why is $0=0$ not possible?

Question

Hi one of my friend showed me one proof, i.e.,

1. $2^2 - 2^2 = 10 - 10$

2. $(2+2) (2-2) = 5 (2-2)$

3. dividing both sides by $(2-2)$

4. $(2 + 2) = 5$

I know this is wrong in first line as both LHS and RHS goes to $0$ and you cannot directly make an equation $0=0$ because $\frac{0}{0} \neq 1$, but I cannot explain this. Can anyone give a perfect reason why we cannot compare $0=0$?

Or, is there any other reason for this to be wrong?

Answer 1

You can't divide both sides by $(2-2)$, because $(2-2)$ is zero, and you cannot divide by zero.

The technical reason for this is that zero does not have a multiplicative inverse in the field of rational numbers (or real numbers, or complex numbers, or any field), because the existence of such an inverse would be inconsistent with the field axioms.

Answer 2

Denominator should be not equal to zero in division. There is no number such that denominator is zero in the set of Quotient Numbers or Real Numbers.

Answer 3

The first two lines are fine, and there's nothing wrong with the equation $0=0$ (or any equation equivalent to it). The wrong step occurs in going from the second to the third line.

This step is, from its structure, the argument that if $ac=bc$ then $a=b$. However this rule does not apply for $c=0$. It does apply for other numbers because those are invertible, that is, there exists a value $c^{-1}$ so that $cc^{-1}=1$. Now there is a general rule that if $x=y$ then $xz=yz$ (and that rule BTW also holds for $z=0$), and therefore for invertible $c$ you can conclude from $ac=bc$ that $acc^{-1}=bcc^{-1}$, that is, $a=b$. However for $0$ this derivation doesn't hold because there's no number $d$ such that $0d=1$, and indeed there are many counter examples showing that for $c=0$ this rule doesn't hold, like $2\cdot 0=3\cdot 0$ but $2\neq 3$, or indeed your fake-proof, where the fact that $c=0$ is (slightly) obscured by writing is as $(2-2)$.