Understanding / by Zero

[Greg H.]

Given

[math]

\frac {x}{0} = x

[/math]

 

[math]

\frac {x}{1} = x

[/math]

 

Then we can say that

[math]

\frac {x}{0} = \frac{x}{1}

[/math]

 

Since we know that x = x, and the two fractions are equal, then 0 must be equal to 1, unless you're changing the definition of equals.

 

(And I know I said I wasn't coming back to this thread, but this thread has reached hitherto unplumbed levels of what comes out the south end of a north facing cow.)

Edited July 4, 2015 by Greg H.

[ajb]

...unless you're changing the definition of equals.

This is exactly my question. Are we loosing the transitivity of the binary relation '='?

[conway]

Ajb, Greg, John

 

I seem to have trouble following you guys here. If...

 

x/0=x

 

x/1=x

 

x/0=x/1

 

 

x=x

 

 

I never arrive at an equation where 0=1.

 

Maybe if I suggest that z2 of zero is =1, and z1 of 1 is = 1, so that in the above equations 0 is only representing z2(of zero), which is = 1.

I think on a more important note that if we apply the above equtions and apply current mathematics we do arrive at an inconsistency

 

 

x/0 = undefined

 

x/1 = x

 

undefined = x

 

This we know is not correct.

Edited July 4, 2015 by conway

[John Cuthber]

 

 

I never arrive at an equation where 0=1.

 

 

I think on a more important note that if we apply the above equtions

 

You don't need to "arrive" at such an expression: we pointed it out for you.

 

You can't apply those equations in maths because division by zero is undefined.

So you don't arrive at any problems.

That's the point.

[Strange]

 

I think the problem is that people are still interpreting your notation as representing numbers, as we understand them. So when you say:

x/0=x/1

People are seeing this as a/b = a/c in which case it is easy to show that b=c (in other words, 0=1).

 

However, your symbols do not represent number but vectors which, if I understand correctly, are defined as follows:

 

All real numbers, r, are represented by a vector R = [a,b]. When r=0 then a=0 and b=1 so R=[0,1]. For all other r, a=r and b=r so R=[r,r]

 

The rules for the basic arithmetic operations are (I think):

X + Y = [a,b] + [c,d] = [a+c, a+c]

X - Y = [a,b] - [c,d] = [a-c, a-c]

X * Y = [a,b] * [c,d] = [a*d, a*d] (which is why it is not commutative)

X / Y = [a,b] / [c,d] = [a/d, a/d]

 

I don't know how other operators would work (e.g. X^Y) or complex numbers.

 

I'm sure someone could work out whether it is possible to have a consistent arithmetic based on this system. If they could be bothered.

I assume equality on these vectors would be defined as: X=y iff a=c and b=d.

Edited July 4, 2015 by Strange

[conway]

Strange

 

Thank you!

 

I have only a few changes to make to your previous post

 

((z1,z2)=A), this only exists in operation of multiplication and division.

 

X * Y = (z1,z2) * (z1,z2) = z1(from X) * z2(from Y).....(so then commutative.) = (z1,z2) = Z

X / Y = (z1,z2) / (z1,z2) = z1(from X, always first) / z2(from Y, always second )......(not commutative) = (z1,z2) = Z

Edited July 4, 2015 by conway

[Strange]

((z1,z2)=A), this only exists in operation of multiplication and division.

 

It seems to me much simpler, and easier to analyse, if you always treat numbers as vectors (pairs) rather than just "magically" converting them when doing multiplication and division.

 

X * Y = (z1,z2) * (z1,z2) = z1(from X) * z2(from Y).....(so then commutative.) = (z1,z2) = Z

 

Using z1 and z2 for all is confusing. How about:

 

X * Y = [x1,x2] * [y1,y2] = [x1*y2, x1*y2]

 

which is not commutative because

Y * X = [y1,y2] * [x1,x2] = [y1*x2, y1*x2] != [x1*y2, x1*y2]

 

Edit ...

No, it seems that multiplication doesn't work like that...

 

Earlier you said that:

0*A=A

and

A*0=0

 

 

Which means:

0 * A = [0,1] * [a,a] = [1*a, 1*a] = [a,a] = A

 

So it looks like the general rule might be:

X * Y = [x1,x2] * [y1,y2] = [x2*y1, x2*y1]

 

But:

A * 0 = [a,a] * [0,1] = ? = [0,1] = 0

 

It may be too late at night, but I can't see any way to fit that into a general rule. In other words, I can't see how you can say what the result is of X * Y without having special cases for X = 0 and/or Y = 0.

 

So it seems to me to be a completely ad-hoc process, rather than part of a formal scheme defining a new number system.

Edited July 4, 2015 by Strange

[conway]

Strange

 

Very well I will use the notions and variables you have suggested. However you still have not quite accurately represented it. So that....

 

 

X * Y = (x1,x2) * (y1,y2) = (x1,y2) = Z

Y * X = (y1,y2) * (x1,x2) = (x1,y2) = Z

X * Y = (x1,x2) * (y1,y2) = (x2,y1) = Z

Y * X = (y1,y2) * (x1,x2) = (x2,y1) = Z

 

commutative.

 

X * Y = Z

 

To have a sum other than a variable then yes, I must know if any variables are or are not zero. This is currently the case anyways.

[Strange]

Strange

 

Very well I will use the notions and variables you have suggested. However you still have not quite accurately represented it. So that....

 

 

X * Y = (x1,x2) * (y1,y2) = (x1,y2) = Z

Y * X = (y1,y2) * (x1,x2) = (x1,y2) = Z

X * Y = (x1,x2) * (y1,y2) = (x2,y1) = Z

Y * X = (y1,y2) * (x1,x2) = (x2,y1) = Z

 

commutative.

 

X * Y = Z

 

I don't think that works. You end up with something which is not a number in your scheme.

 

For example, if we take x=5 and y=4, then X=(5,5) and Y=(4,4) and applying your rules:

X * Y = (x1,x2) * (y1,y2) = (4,5)

Y * X = (y1,y2) * (x1,x2) = (5,4)

X * Y = (x1,x2) * (y1,y2) = (5,4)

Y * X = (y1,y2) * (x1,x2) = (4,5)

 

So you end with two possible values, neither of which is correct - it should be (20,20).

[conway]

Strange

 

The difficulty here is describing what a (5,4) is, or a (4,5). The reality is that x1 for x is value, x2 for x is space. While y1 for y is value, while y2 for y is space. It is that in all cases the value is placed equally into all the spaces then added, if multiplication. It is that in all cases the value is placed equally divisionally into all spaces, than all values subtract but one, if division. A visual example.

 

(5,4)...literally means......

 

5+5+5+5...or The value 5 in the space of 4

 

(4,5)...literally means....

 

4+4+4+4+4...or the vlaue4 in the space of 5.

 

 

It is as you said (4,5) are not numbers, not exactly vectors either. it again is that one is a value, and one is a space.

 

another representation.....

 

x=4 y=5

 

(x1,x2) * (y1,y2) = (x1,y2) = (4+4+4+4+4) = 20

 

or commutative

 

(y1,y2) * (x1,x2) = (y1,x2) = (5+5+5+5) = 20

Edited July 5, 2015 by conway

[Strange]

Strange

 

The difficulty here is describing what a (5,4) is, or a (4,5). The reality is that x1 for x is value, x2 for x is space. While y1 for y is value, while y2 for y is space. It is that in all cases the value is placed equally into all the spaces then added, if multiplication. It is that in all cases the value is placed equally divisionally into all spaces, than all values subtract but one, if division. A visual example.

 

(5,4)...literally means......

 

5+5+5+5...or The value 5 in the space of 4

 

(4,5)...literally means....

 

4+4+4+4+4...or the vlaue4 in the space of 5.

 

 

It is as you said (4,5) are not numbers, not exactly vectors either. it again is that one is a value, and one is a space.

 

another representation.....

 

x=4 y=5

 

(x1,x2) * (y1,y2) = (x1,y2) = (4+4+4+4+4) = 20

 

or commutative

 

(y1,y2) * (x1,x2) = (y1,x2) = (5+5+5+5) = 20

Well, now I am totally confused because earlier, you said:

A = (z1 A , z2 A) = (A as value, A as space) = A

 

2 = (z1 2 , z2 2) = (2 defined values, 2 defined spaces)

 

0 = (z1 0 , z2 1) = (1 undefined value, 1defined space)

 

1 = (z1 1 , z2 1) = (1 defined value, 1 defined space)

And

That is multiplication by zero is relative. Input a function for z1 and z2 where

 

z1 for 0 = 0

z2 for 0 = 1

 

z1 for A = A

z2 for A = A

In other words A = (z1=A, z2=A).

 

Which is what led me to the attempt to formalise your scheme in terms of ... let's call them tuples, rather than vectors as you seem to have a problem with that.

 

So, previously I understood you to say that any non-zero number, n, is represented by the tuple N = (n,n).

Now you seem to be saying that a non-zero number, n, is represented by the tuple N = (q,r), where q and r are factors of n.

 

Either I have totally misunderstood one (or both) of these explanations, or you haven't really thought this thing through. (I am quite happy to admit it may be the former!)

[conway]

Strange

 

Again thank you,

 

 

I must admit I am also confused on the vectors and tuple examples you give. As I have little experience with these things. Suffice to say that I can see relationships between both ways of representing it. I would have to say that ....

 

N = (q,r)...where q and r are factors of N

 

is a much better representation of this idea than

 

N = (n1,n2)

 

So that at this point I can say the following again. Allowing the first above example to be our continued represented form.

 

0 = (q,r) = (0q,1r)

A = (q,r) = (Aq,Ar)

 

Where (q,r) is defined as a value inside of a space.

 

Where

q=value=quantities of values

r=space=quantities of dimensions

 

Such that

 

4=(q,r)=(4q,4r)=(1,1,1,1)=four single values in four single spaces.

 

1=(q,r)=(1q,1r)=(1)=one single value in one single space.

 

0=(q,r)=(0q,1r)=(0)=one single undefined value in one single space.

 

 

so that in multiplication it is that

 

A * X = (Aq,Xr) = Z

Edited July 5, 2015 by conway

[Strange]

I'm afraid that doesn't really make much sense.

 

For example: "A = (q,r) = (Aq,Ar)"

 

The whole point of introducing "q" and "r" was to try and understand how you represent numbers as multiple values. I have no idea what "Aq" and "Ar" are supposed to mean, when the expression was intended to define A.

 

For example, I thought that 12 = (q,r); where q=12 and r=12.

 

But now it seems that 12 = (q,r); where q=2 and r = 6; or q = 3 and r = 4 or q = 12 and r = 1. (How does that work for non-integer values? Can we say that 12 = (q,r) where q=5 and r=2.4?)

 

And it seems we can even say that 12 = (x₀, x₁, ... x_i ... x_n); where [math]\prod_{i=0}^{n} x_i = 12[/math]

 

Except when the value is 0, when it maps on to the unique tuple (0,1).

 

When I have time, I will see if I can write down a general description of how (I think) your system works - and then see if you agree with it. (Although, if you keep bringing this "space and value" nonsense into it, I might not bother.)

 

I am not a mathematician, so I am approaching it from the perspective of defining a new type in a language such as C++. At the moment it still looks a little ad-hoc. It isn't clear how to define even the basic operations operations on these variable-sized tuples in a consistent way. Probably by always mapping them back to the simplest representation. Which is, potentially, going to make it a very inefficient way of doing simple arithmetic. And I still see no real benefit.

Edited July 5, 2015 by Strange

[conway]

Strange

 

"For example I thought that 12 = (q,r), where q=12 and r =12 "

 

This is correct.

 

I didn't mean to imply other wise? I gave examples

 

12 = (12q,12r)

4 = (4q,4r) etc....

 

I was sloppy when I stated it like this

 

A = (q,r) = (Aq,Ar)

 

I meant

 

A = (Aq,Ar)

 

 

 

Also I am prepared to talk of integers and rational and irrationals.

 

Also 0 = (0q,1r) is correct. I will never mention the "nonsense" you spoke of again. Thank you for your time.

Edited July 5, 2015 by conway

[John Cuthber]

 

... I consider the purpose to be "a more accurate" description of reality, which is after all the purpose and intention of mathematics in the first place.

So, you might have come up with something you can call "dividing by zero" but it doesn't apply to numbers.

What reality does it describe?

It doesn't actually answer the reality of dividing a pie.

What might it help with?

What does it describe?

In particular, what does it describe better than ordinary numbers do?

[conway]

John

 

I have come to understand that from your perspective it does not help with dividing by pie, lawyers, and fast cars. But this is not my opinion. As I have stated to you. Any time you write a sentence talking of * and /, the question always is...."how many are in my hand". Such that......

 

 

A lawyer has 120k, and no one to give it to, he still has a 120k.

 

A have a pie, I divided it by zero, I do noting, I still have a pie.

 

I have a fast car, I have no garages to divide it into, I still have a fast car.

[studiot]

 

I will never mention the "nonsense" you spoke of again.

 

 

Has Christmas come early this year?

 

[conway]

Studiot

 

I must tell you that the link you posted in #17 made this exchange of information between myself and others here possible. I thank you. Also to say that yes, I lacked the proper terminology in my descriptions. It was your link that allowed me to write the axiom, there by putting it in proper terminology....if still not quite pinned down. Also I point out that the axiom I posed was not in my post in "speculations" at all. So truly Studiot, thank you for your time.

[studiot]

Way back at the beginning you were dismissive when I commented that many others had had similar ideas before you and that a great deal of work had already been done and suggested that you could save yourself effort by tapping into this and adding your two penn'rth to the world wide effort.

 

To save Professor Strange some hair note that pairs and larger groupings of numbers are identified with many mathematical objects, each with unique useful properties, such as vectors, complex numbers and quaternions, Bra and Ket, to name but a few. When you have more than two, more complex groupings are possible as with Matrices, Tensors etc.

 

The correct word for this is n-tuple (where n is 2 in your case)

 

https://en.wikipedia.org/wiki/Tuple

 

Since there is already a well developed mathematical framework for these things, it is encumbent upon you to develop your system in line with this framework (extending it as necessary) and all the (mathematical) world will applaud if it turns out that you have genuinely thought of something new.

Edited July 5, 2015 by studiot

[conway]

Studiot

 

Yes I was dismissive. Wrongly so. However I still believe that, the "framework" for vectors and tuple's must in the end bend to the framework of "relative mathematics". It is that everything with in it, is in regards to arithmetic. The "base" of mathematics. So that if it is correct it is the limbs that move. So naturally it seems to me the approach is to find flaws in it's logic when applied to vectors and tuples "higher branches of math". Shortest path to this is to ask people that already know, rather than study to the required amount. Requiring much time. I can honestly say as far as arithmetic/algebra is concerned the idea of "relative mathematics" has no contradictions "that I am aware of".

[John Cuthber]

 

A have a pie, I divided it by zero, I do noting, I still have a pie.

 

Yes, when you do nothing, nothing happens.

Nobody has ever asked about that.

What we have asked is what happens when you man up and actually do something?

In particular, what happens when you divide the pie by zero?

 

Doing nothing and then claiming to have divided it by zero isn't going to work.

 

You are behaving like a school kid saying "I have done my homework: I did nothing because I redefined doing my homework as doing nothing."

It wouldn't work at school and it doesn't work here.

[conway]

John

 

Lol...

 

That's exactly how I feel about those who claim A/0 = undefined

 

That's not an answer!

 

 

 

 

 

 

Strange

 

 

For every A in S there exist , q and r, such that any A in operation of multiplication or division is only representing q or r in any given equation. Allowing that in division q is always first r is always second. Allowing that....

 

q for 0 = 0

r for 0 = 1

 

 

0 = (0,1)

 

 

q for A = A

r for A = A

 

A = (A,A)

Edited July 5, 2015 by conway

[John Cuthber]

John

 

Lol...

 

That's exactly how I feel about those who claim A/0 = undefined

 

That's not an answer!

 

OK if you don't think that division by zero is impossible, why don't you tell us how to divide a pie among zero people.

Here's a hint, if you can't it's because the operation isn't defined.

[conway]

John

 

It would seem I can't help myself....

 

"How do you divide a pie amongst zero plp. Hint, you can't"

 

I take a pie in my hands. I do nothing. Then I have successfully divided by zero. And when I am done. I have a pie in my hands.

[John Cuthber]

You can do nothing, pretend to have divided it by zero and hope that nobody sees that the emperor has no clothes on.