Solving the Fallacy of Zero Divided by Zero

[mubashirmansoor]

Some days ago I read a fallacious algabraic argument which was quite interesting and made me think about such cases, Last night I came up with a technique to make sense out of all those fallacies which include diving by zero... The technique is as follows:

lets say:

[tex]a/b=A[/atex]
[tex]a=bA[/atex]​

If we take 'b' as zero, "a = 0" as well and 'A' can be anything.

As a result: [tex]0/0=A[/atex] where 'A' can be anything.

Concludes to two points:

1) Nothing other than zero is divisible by zero, its only zero itself.
2) Zero divided by zero can be anything.

Whats the use of these points?

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The fallacy I had read :

[tex]x^2-x^2=x^2-x^2[/atex]
[tex](x-x)(x+x)=x(x-x)[/atex]
[tex]((x-x)(x+x))/(x-x)=x(x-x)/(x-x)[/atex]​

which results to 1 = 2

Using the points above and repeating the third step of the falacy we have;

[tex](0/0)(2x)=(0/0)(x)[/atex]​

which means:

[tex]v2x=wx[/atex]​

(where v is A#1 & w is A#2)

as we are to keep the equilibrium between the right and left handside of the equation, the relation between v & w is obvious;

[tex]w=2v[/atex]​

by subsituting:

[tex]v2x=2vx[/atex]
[tex](v2x)/(2v)=(2vx)/(2v)[/atex]​

which means x = x and no more a fallacy.

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Even if we look from the other point of view; as multiplicaton is the inverse process of division, and that something multiplied by zero is zero
so logically zero divided by zero can be anything.


I'd be glad for further comments, I know its forbiden to divide something by zero but its fun
Why can't we do the process mentioned above?

Thanks for giving your time.

 

[matt grime]

Because its nonsense?

There are two superficial mistakes I can see.

1. Why would anyone want an algebraic operation that resulted in 'anything' as the outcome? This is precisely the reason why it is undefined in any extension of the reals.

2. You have something backwards. In a suitable extension of the reals we can divide anything by zero except zero.

 

[mubashirmansoor]

I really like to know the problem with the statement of mine & I couldn't really get the 2nd point of yours;

how can we divide something other than zero from zero;

1)which number when multiplied by zero gives us a real number except zero??
2)which number when multiplied by zero gives us zero?

Well I think the logical outcome of these two questions lead to what I had thought...

I'm sure that there is something behind this way of thinking which makes it all wrong but where is it?

One might like to have such an operation which reults to anything for giving a sense to the known fallacious algebraic equations.

I'll be really thankfull for further response.

 

[radou]

You should look at the field axioms of the real numbers, and at the algebraic properties which can be derived from these axioms.

 

[HallsofIvy]

You are correct that "Zero divided by zero can be anything" but since "zero divided by zero" is not then one specific value it is incorrect to say at all that "zero can be divided by zero". If you accept "anything" as a result for the calculation you have no right to say "v2x= wx" so "w= 2v". That's only saying "in order to get a specific result, we have to force 0/0 to be a specific thing, which we have no right to do".

 

[matt grime]

One might like to have such an operation which reults to anything for giving a sense to the known fallacious algebraic equations.

If the operation may result in *any* answer, then how do you know which is the correct one in any given instance?

 

[Hurkyl]

Recall that the thing that makes a function a function is that for any particular set of inputs, there is exactly one output.


If one so desired, one could define a ternary relation _ @ _ = _ defined by

x @ y = z if and only if yz = x​

but one cannot interpret this as defining @ as a function on pairs of real numbers because, as you know, 0@0=x for every x.

Generally one would not use this infix notation for a relation like this, precisely because it looks like @ is being used as a function.

(Of course, if we restricted y to be nonzero, then this does define a function. In fact, @ would be the same as / in that case)