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

1. Feb 26, 2005

### 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?

2. Feb 26, 2005

### 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)

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.

3. Feb 26, 2005

### 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.

4. Feb 26, 2005

### 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.

5. Feb 26, 2005

### 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?

6. Feb 26, 2005

### dextercioby

I don't know,but i like the signs...Inspired from the Oriental writing,perhaps... :tongue2:

Daniel.

7. Feb 26, 2005

### chronon

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

8. Feb 26, 2005

### 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.

9. Feb 26, 2005

Wasn't Russel rather interested in that question?

10. Feb 26, 2005

### Owen Holden

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.

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.

11. Feb 26, 2005

### AKG

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

12. Feb 26, 2005

### 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 particualr set of axioms associated with some mathematicla object then they have already gone wrong as I beleive 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.

Last edited: Feb 26, 2005
13. Feb 26, 2005

### AKG

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?

14. Feb 26, 2005

### Hurkyl

Staff Emeritus
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:

$$\neg \exists x \exists y: (1, 0, x) \in / \wedge (1, 0, y) \in / \wedge x = y$$

Last edited: Feb 26, 2005
15. Feb 26, 2005

### jcsd

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

16. Feb 26, 2005

### 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.

17. Feb 26, 2005

### jcsd

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?

18. Feb 26, 2005

### 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".

19. Feb 26, 2005

### jcsd

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

20. Feb 26, 2005

### 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?