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

[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!

 

[chronon]

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

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

(Owen Holden) is not a real number

Therefore (Owen Holden=Owen Holden) is false?

 

[Owen Holden]

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?

1/0 = 1/0 is trivially true, is contradictrory for me.

Non-existent things are not self-identical.

x=y =df AF(Fx <-> Fy), only applies when x and y are existent objects.

(the x: Fx & ~(Fx)), does not exist!

Therefore, (the x: Fx & ~(Fx))=(the x: Fx & ~(Fx)), is false.

 

[Hurkyl]

1/0 = 1/0 is trivially true, is contradictrory for me.

And it's neither for me, because "1/0 = 1/0" is not a valid string of symbols in the language of the arithmetic of real numbers.

 

[AKG]

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.

If we let any talk about things like 1/0 in the door, then we can derive all sorts of contradictions, like G(1/0) & ~(G(1/0)). In general, especially in mathematics, we don't treat undefined things as false, but don't treat them altogether, i.e. as far as we're concerned, they're meaningless. There is nothing to be gained by talking about undefined things, or things that are not well-defined, moreover, refraining from talking about them insulates us from whatever contradictions arise from talking about them.

 

[arildno]

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!

If this is what he "proves", then his "proof" is invalid, because it uses an undefined term.
Only terms rigourously defined "exists" (in a mathematical/logical sense) , they certainly cannot pop up of nowhere in some "proof".

 

[AKG]

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.

Suppose we have:

G(x) <-> x=x
F(x) <-> ~(G(x)) <-> ~(x=x)

I can prove ~(F(1/0)) by replacing G with F in the proof, so I can prove:
~(~(G(1/0))) <-> G(1/0) <-> 1/0 = 1/0

Of course, we can prove ~(G(1/0)) <-> ~(1/0 = 1/0)

We get (1/0 = 1/0) and ~(1/0 = 1/0), contradiction. I believe the problem arises when you say z=0 in the second or third step, and then continue on with the deduction as though it were meaningful. z=0 is not a valid substitution, because on the left side you get G(1/0) which, according to you, has a false truth value, but I would refrain from giving it, or any statement which talks about something with no referrent, a truth value. The main reason being that, as I believe I have shown above, we can derive any contradiction we want.

 

[AKG]

If this is what he "proves", then his "proof" is invalid, because it uses an undefined term.
Only terms rigourously defined "exists" (in a mathematical/logical sense) , they certainly cannot pop up of nowhere in some "proof".

Well then, care to point out which step of the argument is invalid, and why?

 

[arildno]

Whenever he uses 1/0 as a meaningful term. It is not.

 

[AKG]

Owen

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

If we try to plug z=0 into this, there is no x that satisfies this, so is it not right to say that you can't plug 0 into this in the first place? Otherwise, we get 1/0 =df (nothing)? Certainly, mathematicians treat it this way, that we don't plug 0 into the denominator at all. We don't say that 1/0 =df nothing, we simply disallow any talk of 1/0 in the first place, it's meaningless in mathematics. So, when we say that we can't substitute z=0 in the first place, then the step of your argument (step 2 or 3, can't remember) which does that very substitution would be considered inadmissible.

 

[jcsd]

1/0 isn't necessarily menaigless in mathematics but it is in the real numbers and other rings. So the staemnt 1/0 = 1/0 doesn't mena anything in the context of ring theory, but 1/0 =1/0 is trivial were it has meaning.