Understanding / by Zero

[studiot]

Coincidentally I have been reading a mediocre science fiction thriller that turns out to be in part about aliens dividing by zero.

 

Deep Storm : Lee Child : 2007

 

 

Edited July 6, 2015 by studiot

[Strange]

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)

 

That appears to be a nice clear definition. Unfortunately, it contradicts what you wrote in post #133:

 

 

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.

 

So if we want to multiply 3 x 4, then we have X=(3,3) and Y = (4,4). According to the above, Z = (3,4) or (4,3) whereas it should equal (12,12).

 

Which is why I suggested the correct representation is something like:

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

[conway]

Studiot

 

Good stuff....I would have thought they would have had a better way than a program talking about subtraction......!

 

John

 

Here in is a good example......you keep saying just like the computer says "The number is 1, and there is nothing to subtract form" I got to loop out man!

 

Im thinking.......Just reprogram the computer to say ok If I have 1, and nothing to subtract, then I have 1.

Strange

 

 

Ok, I suppose, If this is what it takes to make it work. I am then assuming that (a,a)=a and (0,1)=0

 

But your suggesting this in not commutative right......let me see.....

 

x * y = (x1,x2)*(y1,y2) = (x1 * y2, x2 * y1) = z

y * x = (y1,y2)*(x1,x2) = (y2 * x2, y2 * x1) = z

 

 

This doesn't work with zero at all.....actually

 

If I make the following statement

 

(x1 * y2 , x2 * y1).....then I have two actually numbers in the parenthesis. This is saying x * y = m,n so that if I multiplied by zero I will always get...

(x1 * y2 , x2 * y1).....which is actually 0 and 1. "lol but not undefined"

 

z does not equal (x1*y2,x2*y1) = (qr,rq)

z must equal (q,r) or (r,q)

 

addition and subtraction are the only operators used when q and r are put back together to make a number again.

Edited July 6, 2015 by conway

[Strange]

Strange

 

Ok, I suppose, If this is what it takes to make it work. I am then assuming that (a,a)=a and (0,1)=0

 

But your suggesting this in not commutative right......let me see.....

 

That doesn't matter right now. We can see whether it is or not later.

 

This doesn't work with zero at all.....actually

 

I have avoided looking in detail at what happens with 0 for the moment. I am just trying to understand how you can make multiplication of non-zero values work.

 

 

If I make the following statement

 

(x1 * y2 , x2 * y1).....then I have two actually numbers in the parenthesis. This is saying x * y = m,n so that if I multiplied by zero I will always get...

 

You always have two (identical) numbers in the parentheses: either X = (x1,x2) (or A = (a,a) if you prefer to write it that way) or 0 = (0, 1).

 

 

z does not equal (x1*y2,x2*y1) = (qr,rq)

 

z must equal (q,r) or (r,q)

 

I don't know what q and r mean in this context.

[conway]

Strange

 

Awesome were making progress

 

I can concede that your way of representation works perfect. The problem is that each and every result of multiplication leaves us with two sums. And when the variables are not zero, both q and r are the same therefore we can just pick one, and always be correct. This will not be the case with zero. How then do we decided which sum to pick in every case 0 involved or otherwise.

 

I agree we must define q and r. Since clearly they are not numbers but rather compose numbers. Suffice to say that...

 

(q,r) = For any A there is a q divided equally an r number of times then added.

 

7=(q,r)=(7q,7r)=(1+1+1+1+1+1+1)

Edited July 6, 2015 by conway

[John Cuthber]

 

 

John

 

Here in is a good example......you keep saying just like the computer says "The number is 1, and there is nothing to subtract form" I got to loop out man!

 

Im thinking.......Just reprogram the computer to say ok If I have 1, and nothing to subtract, then I have 1.

 

It's a good example of how to program a computer to do the wrong thing.

Why would you do that?

In the mean time, can you tell me the answer to the lawyer's problem, or do you accept that no answer has been defined?

[conway]

John

 

Why? Why then is it John that programming a computer this way is the "wrong" way to program it. I have done my best to answer your lawyer question, pie question, and fast car and garage question. It is that you do not like, understand, and or accept these answers. For you and I to continue is not very productive. I hate to say that but....

[Strange]

7=(q,r)=(7q,7r)=(1+1+1+1+1+1+1)

 

I have absolutely no idea what that is supposed to mean. I will give up at this point. Maybe someone else can make sense of this.

[conway]

Strange

 

No.............! lol

 

I understand. Thank you for your time and effort. I tried to explain what I mean.

 

Q and R are unique that Q is always "placed" into an R. There can be varying amounts of Q and R. Sometimes it is that Q is placed subtraction-ally, sometimes it is placed additionally.

 

hopefully this is clearer, if not, again thank you Strange for your time and effort.

 

 

1 = (1q,1r) = 1q into 1r = 1

2 = (1q,1q,1r,1r) = 2q into 2r = 2

3 = (1q,1q,1q,1r,1r,1r) = 3q into 3r = 3

so on....

 

x=2 y=3

 

X * Y = (Xq,Xr) * (Yq,Yr) = (Xq,Yr) = (Xq into Yr) = (2q into 3r) = (2+2+2) = 6

Edited July 7, 2015 by conway

[John Cuthber]

John

 

Why? Why then is it John that programming a computer this way is the "wrong" way to program it. I have done my best to answer your lawyer question, pie question, and fast car and garage question. It is that you do not like, understand, and or accept these answers. For you and I to continue is not very productive. I hate to say that but....

It's wrong because, like you, it doesn't answer the questions.

[conway]

John

 

Maybe it is entirely wrong John. I wonder if you would count the number of pure equations you posted in effort to make your argument. While I am sure I have far exceeded you. My point is that semantics is never accurate when talking of mathematics. You only ever suggested semantics. At least I did more than that. You may claim none was needed to make your point. If this was so, then why does the link on #17, through axioms, state with out semantics, why you can't divide by zero. So then if I can add to these axioms with out changing these axioms, then I have clearly solved for multiplication and division by zero. Assuming the axiom holds up. But again John for you to test my axiom you would have to post an equation showing the inconsistency of the axiom. Please do. But truly I tire of the semantic arguments between you and I. All the best to you John. Thanks for your time.

[Phi for All]

My point is that semantics is never accurate when talking of mathematics.

 

You owe me a new irony meter.

[conway]

Phi

 

I allowed it was only not accurate, not that... to a large degree that I liked it. Lol. Ill get right on that new irony meter for you.

 

To be fare, I have offered more than my share of equations.

[John Cuthber]

Writing lots of equations doesn't help, especially if they are wrong.

Answering the problems would.

 

You have yet to address my first point; how do you propose to invert a many to one function?

[conway]

John

 

I don't know what you mean by a many to one function. Could you provide an example.

[John Cuthber]

Multiplication by zero is a many to one function, and the inverse would be a one to many function- which isn't really much of a function.

 

Let's start with a better behaved one.

 

If I multiply out

(X-1) (X-2) X-3)

I will get a cubic equation in x and there are clearly 3 solutions to that polynomial being equal to zero.

So, if I tell you that the value of the polynomial is zero, you don't know what the value of X is.

The polynomial itself is a perfectly well behaved cubic that looks like this picture

For any value of x you can find the value of y because there is only 1 value of y for any given x.

 

 

But, as I said, the inverse function which is what you get by transposing the x and y axes isn't well behaved.

it looks like the second picture (sorry for the upside down butchered image).

 

Near x=0 it's a one to many (three to be specific) function and, because of that you can't define the value .

If x is zero is y 1, 2 or 3?

You can't say.

 

Well with division by zero, it's even worse.

If you multiply any number by zero you get zero (of course) and so there's no way of "undoing" the function.

But the inverse of multiplication by zero is division by zero; and it's impossible.

So mathematicians simply accepted that you can't undo multiplication by zero; the process for "undoing" it can't be defined.

 

on several occasions you seem to have beleived that 1/0 =x where x is some undefined number.

That's missing the point.

It's not that x is undefined.

it's that the process of division by zero is undefined.

it's not that there isn't an answer, but that the question makes no sense.

I got bored + checked that I can still do a bit of easy algebra. The polynomial is a cubic

x³ -6x²+11x -6 =0

 

The trouble with your idea is that this cubic, and division by zero and a whole lot of other functions have no inverse.

So, just as you have made up an arbitrary rule to cover inverting one function (division by zero) you would still need to make up an infinite set of arbitrary rules for "inverting" all the others.

 

That's going to take you an infinitely long time.

 

And it still won't share a pie.

[conway]

John

 

I must be missing something. I don't see how anything you stated is relevant. The equation you posted doesn't contain division by zero. A line approaching zero, is not the same thing as division by zero (what your suggesting here). X must have a value for Y to be determined. I agree, what's that got to do with division and multiplication by zero.

 

Just because you have no one to give your pie to, (therefore your action is undefined), does not mean you are not holding that pie. And as I pointed out several times, "semantically" speaking, the question always is, what do you hold. Not "what did you do" "was your action definable".

 

 

Upon further thought....Is not division by fractions the very definition of "one to many." 1/.01=100...

 

Is not this 100 actually pieces of the original 1. (attach units such as $).

Edited July 8, 2015 by conway

[Strange]

I don't know what you mean by a many to one function.

 

I find this very odd, and a bit worrying. You have shown repeatedly that you don't have much knowledge of the most basic maths concepts. And yet you claim to have some new insight. I would suggest you take of the many online math courses that are available; you could try Coursera for example. An introduction to calculus would be useful, but that may be too advanced for you right now.

https://www.coursera.org/

[conway]

Strange

 

 

Yes I could do what you suggest. And you are right about my education level. Then I would have much more knowledge. I could invert your scenario however. You yourself strange said "I don't know what this means"......in regards to (q,r) q, being placed into r. I have in fact read a very thread in this forum in mathematics on vectors, to which you your self replied to it. In short the very idea of vectors, is the idea of "space" with "values" placed into them, q being value, r being space, that was what I was talking about, yet you claim to not understand what I was talking about. Your post was purely indented to insult. I understand your done with this thread. So leave out the "passive aggressive" insults as well.

 

 

I have only ever talked of arithmetic. Arithmetic is the foundation of mathematics. What is truth of the foundation is truth of the rest. Therefore If I have a complete working knowledge of arithmetic, then knowledge of the rest is not NECESSARILY required.

 

If I wish to teach a dog to talk, must I then also know how he talks to other dogs.

If I want to re-invent the wheel, must I then also know all about other shapes.

If I want to say the world is round, do I then need to prove all planets are.

To all

 

I have in several examples admitted what it is that I do not know. This alone is evidence that I may indeed (highest probability)...say "A-ha now I get it, This idea is utterly wrong!". But it may be that there is actually something in this idea. As evidenced by more than one poster in this and the "original" thread. For those that wish to see it come out one way or the other do so at their own will. No one is required to continue reading and or posting.

Edited July 9, 2015 by conway

[Bignose]

I have only ever talked of arithmetic. Arithmetic is the foundation of mathematics. What is truth of the foundation is truth of the rest.

The problem is that you aren't just in artihmetic. You have touched upon number theory. You have touched upon vector theory (and reading one thread on that does not make you an expert on it (I've used vectors for nearly 20 years now and there are many thick volume's worth of stuff I don't know)). You are hitting upon much deeper parts of mathematics than just simple arithmetic.

 

You gripe that '"I don't know what this means"......in regards to (q,r)', but you had never defined that notation. And you used it very inconsistently. We're not mind readers, man. In a proper exposition, every piece of nomenclature and notation is explicitly defined the first time it is used.

 

Lastly, you can take Strange's post in two ways. You can choose to be insulted; I can understand that perspective. However, as an alternative, you could choose it to be blunt straightforward criticism. And a recommendation to take some time to learn the extraordinarily rich language of mathematics to help you present your ideas in terminology and notation that is already familiar with people. You can learn what the current theory says so that you can learn what changes you want to make. And further you can learn why the above posts all drew certain conclusions, and how to address them in the structures employed by mathematics today.

 

Frankly, this overwhelming self-assuredness that you present to the forum smacks of Dunning Kruger effect to me. https://en.wikipedia.org/wiki/Dunning–Kruger_effect I could be wrong, maybe you've stumbled on the greatest concept in number theory to date. But as above pointed out by a few of us now, the communication problems need to be addressed. Learning what the current number theory says will only help with this because you'll learn the language that is used by professional mathematicians every day.

Edited July 9, 2015 by Bignose

[conway]

Bignose

 

 

It does not take skill to have an idea. I can have an idea about bananas but never had ate one. If the dunning Kruger effect is an "unskilled" person, which clearly I am, that has "delusions" of how grand their skill is........which clearly I do not, then I do not suffer from it. I have made NO claims as to the "greatness" of my mathematical ability(skill). I have only claimed to have an idea. Further I did finally define, under Strange 's guidance (thank you Strange) both q, and r. I am the one suggesting what it means. It was Strange that was "gripping" about not understanding. As for the rest of your post, as I have already stated, "yeah I need to learn a lot". Again what's that got to do with the idea. If it does, let's talk about it, like John and I (thank you John.). If you feel that the idea is not worth discussing, has no merit. Is not correct. Then carry on sir.

 

Most importantly you point out the vastness of what I touch upon. I touch upon the philosophy of mathematics. I touch upon the pure abstract that is such that it is free of equations and only exists extensionally. Philosophers gave birth to mathematics. All fundamental advancement in mathematics have come from a philosophical change of perspective of mathematics itself. It will be a philosopher that fundamentally changes mathematics again....not a mathematician....albeit they may add much knowledge.

 

 

 

 

To All

 

It seems this has come down to attacking my credentials, which I have at no time claimed to have. (lol, suppose that makes an easy target). I suggest as reasoning for why I was able to "discover" this idea is because I am a philosopher. The things I studied, for 3 years, were not equations, or axioms (though later I did). The things I study were questions like....

 

"what is the nature of zero"

"what is the nature of a number"

"what is value, what is space"

"what does it mean to put a value into a space"

 

These things are not mathematics, these things are philosophy.

Edited July 9, 2015 by conway

[Bignose]

It does not take skill to have an idea.

Right. But it takes a lot of skill to develop that idea into something meaningful in a scientific sense. Just browse Speculations for a few minutes to realize this. That is all I am asking. Take some time to learn the current language so that you can speak it when people are giving you feedback.

 

And, sure, you didn't explicitly make any claims of 'greatness'. But the reluctance to take time to understand the many objections presented to you implies your feelings about your ideas. And the reluctance to take the time to learn the terminology used by professional mathematics. And the reluctance to answer direct questions. It builds a profile.

 

Again, I could be wrong. I am just voicing an opinion and some advice. It is worth exactly what you paid for it. If you don't want my advice, just ignore it.

[conway]

Bignose

 

You are wrong in all three accounts. I have addressed all "objections" presented to me. Maybe not satisfactory enough to you. I freely gave up my terminology in favor of Strange 's terminology. So clearly I have not refused to adapt to the "proper" terminology. In fact I swore to him I would not use my terminology again, until (as you point out) I perceived his comment as and insult, maybe it wasn't. Quote for me one question that you claim I did not confront, and I will directly confront it.

[John Cuthber]

John

 

I must be missing something. I don't see how anything you stated is relevant. The equation you posted doesn't contain division by zero. A line approaching zero, is not the same thing as division by zero (what your suggesting here). X must have a value for Y to be determined. I agree, what's that got to do with division and multiplication by zero.

 

Just because you have no one to give your pie to, (therefore your action is undefined), does not mean you are not holding that pie. And as I pointed out several times, "semantically" speaking, the question always is, what do you hold. Not "what did you do" "was your action definable".

 

 

Upon further thought....Is not division by fractions the very definition of "one to many." 1/.01=100...

 

Is not this 100 actually pieces of the original 1. (attach units such as $).

The "Many" and "One" don't refer to the values, but how many answers there are.

That polynomial is a many to one function because three different numbers (1,2 and 3) all give the same answer.

The point is that Many inputs give One output.

 

Also you have not addressed the objections raised.

My objection was that you couldn't tell the lawyer what to do.

You can't.

So you have not defined a process for division by zero.

[ajb]

Conway, as an example of a many-to-one mapping let us consider two sets S1 =(a,b,c,d) and S2 = (A,B,C) and f : S1-> S2

 

f[a] = A

f = A

f[c] = B

f[d] = C

 

I now want to invert this map, and I will denote this as F : S2 -> S1. You see I cannot do this properly; F = c and F[C] =d is find, but what about F[A]?

 

Well, just like f, F is not really a function we have 'branch points' here meaning that F[A] = a or b and to make a function I have to decide which one to pick. Without some further reasons I cannot make a meaningful choice. I cannot properly invert f.

 

Now compare this with multiplication by zero. You see there is no proper way to invert this operation; not without making some artificial choices and loosing some basic rules of 'numbers'.

Edited July 9, 2015 by ajb