How do I explain the fallacy $\frac00=2$ in this case?

Question

$$ \begin{align} \frac00 &= \frac{x^2-x^2}{x^2-x^2}\\ &= \frac{(x-x)(x+x)}{x(x-x)}\\ &= \frac{x+x}x\\ &= 2 \end{align} $$
Thus $\frac00=2$

How do I explain this fallacy?

Answer 1

You can't cancel $(x-x)$ factor in your third step as that would mean you are assuming $\dfrac{0}{0}=1$ in the first place.

Answer 2

Argument 1 :

zero/ zero itself is indeterminate . you cannot perform operations to it . I will tell why it is indeterminate then we can understand why we cannot operate on that

Try $x^2/x$ or $x/x^2$, as x goes to 0. In both cases, you have 0/0, but the first limit is 0 and the second is infinite. so this results in ambiguity , so your opration

$$\frac{0}{0} = \frac{x^2 - x^2}{x^2 - x^2}$$

you have assumed that 0/0 is 1

, which itself is wrong so whatever we prove with the wrong assumption is also wrong .

Argument 2:

You cannot cancel 0's or terms that give zeros .If that is possible then any number is equal to any number

$$ 0 = 0 $$ $$ 0 \times 2 = 0 \times 3 $$ cancel zero on both sides $$ 2 = 3 $$ This by itself will support why we should not cancel zero's. If you cancel it means you assumed the indeterminate 0/0 is equal to 1 , which is wrong as I proved in argument 1 .

Answer 3

If in a fraction, both numerator and denominator have a non-zero common factor, say $a$, then technically we're not just cancelling $a$ out, we are replacing $\frac aa$ by 1, because if we divide $a$ by $a$ in long division format. we would be looking for a multiple of $a$ which would give back $a$, so we choose 1.

But in case, we are diving $(x-x)$ by $(x-x)$ in long division format, we try to look for a multiple of $(x-x)$ which gives back $(x-x)$. Now this multiple is not just $1$, it can be $2$ $((x-x)*2=0*2=0=(x-x))$, or it can be $10$ or $2000$ or anything. That's why $\frac{x-x}{x-x}$ has many answers, in short, it's indeterminate or undefined. Hence the fallacy.

Answer 4

A good way to think of this is by looking at the equation, 0/0 = x. Multiply both sides to obtain 0x = 0. This equation is satisfied by all real numbers, and 2 is contained within the set of real numbers, so the proof's conclusion was not incorrect. It's just that one could use a similar proof to prove that 0/0 equals any other real number.