@guigus,
guigus wrote:
If we ask whether dividing zero by zero equals one, then we must answer yes, because if we multiply one by zero then we recover zero. However, if we ask whether it equals two rather than one, then we must still answer yes, because if we multiply two by zero then we also recover zero. Likewise, the result of dividing six by two is three only because multiplying three by two results in six. Hence, the result of dividing zero by zero is any number, as multiplying any number by zero results in zero. So dividing zero by zero makes any number the same as any other number. And if all numbers are the same, then all operations between them are also the same. Yet still, in taking a single division of zero by zero with any single quotient, we must recognize it as a perfectly valid operation, since this quotient multiplied by zero results in zero. Nothing at each single division of zero by zero alone makes it invalid: each one is a place where all numbers falsify themselves. Conversely, these operations are only possible if any number is different from any other number, hence true, so every number must be true to be false -- in the division of zero by zero -- and false to be true -- since the division of zero by zero remains valid.
Likewise, whenever I say, "right now, I am lying," if what I am saying is true, then it must be false, and if it is false, then it must be true.
You are correct, if we define x/y as (a number z such that: x=y*z).
That is to say, 0/0 = (a number z such that: 0 = 0*z).
But (0*z = 0), is true for all numbers z.
Therefore,
0/0 = 0 and 0/0=1 and 0/0=2 etc..
This proceedure leads to the absurdities that you mention.
Logic/mathematics abhors contradictions..Bertrand Russell.
For example..
0/0 =df (a z: 0=0*z) -> 0=1.
~(0=1).
Therefore,
~(0/0 =df (a z: 0=0*z)).
You are wrong to define x/y =df (a z: x=y*z).
(Because it leads to contradictions.)
The correct definition of x/y, which does not lead to your absurdities is:
x/y =df (the z: x= y*z).
Your failure to distinguish between the indefinite article (a) and the
definite article (the), is the source of your errors here.
See: On Denoting..B. Russell.
http://www.cscs.umich.edu/~crshalizi/Russell/denoting/
D1. G(a z: Fz) =df Ez(Fz & Gz).
D2. G(the z: Fz) =df Ey(Az(z=y <-> Fz) & Gy).
D3. x/y =df (the z: x= y*z).
~(0/0=0).
Proof:
1. 0/0 = 0 <-> Ey(Az(z=y <-> 0=0*z) & y=0).
By: D3, D2.
2. 0/0=0 -> Ey((0=y <-> 0=0*0) & (1=y <-> 0=0*1)).
By: 1, AxFx -> (F0 & F1).
3. 0/0=0 -> Ey((0=y & 1=y).
By: 2, 0=0*0, 0=0*1.
4. ~Ey(0=y & 1=y). elementary theorem.
5. 0/0=0 -> contradiction.
By: 3, 4.
Therefore,
6. ~(0/0=0).
By (p -> contradiction) <-> ~p).
QED.
Similarly...~(0/0=1) and ~(0/0=2) and ~(0/0=3) etc..
That is to say, there is no number that is equal to 0/0.
~Ex(x=0/0) is a theorem of elementary arithmetic.
0/0 does not exist!
There is no thing that is equal to 0/0, including itself.
~(0/0=0/0), is tautologous.