speed of light

Originally posted by mtthw
It would have to be defined within a particular number system, which wouldn't be the number systems we use (R, Q etc), so any results wouldn't apply to these systems. So maths as we know it would still be fine.

Whether the new system would have any practical applications is debatable, but as long as division in this new system behaved in the same way as ...[text shortened]... ion by zero is fine then!). You can say it's possible in a particular framework.

But then we play with words. If you define division as subtraction and vise versa, then division by zero in this new definition is as possible as subtraction with zero in the standard one.

Let me rephrase like this: Nowhere, with the standard definitions of divizion and zero, in any number system, defined systematically with the operator of multiplication, is there an multiplicative inverse of zero.

But of course we can always define whatever we want, even division by zero, with a value of whatever we want, for example 1/0=5. But with no mathematical use whatsoever.

But at the very point of us wanting to be serious, it is not possible.

Originally posted by FabianFnas
But at the very point of us wanting to be serious, it is not possible.

As I said, though, I remember seeing a paper posted here that did it.

In that, the system behaved exactly as you would expect the number system to when not dividing by zero (or adding by infinity, or any of the other troublesome operations). And yet division by zero was allowed, and the system was consistent. It did this by introducing additional numbers to the system.

Even if I am misremembering, I would be impressed if you proved such a system was impossible.

Originally posted by mtthw
As I said, though, I remember seeing a paper posted here that did it.

In that, the system behaved exactly as you would expect the number system to when not dividing by zero (or adding by infinity, or any of the other troublesome operations). And yet division by zero was allowed, and the system was consistent. It did this by introducing additional numbers ...[text shortened]... if I am misremembering, I would be impressed if you proved such a system was impossible.

Yes, I saw a link somwhere too that a mathematician said that division by zero was possible. His article was turned down by others beeing of no use whatsoever. The proof, or method, was of the same kind of proving there is a perpetum mobile, time machine, infinit energy in water, and such. If I find the article, I post the link.

No, I can't proove in this kind of forum that it is impossible to divide by zero. I asked my math teacher and he proved it very elegantly that it is not possible. I advice you to ask yours.

Originally posted by FabianFnas
No, I can't prove in this kind of forum that it is impossible to divide by zero. I asked my math teacher and he proved it very elegantly that it is not possible. I advice you to ask yours.

Division is the inverse operation to multiplication. For an inverse to exist you need the product to be unique (in other words given a, and c there is only one value of b for which ab=c). 0x1 = 0, 0x2 = 0, ... , 0xN = 0, ... and so we can't find an inverse - or just as bad we can find an infinite number of inverses.

Originally posted by FabianFnas
Yes, I saw a link somwhere too that a mathematician said that division by zero was possible. His article was turned down by others beeing of no use whatsoever. The proof, or method, was of the same kind of proving there is a perpetum mobile, time machine, infinit energy in water, and such. If I find the article, I post the link.

No, I can't proove in t ...[text shortened]... . and so we can't find an inverse - or just as bad we can find an infinite number of inverses.

If is was possible you would have to extend the number system, with some other 'number' X. At which point clearly 0xX would not be 0 for consistency.

It might be provable that it's not possible. But proving that division by zero is impossble in a specific number system is one thing, proving that no possible number system exists where it can be allowed is something completely different.

(And I never claimed it would be useful!)

Originally posted by mtthw
If is was possible you would have to extend the number system, with some other 'number' X. At which point clearly 0xX would not be 0 for consistency.

It might be provable that it's not possible. But proving that division by zero is impossble in a specific number system is one thing, proving that no possible number system exists where it can be allowed is something completely different.

(And I never claimed it would be useful!)

I'm trying to understand what you mean. Is it the following:

Take the set of real numbers R and associate a new number E(y) with each element of y \in R. Then we have that 0 x E(y) = y. ay + bz = 0 x aE(y) + 0 x bE(z) = 0 x (E(ay + bz)) so that E(R) is a vector space.

The problem is that we also have that 0 x y = 0, substituting our definition of E(y) into this we get 0 x 0 x E(y) = 0 => 0 x E(y) = 0 which contradicts 0 x E(y) = y. So this attempt to extend the number system in a non-trivial way fails quite quickly. I really don't think that other methods are going to fare much better.

Originally posted by DeepThought
I'm trying to understand what you mean. Is it the following:

Take the set of real numbers R and associate a new number E(y) with each element of y \in R. Then we have that 0 x E(y) = y. ay + bz = 0 x aE(y) + 0 x bE(z) = 0 x (E(ay + bz)) so that E(R) is a vector space.

The problem is that we also have that 0 x y = 0, substituting our definition o ...[text shortened]... fails quite quickly. I really don't think that other methods are going to fare much better.

No. What I mean is this.

The real numbers are defined by a set of axioms (12 of them, if I remember correctly). These axioms specifically disallow division by zero, as if it was allowed it would lead to a contradiction.

Prove that no set of axioms exist such that division by zero can be allowed consistently (while still having a definition of division that we recognise as having a similar meaning to what we are used to).

I am specifically not talking about the real numbers and vector spaces here.

Originally posted by mtthw
The real numbers are defined by a set of axioms (12 of them, if I remember correctly). These axioms specifically disallow division by zero, as if it was allowed it would lead to a contradiction.

Prove that no set of axioms exist such that division by zero can be allowed consistently (while still having a definition of division that we recognise as having a ...[text shortened]... used to).

I am specifically not talking about the real numbers and vector spaces here.

It follows from the field axioms; in fact, from a sub-set thereof.

Multiplication is defined as a function (the nature of which is really not that material to the point) over a set of objects (numbers, vectors, you name is) S, from (S,S) to S. I.e., it's a function which takes one element e1 from S, and another element e2 from S, and returns a single element e3 from S, designated as e1*e2. Multiplication is complete; i.e., for all e1 and e2 there is a resulting e1*e2.

0 is defined as that element from S for which 0*e=0, for all e from S.

Division is defined as another function from (S,S) to S, which is the inverse of multiplication; i.e., if e1*e2=e3, then e3/e1=e2, and vice versa. Division is not defined as necessarily complete, though it may turn out to be; but your question is for a proof that it cannot be.

I think you would agree that all the above is necessary for division to be defined as we would recognise it, right?

Well then, in division by zero, we assume that e1 is 0, and initially also that e3 is non-zero, and we get:

e3/e1=e2 => e1*e2=e3
e3/0 =e2 => 0 *e2=e3

But also, by the second definition (that of zero itself),

0*e2=0

Thus, our initial assumption that e3 was non-zero is false. Well, then, what if we only allow dividing zero by zero? Ah, but then we get:

e3/e1=e2 => 0/0=e2 => 0*e2=0

and this last equation is (again by zero's definition) true for all elements e2 from S.

So, for e not equal to 0, e/0 has no possible value which satisfies the three definitions above; while for e equal to 0, e/0 = 0/0 leads to an equation which is satisfied by every value, and therefore cannot be consistently said to have a single value, either. Therefore, one can not divide by zero consistently while still assuming a meaning for "division" which is reasonably close to what we normally assume it to be; quod erat demonstrandum.

Richard

Originally posted by Shallow Blue
It follows from the field axioms; in fact, from a sub-set thereof.

You can stop there. I'm asking the question, is it possible to use a different set of axioms to get a particular result? You can't rely on any specific set of axioms to prove it impossible.

Anyway, this is fairly pointless, so I'll stop now 🙂

Originally posted by mtthw
You can stop there. I'm asking the question, is it possible to use a different set of axioms to get a particular result? You can't rely on any specific set of axioms to prove it impossible.

Anyway, this is fairly pointless, so I'll stop now 🙂

Except for the trivial case where you only have zero in your algebra S = {0}, so that 0 x 0 = 0, 0/0 = 0, 0 + 0 = 0 and 0 - 0 = 0, I don't think it's possible to find a set of axioms that will do it. Both Shallow Blue's argument and mine hinge on the definition of zero. Basically that 0 * x = 0 for all x in S. This is a many to one relation and you aren't going to be able to define an inverse, or even control it reasonably well like with square roots. Basically you need to find a way of setting up zero so that multiplication by it isn't many to one, and then it stops being zero.

Originally posted by mtthw
You can stop there. I'm asking the question, is it possible to use a different set of axioms to get a particular result? You can't rely on any specific set of axioms to prove it impossible.

Anyway, this is fairly pointless, so I'll stop now 🙂

No, you were asking whether it was possible to prove or disprove the impossibility of dividing by zero, using different axioms, while still keeping division something we would recognise as division. And the answer is: no, because unless you keep to the field axioms, division is not reasonably close to what we would call division.

Of course you can ditch any amount of axioms and then say, see, I can divide by zero, because I have no rules any more that prevent me from doing so. To which the proper answer is: yes, and under the same rules, I can divide fourteen by boing-boing and get purple as the result, because you've thrown away the nature of division in your quest for division by zero.

And if that doesn't satisfy your demands, here's one of my own to counter yours: you define division as you think it should be. Post your definition. We'll pick holes in it large enough to prove either that your "division" is not division, or that you still can't divide by zero.

Richard

Originally posted by Shallow Blue
No, you were asking whether it was possible to prove or disprove the impossibility of dividing by zero, using different axioms, while still keeping division something we would recognise as division. And the answer is: no, because unless you keep to the field axioms, division is not reasonably close to what we would call division.

Of course you ...[text shortened]... that your "division" is not division, or that you still can't divide by zero.

Richard

It might be possible to construct a viable arithmetic by having products of /0 take off in another dimension, in math you can just add dimensions willy nilly so suppose we just have a rule that says
/0 causes a shift of the result to be in an dimension X and have that dimensional referance be the place where such results reside. I am thinking here of the fact that there are many forms of infinity, some infinities greater than other infinities, so if the results are shot off into another dimension, it would be saying we recognize /0 in OUR dimensional framework is undefined but leaving open the idea of putting such results in another framework where infinities can be compared and so forth. So it would be like an extra vector pointer besides the three or four we normally think of.