by the undefined expression $\displaystyle\frac{0}{0}$. 
A: $1.$ You cannot start with $\frac00$. This is not a number.
$2.$ You are trying to simplify the fraction by dividing numerator and denominator with $(10-10)$. This cannot be done.
Assume beginning from the end going towards the start. The wrong part is when you multiply with $(10-10)$. Because then denominator would be $0.$
With your logic $\frac00=\frac{0*a}{0*b}=\frac{a}b$.
In general: In almost all these kind of problems where someone proves that $1=2$ (or something similar), most of the times, somewhere the denominator will be $0.$
A: You do some thing like this
$$\frac{0\cdot 5}{0\cdot 9}=\frac{5}{9}$$you haven't to divide over zero
A: Well, the first expression in the argument, $0/0$, is meaningless. Everything after that is only connected to the first expression by whimsy, not by mathematics.
What you're seeing here is that, if you try to work with meaningless expressions, then absurd things can happen.
A: The answer is very simple. Look at the denominator of the fraction, especially on $(10−10)$. This is exactly zero! You mustn't divide any number (including zero) by zero, even if in the numerator you have zero in identical expression as in the denominator (I mean of course $(10-10)$). So the problem is that you have above and below the line the same expression, but at first glance you don't notice fact, that it is zero, especially in the denominator and in this way you forget about not dividing by zero. That's very fantastic paradox.
A: Your mistake is that you thought, $100-100$ is not a sum. But these $100s$ are still waiting for you to add them. When you thought like that, you began to expand and contract the expression  so that $100-100$ takes the shape of a non zero number through which you could carry out the division operation freely. When it became $\frac{20}{10}$, you executed the operation. $100-100$ is already a sum with just having another look but mathematically, it is same as $0$.
By the way, you have created a good explanation that how dangerous may be the "division by zero" for mathematics. 
A: The problem of this proof is your first proposition, that the number $\frac00$ exists, or, in other words, that the operation of division is defined for the value of $0$. In fact, the operation $(x,y)\mapsto \frac{x}{y}$ is defined on $$\mathbb R\times (\mathbb R\setminus \{0\}).$$
The operation is defined like so: the number $a=\frac{x}{y}$ is the unique number for which $a\times y = x$. This uniqueness allows you to do all the algebraic operations on these numbers, so it is quite necesary.
In the case of $\frac00$, this would mean that $\frac00$ is the unique number $a$ for which $0\cdot a = 0$. However, since $0\cdot a = 0$ for any value of $a$, such a unique number does not exist, therefore we cannot sensibly define $\frac00$.
A: That's a nice one; I haven't seen it before. It would be nicer backwards.
The issue is in the very first step. $0/0$=nothing ever, and for exactly this reason. If $0/0$ were a number, then we could do exactly what you've done here and obtain $0/0=2$. But now multiplying both sides by $1/2$, we get $\frac{1}{2}\times\frac{0}{0}=1$ But now, multiplying together the left side, we end up with $\frac{1\times 0}{2\times 0} = \frac{0}{0}$. So now we have $\frac{0}{0}=1$ and $\frac{0}{0}=2$, which implies $1=2$. That's clearly wrong. If we keep riding this train we can eventually prove that every number equals every other number, up is down, cats are dogs, and Bertrand Russell is the pope. But since division by 0 is not an operation, it is all meaningless.
A: What you provide is an equality statement
$$
e_1 = e_2
$$
of two expressions and the infix equality operator.
The expression $e_1$
$$
\frac{0}{0}
$$
thus the result of dividing 0 by 0 (which is an instance of dividing a number by zero), is usually considered as undefined, e.g. there is no real number that could be used as result and not cause trouble elsewhere. 
The second expression $e_2$ is
$$
\frac{100-100}{100-100}
$$
We could now argue that both expressions are equal for example by one of these arguments:


*

*fractions are equal if they have equal nominators and denominators and this is the case as reduction by partial evaluation of nominator and denominator individually is unproblematic

*each expression is an instance of division by zero and thus the same failure

*each expression is evaluated to the same special undefined value e.g. div, $\bot$, false (used for partial functions) or undefined (e.g. JavaScript) and thus equality of values occurs


Where the first argument would probably be objected by the argument that these expressions are constructed like fractions (syntactic equality $e_1 = e_2$, the expressions are the same) but are not proper fractions, because zero denominators are not allowed there, thus equality by value ($\sigma(e_1) = \sigma(e_2)$) can not apply because lack of values due to evaluation not happening (semantic $\sigma$ which assigns a value to the expression is not defined).
The second might raise the question if those two are different from other failures like $\infty - \infty$ or $1/-$ or $(1,2,3)+(1,2)$.
Nonetheless we might get convinced that equality holds and then continue boldly with $e_3$
$$
\frac{10^2-10^2}{10(10-10)}
$$
and $e_4$
$$
\frac{(10+10)(10-10)}{10(10-10)}
$$ 
and then try it again with $e_5$
$$
\frac{10+10}{10}
$$
This is not equality by argument 1, 2 or 3 but by argument


*two fractions are equal if nominator and denominator are each the same multiple of the other one $$\frac{n_1}{d_1} = \frac{c \, n_1}{c \, d_1}$$


We note that $e_5$ is now a number while $e_1,\ldots,e_4$ were not (be it $./0$, $\bot$, false but certainly no real number). 
So inequality because of different type applies (comparing apples with pears).
Thus argument 4 was invalid, reduction by common division by $(10-10)=0$ is not valid to derive a fraction with the same value, as it changes the object from being not a number into a number.
We rather have
$$
e_4 \ne e_5
$$
Summary: Division by zero is the culprit but only at the fourth equality (see above) and not before. And there it is not a single division but a common division to reduce a fraction.
