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 u can't directly make a equation 0=0 because 0/0 != 1 but i can't explain this can anyone give a perfect reason why can't we compare 0=0

or is there any other reason also for this to be wrong?????


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.

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    $\begingroup$ That's not so much a "technical reason" why one cannot divide by zero, as just another way to state in fancier words the fact that one cannot divide by zero. $\endgroup$ – Henning Makholm Aug 6 '12 at 12:02
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    $\begingroup$ Well, I thought it might be appropriate to allude to the fact that zero having a multiplicative inverse would be inconsistent with the other field axioms, instead of just repeating "You can't divide by zero!" like we all hear in grade school. What do you think the best way to do this is? $\endgroup$ – Potato Aug 6 '12 at 12:10
  • $\begingroup$ Well, then you seem to have forgotten actually mentioning field axioms in the answer, instead of just repeating the fact that $\mathbb Q$ (or $\mathbb R$, or $\mathbb C$) don't contain muliplicative inverses of 0. There didn't seem to be any allusion to this being a general property of all fields rather than just something that happens to be true for those fields you mention. $\endgroup$ – Henning Makholm Aug 6 '12 at 12:47
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    $\begingroup$ I have edited the answer. Thanks for your help. $\endgroup$ – Potato Aug 6 '12 at 20:45

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 $(2-2)$.

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    $\begingroup$ +1, this is a great answer because it explains everything perfectly to a level close to that of the question (without getting higher level definitions such as fields or axioms involved). I just wish every answer in this site was like this one. $\endgroup$ – JMCF125 Jul 13 '13 at 14:52

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.

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    $\begingroup$ I downvoted this answer as it is simply too lazy and not very helpful. $\endgroup$ – Tobias Kildetoft Aug 6 '12 at 8:35
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    $\begingroup$ i have upvoted because it is correct $\endgroup$ – dato datuashvili Aug 6 '12 at 8:36
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    $\begingroup$ This addresses the revised version: what do you call the denominator of a Real Number? $\endgroup$ – Did Aug 6 '12 at 13:27
  • $\begingroup$ @dato: Why did you choose to delete your own answer? $\endgroup$ – Did Aug 6 '12 at 13:33
  • $\begingroup$ I upvoted this answer because it explains why $\mathbb Q$ are the rational numbers. $\endgroup$ – JMCF125 Jul 13 '13 at 14:47

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