Proof that, (1/0 = 1/0) is false.

[Owen Holden]

~(1/0 = 1/0)

Proof:

D1. 1/z =df (the x: 1=x*z & ~(z=0))

D2. G(the x: Fx) =df Ey(Ax(x=y <-> Fx) & Gy)


T1. ~(1/0 = 1/0)

Proof:

1. 1/z = 1/z <-> Ey(Ax(x=y <-> (1 = x*z & ~(z=0))) & y=y)
By, D1 and D2.

2. 1/0 = 1/0 <-> Ey(Ax(x=y <-> (1 = x*0 & ~(0=0))) & y=y)
By: 1, z=0.

3. 1/0 = 1/0 <-> EyAx(x=y <-> (1 = x*0 & ~(0=0)))
By: 2, y=y.

4. 1/0 = 1/0 <-> EyAx(x=y <-> contradiction)
By: 3, 0=0.

5. 1/0 = 1/0 <-> EyAx~(x=y)
By: 4, (x=y <-> contradiction) <-> ~(x=y).

6. 1/0 = 1/0 <-> ~AyEx(x=y)
By, 5, EyAx(~Rxy) <-> ~AyEx(Rxy)

7. 1/0 = 1/0 <-> contradiction
By: 6, AyEx(x=y).

8. ~(1/0 = 1/0)
By 7, (p <-> contradiction) <-> ~p.

Q.E.D.

Since, Exists(the x: Fx) <-> (the x: Fx)=(the x: Fx) is a theorem.

(See: Principia Mathematica *14.28 p 184)

T2. ~(Exists(1/0)), is also proven true.


If we can assert that Ax(x*0=0), then D1. is simplified.

D1a. 1/z =df (the x: 1=x*z), and the proof still works.

Any opinions?

 

[AKG]

1. G(1/z) <-> Ey(Ax(x=y <-> (1 = x*z & ~(z=0))) & G(y))
By, D1 and D2.

2. G(1/0) <-> Ey(Ax(x=y <-> (1 = x*0 & ~(0=0))) & G(y))
By: 1, z=0.

3. G(1/0) <-> Ey(Ax(x=y <-> contradiction) & G(y))
By: 2, 0=0.

4. G(1/0) <-> Ey(Ax~(x=y) & G(y))
By: 3, (x=y <-> contradiction) <-> ~(x=y).

5. G(1/0) -> EyAx~(x=y) & EyG(y)
By: 4, Ey(P(y) & Q(y)) -> EyP(y) & EyQ(y)

6. G(1/0) -> ~AyEx(x=y) & EyG(y)
By: 5, EyAx(~Rxy) <-> ~AyEx(Rxy)

7. G(1/0) -> contradiction
By: 6, AyEx(x=y).

8. ~G(1/0)
By: 7, (p <-> contradiction) <-> ~p.

Therefore, anything can be said of (1/0). In your case, G(x) <-> x=x. Well, we could have replaced it with G(x) <-> ~(x=x), and thus concluded that x=x. So we get a contradiction (since it contradicts your proof that ~(x=x)). I believe that's because you're using something undefined, namely 1/0, and assuming it makes sense to say G(1/0), specificaly, 1/0 = 1/0. The fact that we can derive any property of 1/0 (assuming my proof above is valid) we want, even contradictory properties, suggests that it really doesn't make sense to talk about its properties. So statements like G(1/0) are not false, but meaningless, as they lead to absurdities.

 

[dextercioby]

Okay,what are you trying to prove...?That:
\frac{1}{0}\neq \frac{1}{0} ?

First of all define the symobols appearing in the sides of the nonequality...

Daniel.

 

[AKG]

Well, he has shown that they are not defined. Or rather, he has defined 1/z such that it is undefined if z = 0.

 

[arildno]

Sorry for being dumb and uneducated, but I can't see the point of this.
By the aid of the laws of arithmetic, we may show that z*0=0 for any number z.

Hence, the multiplicative inverse of 0, 1/0, can't exist .

What is lacking here?

 

[dextercioby]

I don't know,but i like the signs...Inspired from the Oriental writing,perhaps...

Daniel.

 

[chronon]

D1. 1/z =df (the x: 1=x*z & ~(z=0))

i.e. 1/z is only defined for z<>0. Aren't you assuming what you're trying to prove?

 

[AKG]

It is not a mathematical matter, really, but one regarding how we speak of things that don't exist. Equality is a reflexive relation, so we might say that "Santa = Santa", but Santa doesn't exist, so does that make the above sentence meaningless or false? Or should we speak of Santa existing in a different sense?

Consider the theorem mentioned in the first post:

Exists(the x: Fx) <-> (the x: Fx)=(the x: Fx) is a theorem.

Let Fx <-> x*0 = 1. We know ~E(the x: Fx), so ~((the x: Fx)=(the x: Fx)). This would suggest that "the multiplicative inverse of 0 is equal to the multiplicative inverse of 0" is false, rather than meaningless. But as demonstrated, we can also show it to be true (we can show it to have any property we want, in fact). So is the statement both true and false? That can't be. So what exactly is the best way to talk about such things? I think that any statement which can be proven both true and false suggests not that the rules of inference and axioms of the logic are contradictory, but that the statement is not a proper statement of in that logic, i.e. with respect to that logic, it is meaningless, even though prima facie it may seem to make sense. "This sentence is false", thus, is not both true and false, it is rather something meaningless, something whose truth value can't be determined, although, prima facie, it looks like any other sentence whose truth value can be determined.

 

[DeadWolfe]

It is not a mathematical matter, really, but one regarding how we speak of things that don't exist. Equality is a reflexive relation, so we might say that "Santa = Santa", but Santa doesn't exist, so does that make the above sentence meaningless or false?

Wasn't Russel rather interested in that question?

 

[Owen Holden]

1. G(1/z) <-> Ey(Ax(x=y <-> (1 = x*z & ~(z=0))) & G(y))
By, D1 and D2.

2. G(1/0) <-> Ey(Ax(x=y <-> (1 = x*0 & ~(0=0))) & G(y))
By: 1, z=0.

3. G(1/0) <-> Ey(Ax(x=y <-> contradiction) & G(y))
By: 2, 0=0.

4. G(1/0) <-> Ey(Ax~(x=y) & G(y))
By: 3, (x=y <-> contradiction) <-> ~(x=y).

5. G(1/0) -> EyAx~(x=y) & EyG(y)
By: 4, Ey(P(y) & Q(y)) -> EyP(y) & EyQ(y)

6. G(1/0) -> ~AyEx(x=y) & EyG(y)
By: 5, EyAx(~Rxy) <-> ~AyEx(Rxy)

7. G(1/0) -> contradiction
By: 6, AyEx(x=y).

8. ~G(1/0)
By: 7, (p <-> contradiction) <-> ~p..

Precisely so. But, I would say 8. ~(G(1/0)).

~G(1/0) is ambiguous, if (1/0) does not exist.

~(Fx) <-> (~F)x iff E!x.

G(1/0) is false for every G.
[~G](1/0) is false for every G.

Therefore, anything can be said of (1/0).

Not so. AG~(G(1/0)) <-> ~EG(G(1/0)).
i.e. there is no positve (primary) predicate of (1/0) that is true.
i.e. It is defined and it does not exist.

 

[AKG]

Precisely so. But, I would say 8. ~(G(1/0)).

~G(1/0) is ambiguous, if (1/0) does not exist.

~(G(1/0)) <--> ~(~(1/0 = 1/0)) <--> 1/0 = 1/0, right? Where's the ambiguity?

 

[jcsd]

Of course 1/0 = 1/0 that's trivial, the good thing about logic is thta you don't even have to ask yourself what 1/0 means to show it.

The proof that 1/0 is undefined in any ring is that for any a and b a ring (as soon as someone tries to prove something like 1/0 is undefined without referring to any particular set of axioms associated with some mathematicla object then they have already gone wrong as I believe for example in the affinely extended reals 1/0 is or at leats can be defined):

0*a = 0*a + 0 = (b + -b)*a + ba + -(ba) = (b+ -b + b)a + -(ba) = ba + -(ba) = 0

yet for if 1/0 is defined then: 1/0 = 0*1/0 = 1, so clearly 1/0 is not the member of any ring.

 

[AKG]

Of course 1/0 = 1/0 that's trivial

You mean that it is trivially true? Well then, what do you make of Owen's proof that ~(1/0 = 1/0). Is it, then, also false? Or is it meaningless altogether, i.e. is 1/0 = 1/0 trivially true, or meaningless since 1) it is an equation using undefined terms and 2) it seems to be both true and false if treated like a meaningful equation?

 

[Hurkyl]

I would say that it's meaningless myself: the function application notation absolutely requires that the arguments be elements of the domain. Since (1, 0) is not in the domain of /, 1/0 is not a valid logical term.

Now, it would be correct to say:

<br /> \neg \exists x \exists y: (1, 0, x) \in / \wedge (1, 0, y) \in / \wedge x = y<br />

 

[jcsd]

It's menaingless as it uses a term (1/0) that can't be defined by D1.

 

[AKG]

So what about the statement, "The present King of France is bald?" (France is a republic, it has no king). Is it false or meaningless? If we want to logically assign it a truth value, I too would consider it meaningless (and thus claim that there is no truth value), but there is disagreement, and many people would categorize it as false.

 

[jcsd]

So what about the statement, "The present King of France is bald?" (France is a republic, it has no king). Is it false or meaningless? If we want to logically assign it a truth value, I too would consider it meaningless (and thus claim that there is no truth value), but there is disagreement, and many people would categorize it as false.

To me at leats you need a few more premises as for example if we add the premise "kings who are not bald have hair" (i.e. X has hair is the negation of X is blad), then does the king of france have hair or not?

 

[AKG]

The problem is "X has hair is the negation of X is bald" is not necessarily true. "X has hair" is the negation of "it is not the case that X has hair".

 

[jcsd]

"X has hair" is the negation of "X is bald" simply because we say it is, it's a premise.

 

[AKG]

We say it is when X exists. If X doesn't exist, then "X is bald" may be false by virtue of X not existing, so we get ~(X is bald), but this doesn't mean X has hair. Remember, X doesn't exist, so it is wrong to say that X has hair.

(X is bald) is false
~(X is bald)
If X exists, we can go further with this and conclude (X has hair)
If X doesn't exist, and we try to go further with this, we get (X has hair) but X doesn't exist, so that's absurd. This suggests that we can't go further with this, and so to say, as you have, ~(X is bald) <--> (X has hair) gives us problems if X doesn't exist. So what do we do in this case? Do we retract the claim that ~(X is bald) <--> (X has hair)? Or do we say that the claim only applies when X exists?

 

[jcsd]

No, I am saying we are adding the premise ~(X is bald)<-->(X has hair), which on the face of it seems like a reasonable premise.

 

[AKG]

Prima facie, yes, it is a reasonable premise. Adding that premise, (X is bald) becomes meaningless if X doesn't exist. Only without adding that premise can we safely say (X is bald) is false. The issue is that many people would consider a statement (X is bald) to be false where X doesn't exist, meaning that they would deny your premise. Personally, I accept your premise, and what it entails (that "X is bald" is meaningless if X doesn't exist).

 

[honestrosewater]

So what about the statement, "The present King of France is bald?" (France is a republic, it has no king). Is it false or meaningless? If we want to logically assign it a truth value, I too would consider it meaningless (and thus claim that there is no truth value), but there is disagreement, and many people would categorize it as false.

I don't know how you're using "statement", but if "The present King of France is bald" can have a truth-value, say, if can be a proposition, how do you decide which propositions have truth-values and which propositions are meaningless? You do need some way of deciding, yes? If there is some way of deciding, why not make the decision outside of the system and just not allow meaningless propositions to be propositions in the first place?

 

[AKG]

I think you just misinterpreted my use of the word "meaningless." Propositions have truth values, so I would say that "The present King of France is bald" is simply not a proposition.

 

[honestrosewater]

I think you just misinterpreted my use of the word "meaningless." Propositions have truth values, so I would say that "The present King of France is bald" is simply not a proposition.

Okay, that makes sense. Why don't you think such meaningless statements shouldn't be allowed in as false propositions? Edit: Sorry, I meant while making any needed modifications to your axioms or rules of inference.

 

[AKG]

Various reasons. For one, if we don't let such statements in, then we can safely say (X is bald) <-> ~(X has hair). Also, treating statements that make reference to something that doesn't exist as meaningless allows us to nicely deal with paradoxical statements (like the Barber Paradox or Russel's Paradox, perhaps even the Liar Paradox). Thirdly, as was shown before, if we deal with statements involving things like 1/0 as meaningful, then we can derive G(1/0) and ~(G(1/0)), a contradiction. However, if we say that G(1/0) is meaningless, then if we derive a meaningless statement and the negation of that meaningless statement, we don't have a contradiction, just a bunch of meaninglessness, which isn't problematic.

 

[honestrosewater]

Various reasons. For one, if we don't let such statements in, then we can safely say (X is bald) <-> ~(X has hair). Also, treating statements that make reference to something that doesn't exist as meaningless allows us to nicely deal with paradoxical statements (like the Barber Paradox or Russel's Paradox, perhaps even the Liar Paradox). Thirdly, as was shown before, if we deal with statements involving things like 1/0 as meaningful, then we can derive G(1/0) and ~(G(1/0)), a contradiction. However, if we say that G(1/0) is meaningless, then if we derive a meaningless statement and the negation of that meaningless statement, we don't have a contradiction, just a bunch of meaninglessness, which isn't problematic.

But that's why I added that you would make any needed modifications to the rules. Do you know that the resulting system would not be better (by your own standards) than the original?

Edit: Wait, if you let G(1/0) be meaningless, you can't derive anything from it- it isn't a proposition and doesn't get in the door. That's what I don't understand- why limit the things you let in the door? Seriously, that's not meant as a rhetorical question.

 

[chronon]

What worries me about this 'proof' is that it seems to lead to a stronger result than it starts with - you start with 1/0 being undefined and end up with 1/0 being fundamentally contradictory. This argument is wrong. Suppose you did the same thing with 2/3. This is not an integer. You can define division over the integers using

D1. a/b =df (the x: a=x*b & (a is a multiple of b))

Since 2 is not a multiple of 3 this leaves 2/3 undefined. As far as I can see you could follow the same argument to say that 2/3<>2/3, which would suggest that any attempt to give a meaning to 2/3 would be contradictory.

 

[Owen Holden]

~(G(1/0)) <--> ~(~(1/0 = 1/0)) <--> 1/0 = 1/0, right? Where's the ambiguity?

Wrong. I said ~G(1/0) is ambiguous, ~(G(1/0)) is not ambiguous.

Why do you claim ~(G(1/0)) <-> (1/0=1/0) ?

G(1/0) is false for all G, therefore ~(G(1/0)) is true.
(1/0=1/0) is false, therefore, the equivalence fails.

 

[Owen Holden]

Thirdly, as was shown before, if we deal with statements involving things like 1/0 as meaningful, then we can derive G(1/0) and ~(G(1/0)), a contradiction..

No, you cannot derive G(1/0) and ~(G1/0)).

Your post, #2, shows that ~(G(1/0)) can be asserted, but it does not show that G(1/0) can also be asserted.

However, if we say that G(1/0) is meaningless, then if we derive a meaningless statement and the negation of that meaningless statement, we don't have a contradiction, just a bunch of meaninglessness, which isn't problematic.

But, it is problematic to infer any proposition from a meaningless expression, as honestrosewater has said.

 

[Owen Holden]

What worries me about this 'proof' is that it seems to lead to a stronger result than it starts with - you start with 1/0 being undefined and end up with 1/0 being fundamentally contradictory. This argument is wrong. Suppose you did the same thing with 2/3. This is not an integer. You can define division over the integers using

D1. a/b =df (the x: a=x*b & (a is a multiple of b))

Since 2 is not a multiple of 3 this leaves 2/3 undefined. As far as I can see you could follow the same argument to say that 2/3<>2/3, which would suggest that any attempt to give a meaning to 2/3 would be contradictory.

Agreed. By your D1. 2/3 is defined.

D1. 2/3 =df (the x: 2=x*3 & (2 is a multiple of 3)).

(2 is a multiple of 3)) is a contradiction, therefore (2/3) does not exist among the integers.


I did not start with 1/0 being undefined, my D1 includes 1/0.

D2. ~(x=0) -> 1/x =df (the y: x*y=1), is a conditional definition which leaves 1/0 undefined.

D1. 1/x =df (the y: x*y=1), defines 1/0 as (the y: 0*y=1).
But, 0*y=0 for all y, is a theorem.
Therefore, (the y: 0*y=1) does not exist.

 

[Owen Holden]

We say it is when X exists. If X doesn't exist, then "X is bald" may be false by virtue of X not existing, so we get ~(X is bald), but this doesn't mean X has hair. Remember, X doesn't exist, so it is wrong to say that X has hair.

Yes.
If X does not exist, 'X is bald' is false, and, 'X has hair' is false.
There is no-thing that X has or is.

(X is bald) is false
~(X is bald)
If X exists, we can go further with this and conclude (X has hair)
If X doesn't exist, and we try to go further with this, we get (X has hair) but X doesn't exist, so that's absurd. This suggests that we can't go further with this, and so to say, as you have, ~(X is bald) <--> (X has hair) gives us problems if X doesn't exist. So what do we do in this case? Do we retract the claim that ~(X is bald) <--> (X has hair)? Or do we say that the claim only applies when X exists?[/QUOTE]


~(X is bald) <--> (X has hair), is contradictory if X does not exist.
X is bald, and, X has hair, are both false.
~(X is bald), and, ~(X has hair), are both true.

The present king of France is bald, is false.
The present king of France is non-bald, is false.

~(The present king of France is bald), is true.
~(The present king of France is non-bald), is true.

E!x ->. ~(Fx) <-> (~F)x, for all x's.

 

[arildno]

I'm sorry, but I still can't see the point in any of this hair-splitting business.
Could someone please explain why this "logical" exercise is something else than worthless nonsense?

 

[jcsd]

~(X is bald) is just as valid a statement as (X is bald) (that is to say logically there is no difference bwteen stements which assert something postively and those which assert something negatively) so by your logic ~~(X is bald) is also true for objects which don't exist clearly disallowed by the law of non-contardiction.

 

[chronon]

But, 0*y=0 for all y, is a theorem.

Which you do not use in your original proof. Since you don't specify what system your proof applies to the implication is that 1/0 is fundamentally undefinable. This is not true. For instance your computer gets along perfectly consistently but the FPU probably gives a result of +INFINITY for 1.0/0.0 . When you think of what a computer might do then you get more insight into the problem

e.g.

if 1/0=UNDEFINED then presumably (1/0=1/0)=UNDEFINED (not FALSE) and 0*(1/0)=0*UNDEFINED=UNDEFINED (not 0)

You could have 1/0=+INFINITY and (1/0=1/0)=UNDEFINED as well as 0*(1/0)=0*INFINITY=UNDEFINED

Alternatively 1/0=+INFINITY and (1/0=1/0)=TRUE (contradicting the result of your proof), but you still have 0*INFINITY=UNDEFINED. I think that this is what the FPU actually does. See http://www.website.masmforum.com/tutorials/fptute/

 

[jcsd]

Yep 0*a = 0 is a theorum in ring theory (though infact Owen you've used the theorum that a*0 = 0 which is also a theorum in ring theory, but is distinct due tot he fact that muplication is not necessarily commutative), it's not a general truism.

 

[Owen Holden]

~(X is bald) is just as valid a statement as (X is bald) (that is to say logically there is no difference bwteen stements which assert something postively and those which assert something negatively)

Wrong. Please demonstrate why ~(X is bald) <-> (x is bald)??

so by your logic ~~(X is bald) is also true for objects which don't exist clearly disallowed by the law of non-contardiction.

Wrong again, (X is bald) is false, ~(X is bald) is true, therefore, ~~(X is bald) is false!

 

[jcsd]

Wrong. Please demonstrate why ~(X is bald) <-> (x is bald)??

I don't believe I said ~(X is bald) <--> (X is bald) whta I said is that in logic they are both equally valid statements (And by that I'm not assigning them turth values) i.e. the negation of a statement is also a statement.

Wrong again, (X is bald) is false, ~(X is bald) is true, therefore, ~~(X is bald) is false!

I think you've missed th epoint, what you are essientially saying is that for any arbitary stament P about an object that doesn't exist ~P is true, therefore for two arbiatry statemnts Q and R for a non-existant object, by what you have said, we know that ~Q and ~R are both true yet if R is infact ~Q this means that ~Q and ~~Q are both true!

 

[Owen Holden]

I don't believe I said ~(X is bald) <--> (X is bald) whta I said is that in logic they are both equally valid statements (And by that I'm not assigning them turth values) i.e. the negation of a statement is also a statement.

If you are saying that ~(X is bald) and (X is bald) are both propositions, then I agree.

I think you've missed th epoint, what you are essientially saying is that for any arbitary stament P about an object that doesn't exist ~P is true,

No. ~(F(X)) is true and F(X) is false.
It is false to say, any arbitrary statement about X is false.

therefore for two arbiatry statemnts Q and R for a non-existant object, by what you have said, we know that ~Q and ~R are both true yet if R is infact ~Q this means that ~Q and ~~Q are both true!

Not so.

 

[jcsd]

Yes the term statement and propostion are used synomously.

The probelm is thta you seem to think there is a fundmanetal distinction difference between staemnts expressed as F(X) and those expressed as ~G(X) where it's perfectly valid to express a statement F(X) as ~H(X) and a statement ~G(X) as J(X).

therefore for two arbiatry statemnts Q and R for a non-existant object, by what you have said, we know that ~Q and ~R are both true yet if R is infact ~Q this means that ~Q and ~~Q are both true!

Not so.

Yes I know it is not ture if you'd actually bother read what I say rtaher than trying obscure it you'd see that, it is howvere a cosnequence of what you have siad.

Let me ask you a question is the arbitary statemnt P true or flase for a non-existant object (as so far the discussion of whethr statemnts are truye or false about non-existant objects has been independt of the premises and one of ther things about logic is that we don't ahev to actually have to necessarily know what a staemnt is to assign it a truth value, you should be able to tell me)>

 

[matt grime]

You mean that it is trivially true? Well then, what do you make of Owen's proof that ~(1/0 = 1/0). Is it, then, also false? Or is it meaningless altogether, i.e. is 1/0 = 1/0 trivially true, or meaningless since 1) it is an equation using undefined terms and 2) it seems to be both true and false if treated like a meaningful equation?

But it isn't an equation in the sense of dealing with real numbers, is it? As mathematical objects, what ever those two symbols stand for, they will be equal if it makes sense to talk of the objects they represent being equal. So, it is either trivially true, or it is vacuous.

 

[arildno]

I still don't get what the OP is doing.
Has he shown: {?]->£/% or ["\\&<->!" ?

 

[Owen Holden]

I don't believe I said ~(X is bald) <--> (X is bald) whta I said is that in logic they are both equally valid statements (And by that I'm not assigning them turth values) i.e. the negation of a statement is also a statement.





I think you've missed th epoint, what you are essientially saying is that for any arbitary stament P about an object that doesn't exist ~P is true, therefore for two arbiatry statemnts Q and R for a non-existant object, by what you have said, we know that ~Q and ~R are both true yet if R is infact ~Q this means that ~Q and ~~Q are both true!

Nonsense.

There is no case in which p and ~p is true.

 

[Owen Holden]

I still don't get what the OP is doing.
Has he shown: {?]->£/% or ["\\&<->!" ?

It seems sad to me that you do not understand symbolic reasoning.

 

[arildno]

The point is that you haven't given any justification of that your so-called "proof" is nothing else than a meaningless jumble of symbols.

1/0 cannot be defined as a real number, by a trivial use of the axioms of arithmetic.
So what's your damn point?

 

[Owen Holden]

The point is that you haven't given any justification of that your so-called "proof" is nothing else than a meaningless jumble of symbols.

1/0 cannot be defined as a real number, by a trivial use of the axioms of arithmetic.
So what's your damn point?

The damned point is that (1/0 =1/0) is false.

Of course (1/0) is not a real number, who said it was?

Your abusive attitude is not interesting to me!

 

[arildno]

Of course (1/0) is not a real number, who said it was?

Which proves your shenanigans were worthless in the first place.

The damned point is that (1/0 =1/0) is false.

Well, since 1/0,or for that matter "=", by your own admission doesn't mean anything any longer, it's not very interesting, is it?[/QUOTE]

Your abusive attitude is not interesting to me!

Just about the only meaningful statement you've made so far.

 

[Owen Holden]

Which proves your shenanigans were worthless in the first place.

Clearly, you must be specific, if you intend to make sense.

Well, since 1/0,or for that matter "=", by your own admission doesn't mean anything any longer, it's not very interesting, is it?

Nonsense.

 

[jcsd]

Nonsense.

There is no case in which p and ~p is true.

Exactly and I never claimed otherwise! Now why don't you try actually answering the points in my post

 

[jcsd]

Which proves your shenanigans were worthless in the first place.

Well, since 1/0,or for that matter "=", by your own admission doesn't mean anything any longer, it's not very interesting, is it?

Just about the only meaningful statement you've made so far.

The actual argument is vacous in the extreme, it relies on the propostion that z*1/z = 1 when z is not equal to zero, it then uses this propostion to conclude that if z = 0 then z*1/z = 1 (!) and 0 is not equal to 0!