What is wrong in this proof?

\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.

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You are never allowed to divide by zero from elementary school... –  TTY Mar 13 at 7:50
Thanks for your feedback. :) I know that but thenIf I assume this proof is correct => 0/0=2 or 0=0 (on cross multiplying), does this mean 0/0 can be equated to anything we want? –  Harshal Gajjar Mar 13 at 7:55
no, "equality" is a relation which hold only for numbers in this case, the expression $0/0$ is not a real number, you cannot equate it to anything. –  TTY Mar 13 at 7:58
Everything..... –  Awesome Mar 13 at 8:02
@Awesome. For me, you made the best and shortest answer to a question which is not a question -> +1 –  Claude Leibovici Mar 13 at 8:11

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 $\frac00$.

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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 $\displaystyle\frac{0}{0}$.

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You don't need to work backwards at all, the proof literally starts with $0/0$. –  Najib Idrissi Mar 13 at 11:17
@nik Nowhere did I indicate that one needs to work backwards. I simply point out that if OP is confused by this "equation", then he or she might consider working backwards to see why it doesn't make sense. Obviously there are many reasons why this equation is invalid (as indicated by the seven answers here). –  Brian Fitzpatrick Mar 13 at 18:16
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.

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or a square of a negative number; the logarithm of a negative number, those are the most common but any multivalued function can be used –  ratchet freak Mar 13 at 10:45

Everything.            

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First time. May be, the next $10^{99}-1$ will follow from the community. Cheers. –  Claude Leibovici Mar 13 at 8:30
Step 1 : Make a new ID 2. Get enough points to upvote. 3. Upvote. 4. Go to step 1. –  Awesome Mar 13 at 8:35
@5xum But the questions is "$what$ is wrong in this proof" not "please explain $why$ it is wrong". –  Awesome Mar 13 at 9:41
@5xum Wrong start makes it all wrong in spite of the facts you mention. –  Awesome Mar 13 at 9:49
Finally...Someone has a sense of humour here. –  Awesome Mar 16 at 10:42

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.

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Thanks for the feedback. :) If I assume this proof is correct => 0/0=2 or 0=0 (on cross multiplying), does this mean 0/0 can be equated to anything we want? –  Harshal Gajjar Mar 13 at 7:55
I guess. More sensibly, it can be equated to nothing, because it leads to nonsense, and math stops working. The correct approach is to recognize it as an undefined expression that doesn't equal anything, ever. –  G Tony Jacobs Mar 13 at 7:57
@HarshalGajjar $\frac00$ cannot be equated to anything. The symbol $\frac00$ makes no sense, as division with $0$ is not defined. Just because you can write it, does not mean it has any meaning. I can write the phraze "what is the color of one mile?", but that does not mean in itself that the question has any meaning. Miles do not have colors, just as the number $0$ has no inverse. –  5xum Mar 13 at 8:03
It might help to note that "division" is really just multiplication by reciprocals. The number $0$ simply hasn't got a reciprocal, because the equation $0\cdot x=1$ has no solution. –  G Tony Jacobs Mar 13 at 8:05

You do some thing like this $$\frac{0\cdot 5}{0\cdot 9}=\frac{5}{9}$$you haven't to divide over zero

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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.

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Actually you don't need to keep riding this train. Let A="any statement". B="1=2". Now we know either A or B is true, (vacuously, since B is true). But we know $1\ne2$. So, either A or B is true, but B is false. So A must be true. Some boolean algebra never hurt anybody. :D –  Sabyasachi Mar 16 at 10:44