I'm very interested in the number 0
And I was not very satisfied when divison of zero is undefined (except in projective geometry, but the "division sign" does not mean the division in a simple sense)

One day I came up with this
Given
0N=0 ---(*) (where N is any real number)
(But we will consider N=0 separately below unless specified as it seemed to does not work)

If we define a new relation (0-1) which act as the inverse of "multiplied by 0" (Note 0-1 =/= 1/0), then the above becomes
N=0-1 ---(1)

Now given
0/N=0 (where N is any real number) ---(2*)
(However unless specified, we only consider N=/=0)

(0)(1/N)=0
Since 1/N=/=0, it is just a constant
Using (1), it now becomes
1/N=0-1 ---(2a)

Furthermore using (1)

Therefore (1)=(2a)
and ()

Therefore I get two bizarre results
1/N=N
1/0-1)=0-1)
(N=/=0)

Now consider
1/0
By (*) we get
1/0=1/0N
=(1/0)(1/N)
Using (3), it now becomes
(1/0)(N)
=N/0
Therefore another bizarre result
1/0=N/0---(4)
Since N is any real number therefore
1/0=2/0=3/0=1.5/0=root(2)/0 ... EXCEPT 0/0

Now consider when N=0 in (*), i.e.
(0)(0)=0
Using the relation invented in (1)
0=0-1 (N=0) ---(5)

Unfortunately, I still get nowhere when dealing with 0/0
0/0
=0(1/0)
Using (4)
=0(N/0)
When N=0
Using (5)
=0-1(N/0-1)


For Tl;dr
Red means some possibly flawed steps and green means steps that are proved true (NOTE if one step is flawed that means the subsequent steps are also flawed, however I only highlighted the steps that seemed to assume something)

Please tell me what's wrong in this "proof" and debunk if necessary

You started with

0 \times N = 0

But to retain this equation when you introduce 0^{-1} you need to multiply both sides by the inverse:

0^{-1} \times 0 \times N = 0^{-1} \times 0

It does not follow from this that N=0^{-1}.

You haven't specified what the value of 0^{-1} \times 0 is.

You can't have 0^{-1} \times 0 = 1, because then you'd have N=1, but since N can be an arbitrary number this is a contradiction that tells you that your definition of the "multiplicative inverse of zero" must be wrong.

This is why 0/0 is undefined, by the way.

I'm not sure whether I've got the right knowledge, or interpret correctly of the properties of inverses

What I know about inverses:
They are symmetrical about y=x
f-1(f(x))=f(f-1)(x))
Only 1-1 f(x) have inverses as functions (without restricting domains)

By this bolded property, is it correct to assume (0)0-1 just cancel each other out?

EDIT: Nvm, reading your edited statement
EDIT 2: Ah I understand now, if I tried to use the inverse on both sides, then N=1. I can't just left that zero on the other side untouched

This also raise more questions related to 0

1. If I say f and f-1 cancel each other's actions out, is it means that I am actually saying f(f-1)=1?

2. Actually what is the detailed prove of 0N=0 ,especially 0x0=0?

3. 0/0, 1/0, 2/0 etc. are undefined (in standard definition of "division"). They LOOKED different, is it because of how we WRITE them? (compare with 1=0.9999999...)

4. Is other undefined e.g. d/dx|x| when x=0, sin-1(3),log0 etc. differ from the divisions by zeros i)algebrically, ii)geometrically?

5. Is it possible to invent a new no. line where divisions by zeros are defined, in a similar way when mathematicians imagine/define the no. i=root(-1) (where complex numbers turns out to have practical applications)?If yes, where will this line/number system be located relative to other lines/number systems? (e.g. the imaginary line is perpendicular to the real line)

6. I understand why 0/0 is undefined, but why 00 is however, commonly defined to be 1 yet it cause no problems (even it is technically undefined)?

7. Is it possible to have a number system of base 0 or negative (or other non interger numbers)?

8. a/0 where a=/=0 can be defined as +-infinity or unsigned infinity depending on the maths field you are considering. But for 0/0 is it undefined for all known fields of maths?

I got this...

ON/1=0/N

Added it to some soup, and it was very tasty!

1/0 doesn't exist in the Reals, so treating it like a number is like saying "What is the x which satisfies x + elephant = table?", it's nonsense. If you treat it like a normal number then you can 'prove' all sorts of false things.

Remember, a number like 1/3 is not "1 divided by 3", it is "The UNIQUE number which when multiplied by 3 gives 1", 1/3 is just a nice notation. Likewise with any rational number a/b, it is the UNIQUE number which when multiplied by the integer b gives the integer a. For instance, 1/3 and 3 form a pair, where one is the unique number which multiplied by the other gives 1. Does this happen with 1/0? Suppose you said 0 multiplies 1/0 to give 1. What multiplies 2/0? Well 2/0 = 1/(0/2) = 1/0 so it's partner is 0 again. But then that's not a unique pairing. So allowing there to be a partner to 0 makes for all sorts of issues. That is why there is no multiplicative inverse of an additive identify in such types of fields.

I'd suggest reading up on rings, groups, principle ideal domains and fields. There's all sorts of fundamental reasons why the Reals are what they are and don't include some 1/0 ~ infinity element (for one that would make them compact under the usual topology, ie morphic to a sphere, and there's a can of worms you don't want to let out of the bag).

This also raise more questions related to 0

1. If I say f and f-1 cancel each other's actions out, is it means that I am actually saying f(f-1)=1?

In general

f(f^{-1}(x)) = f^{-1}(f(x))=x

But that's not exactly what you're doing here. We already have as a given that the inverse function of multiplication is division.


2. Actually what is the detailed prove of 0N=0 ,especially 0x0=0?

I think it's actually an axiom of arithmetic that requires no proof. It's a definition of what we mean by "0". That is, it is a given that:

0 + x = x + 0 = x
0x = x0 = 0


3. 0/0, 1/0, 2/0 etc. are undefined (in standard definition of "division"). They LOOKED different, is it because of how we WRITE them? (compare with 1=0.9999999...)

If they are undefined, then there's no way to compare them to say if they are different or the same.


4. Is other undefined e.g. d/dx|x| when x=0, sin-1(3),log0 etc. differ from the divisions by zeros i)algebrically, ii)geometrically?

Geometry is approximately reducable to algebra, so if something is algebraically undefined then it is probably geometrically undefined too.

You can't compare two undefined things to say whether they are the same or different.


5. Is it possible to invent a new no. line where divisions by zeros are defined, in a similar way when mathematicians imagine/define the no. i=root(-1) (where complex numbers turns out to have practical applications)?If yes, where will this line/number system be located relative to other lines/number systems? (e.g. the imaginary line is perpendicular to the real line)

It is only a convention to graph the imaginary axis perpendicular to the real axis (that's called an Argand diagram). There's no sense in which imaginary numbers are "really" perpendicular to real numbers.

I'm not sure about defining systems where division by zero is allowed.


6. I understand why 0/0 is undefined, but why 00 is however, commonly defined to be 1 yet it cause no problems (even it is technically undefined)?

In general x^0 = 1 for all x. As you say, 0^0 is undefined, by you can define it to remove the discontinuity in the function f(x) = x^0 if you want to.


7. Is it possible to have a number system of base 0 or negative (or other non interger numbers)?

A base 0 number system is impossible. I'm not sure what use a negative-base number system would be.


8. a/0 where a=/=0 can be defined as +-infinity or unsigned infinity depending on the maths field you are considering. But for 0/0 is it undefined for all known fields of maths?

Probably not. Not my area of expertise, though.

Secret,

Divide by zero equal with :spank:

There is a "Point" often missed. Zero was actually a number created to fill the void, initially a "radix point" was used to reference the absence of a value. So really it shouldn't be dividing by zero, but dividing by a radix point.

In the realm of positive integers, x/y means to count the number of times that y can be subtracted from x before x is reduced to 0 (including any fractional part of y that must be subtracted to reach 0).

Thus, x/0, where x>0, equals ∞, for the subtactions will never reduce x to 0 despite ∞ number of subtractions.

And, x/0, where x=0, equals 1, for we will count one required process of subtraction and the remainder will equal 0.

In these two contexts, I have no trouble with division by 0.

And, x/0, where x=0, equals 1
No.

This is why 0/0 is undefined, by the way.


0/0 is undefined yet 0*0=0 ?

1/0 doesn't exist in the Reals, so treating it like a number is like saying "What is the x which satisfies x + elephant = table?", it's nonsense. If you treat it like a normal number then you can 'prove' all sorts of false things.

Remember, a number like 1/3 is not "1 divided by 3", it is "The UNIQUE number which when multiplied by 3 gives 1", 1/3 is just a nice notation. Likewise with any rational number a/b, it is the UNIQUE number which when multiplied by the integer b gives the integer a. For instance, 1/3 and 3 form a pair, where one is the unique number which multiplied by the other gives 1. Does this happen with 1/0? Suppose you said 0 multiplies 1/0 to give 1. What multiplies 2/0? Well 2/0 = 1/(0/2) = 1/0 so it's partner is 0 again. But then that's not a unique pairing. So allowing there to be a partner to 0 makes for all sorts of issues. That is why there is no multiplicative inverse of an additive identify in such types of fields.

I'd suggest reading up on rings, groups, principle ideal domains and fields. There's all sorts of fundamental reasons why the Reals are what they are and don't include some 1/0 ~ infinity element (for one that would make them compact under the usual topology, ie morphic to a sphere, and there's a can of worms you don't want to let out of the bag).

interesting post there sport . Likey Likey . Stupid people can understand that one . Very good

0/0 is undefined yet 0*0=0 ?

x / 0 is undefined for any x because x * 0 = 0 for any x.

x / 0 is undefined for any x because x * 0 = 0 for any x.

Anyone who at least passed an algebra class knows that you are correct, but the real question is whether you are correct by decree rather than by necessity. Specifically, let's recognize that the numbers zero and infinity are unlike any other numbers, one with no count and the other with no countable end. So, why should they obey all the rules of other numbers?

I think it's important to experimentally and logically work out what various operations with these numbers might mean in each individual context in which they occur. The word "undefined" is merely one such view. Other views might someday prove profitable.

Zero is a man made number on its own. All things have an opposite, and the Universe is made from opposite particles. Zero is a result of two opposites, but on its own it is being used as a cause, and not an effect.. zero is never a cause, it is always an effect. So zero on its own doesn't exist.

Infinity however isn't a number, so infinity doesn't have the same problem. If infinity is just used as an approximation of top end, with no known top end.. so infinity is ?. You are allowed to use infinity on its own. There are some mistakes made with infinity however.. infinite black holes.. black holes, aren't infinite, they have a top end which isn't really that high at all. Some other mistakes that I can't think of at the moment.

0/0 is undefined yet 0*0=0 ?Because the integers form a ring but not a field.

"0" is only a "place holder" I was taught (me thinks) - not actually a "number". Possibly the "Radix" mentioned earlier means the same. I have no real number education, heck I still have not figured out "interest on a loan" - I mean how can it be worth more, yet buy less ?!! Geezey Peezey, I digressed!

questions of mathematical rigor for me are best approached by the two interminable mathematical applications of my life, food and money, always gnawing at me, always forcing me to stop hoping the obvious and the impossible will merge and I will wake up as the center of the universe instead of its edge.

Food works best for division problems, money for subtraction.

So if I start with a pie, I prefer pecan, but they are hard to slice, so maybe leave that for integer division. say I have a key lime pie and I wanna divide it by 0. okay I'm gonna start with my infinitely thin knife so I don't have to account for truncation error. so of course this is very theoretical since I cant afford the key lime anyway so the infinitely thin knife is completely reasonable. I will for a moment at least temporarily believe that I have created a monster because such a device will never impart the force required but wait there's hope because I would use infinite force.

now you're sitting here at the table with me ogling my pie and so I feel kind of obligated to ask you if you want some. you sense a discourtesy of some kind, and say, well, just a teeny tiny slice. More out of decorum than actual magnitude of your hunger.

So I gleefully set off in pursuit of the slicing operation that produces the infinitessimal for you, leaving me with an amount approaching unity. You notice maybe that every slice I remove I then slice in half again. It occurs to you that this pie will become radioactive with age before you get a chance to savor it.

So, tapping at your Rolex, you inquire into the status of your slice. Am I done yet? Well let's see, let me add up my remainders and see if I arrive at unity. So I go: 1/2 for me 1/2 for you, slice, 1/2 + 1/4 for me, 1/4 for you, slice, etc, pretty soon the band has packed and left, then the barflies, so you say hey, I won't live to eat it, this won't work.

So we simply agree upon the possibility of an infinitely small piece, somewhere out in the infinite future. So then you say, man, it would be much easier to start with two infinitely thin knives, applied with infinite force, separated by an infinitely small gap. I say, sure, why not, so I give you nothing in one operation, and gladly consume the rest.

Over coffee, the elixir of ponderment, you ask, how many pieces of infinitely small pie could I get with that little gadget, you know, being a patent attorney, so you get these dollar signs in your eyes, you might just buy Rolex Corp. cuz you'll be sitting on an infinite stack of Benjamins.

So I ponder this, and conclude that indeed you will indeed get an infinity of zero width slices but i'm not sure of the par value in, say, pennies per slice. Or gold atoms per slice. Whatever.

So you then stumble upon Ponzi squared, which goes like this: for each zero width slice you can subdivide them again, each producing an infinite subset of infinitely infinitely small slices. But then you can do Ponzi to the power infinity, then that to infinity, and so on and so on...

So yeah, I think you're onto something there. *burp*

"0" is only a "place holder" I was taught (me thinks) - not actually a "number". Possibly the "Radix" mentioned earlier means the same. I have no real number education, heck I still have not figured out "interest on a loan" - I mean how can it be worth more, yet buy less ?!! Geezey Peezey, I digressed!


So then 0 doesn't represent nothing in itself (definitely not 0 is a figure), perhaps then {nothing/nothing}\,\neq\,{undefined}.

So then 0 doesn't represent nothing in itself (definitely not 0 is a figure), perhaps then {nothing/nothing}\,\neq\,{undefined}.

Manipulating quotes is not the cleverest of options on a forum.

Please take a look at the history of Zero before responding.
http://yaleglobal.yale.edu/about/zero.jsp

Does 0 represent nothing when it's a mathematical abstraction - yes or no?

Abstractions e.g. x/0=undefined, 0/0=undefined, x is a variable.

Thanks for the article. :bugeye:

0 = nothing/ zero/ none.
What did you think it represents?

If I have 0 apples it means there is nothing in my current stock of apples.

I asked does it represent 'nothing' in of itself or always e.g. in mathematical abstractions.

Er, 0 is a mathematical term.

I know that and that's not what I asked x=variable so if x/0=undefined does that mean nothing/nothing=undefined?

I know that and that's not what I asked
Really?

I asked does it represent 'nothing' in of itself or always e.g. in mathematical abstractions.


x=variable so if x/0=undefined does that mean nothing/nothing=undefined?
0/0 is undefined.
As has been pointed out in this thread (post #2) which YOU have quoted and replied to.

So? I see you have avoided answering my question.

If I'm "avoiding" it it's because I have no idea what you're talking about.
0/0 is undefined.

I know that 0/0 is undefined but I'm asking about 'nothing' since it represents 'nothing' what about nothing/nothing?

Oh, okay.
What's fish/ cabbage?
What's sand/ steel?
What's water/ water?

You're applying mathematics to "objects". It doesn't work. It's meaningless.

No integer represents those "objects" the way 0 represents 'nothing'.
'Nothing' itself is a lack of quantity.

No single digit number represents those "objects" the way 0 represents 'nothing'.
It doesn't matter.
0 in mathematics is a mathematical term.


'Nothing' itself is a lack of quantity.
"Nothing" is a lack of existence. (Unless speaking mathematically, then it's a lack of quantity).

What's nothing * sand?
Meaningless.

Quantity isn't limited to mathematics it's also in philosophy.

Really? As a non-mathematical term?

Are you saying 'nothing' cannot be used in logic which is a subset of mathematics and vice-versa.

Are you saying 'nothing' cannot be used in logic which is a subset of mathematics and vice-versa.
Nope.
But I am saying, and have said, that mathematics doesn't apply to non-mathematical instances.

How does your question act as a reply to mine?

Really? As a non-mathematical term?

Nope.
But I am saying, and have said, that mathematics doesn't apply to non-mathematical instances.

How does your question act as a reply to mine?


0/0 is undefined but only as an abstraction perhaps one of the possible ways or the only way it can represent nothing.

Then that's your problem.
Since mathematics is what deals with quantity. (The philosophical term meaning something else entirely).

Ok, I have edited my post.


0/0 is undefined but only as an abstraction perhaps one of the possible ways or the only way it can represent nothing.

No, 0/0 is undefined as a mathematical term.
(Which is about the only place it can occur).

Yes, it is undefined but undefined could be one of possible answers or 0 can't represent 'nothing' in all cases (not talking about philosophy.) It's afterall only an abstraction.

Undefined is the answer.


0 can't represent 'nothing' in all cases (not talking about philosophy.) It's afterall only an abstraction
Huh?
And it's an abstraction as much as "1" is.

Sure, yet 1 can represent something physical e.g. 1 dog and 0 can represent no dogs but what does x/0 represent where x is a variable or 0/0?

Sure, yet 1 can represent something physical e.g. 1 dog and 0 can represent no dogs but what does x/0 represent where x is a variable or 0/0?
Um, the "1" represents the quantity (non-physical) of dogs. The word "dog" represents the physical dog.

What does 3/2 represent?

Are you saying quantities cannot represent physical objects.

3/2 can represent one and a half of something.

Quantities do NOT represent physical objects.
They represent the number of physical objects.

3/2 represents NO physical object until you add a noun (and it's the noun which, as stated, actually represents the object itself).
3/2 fish.
3/2 dogs.
3/2 doughnuts.
What physical object is represented by the "3/2" in all of those sentences at the same time? None. Not a one.

Agreed, yet they can consistently be used with the nouns that represent something physical. How can x/0 or 0/0 be used consistently with a noun to represent something physical? Let there be 8 dogs, if I were to multiply 8 by 2 then i get 16 lets let there be 16 dogs if I multiply 0 by 8 I get 0 let there be no dogs.

Agreed, yet they can consistently be used with the nouns that represent something physical. How can x/0 or 0/0 be used consistently with a noun to represent something physical?
There are many mathematical terms that can't be used with something physical, why should this be anything special?

Can I ask for 2i+3 gallons of petrol?
How about 0/1 lb of beef?

I can easily 'use' your examples for something physical.

Go ahead.

There are many mathematical terms that can't be used with something physical, why should this be anything special?

Can I ask for 2i+3 gallons of petrol?
How about 0/1 lb of beef?

i is a variable why shouldn't I be able to 'use' it, and 0/1=0 so 0lb of beef.

i is not a variable.
It's the square root of -1.

And where, exactly, would you ask for 0lb of beef?

Why should I have to ask for 0lb of beef I can picture there is no beef. ;)

Yet you wouldn't use the term 0/1 lb of beef.
Neither would you ask for 0 lb of beef.

Does that equate it to being 'unusable' for a physical object? I think not, I can still imagine it.

I can imagine green unicorns with elephant ears and a shotgun.
Does that make them "usable" at all?

By 'usable' I mean it's conceivable. i is clear in that it doesn't beg any question from one but 0/0 "makes" one ask questions about 'nothing'.

I still don't quite the "problem".
Unless it's just that you have some desire for it to "make sense" in a "real" fashion.

Shit happens.
If infinity is the biggest "possible number" you can have how is it possible (or even conceivable) to have something larger than infinity?
Yet we do and it's a solid part of mathematics...

I still don't quite the "problem".
Unless it's just that you have some desire for it to "make sense" in a "real" fashion.

It's more I am curious whether the mathematical statements can be realized to be physically 'usable' hypothetically. I would like to know it either can be realized or cannot be realized or 'other'.


Shit happens.
If infinity is the biggest "possible number" you can have how is it possible (or even conceivable) to have something larger than infinity?
Yet we do and it's a solid part of mathematics...

The "biggest possible number" is in question can it be realized hypothetically whether there is a "biggest possible number", it seems more that one can always add 1 to a number no matter how big it is hypothetically.

It's more I am curious whether the mathematical statements can be realized to be physically 'usable' hypothetically. I would like to know it either can be realized or cannot be realized or 'other'.
But numbers aren't required to be usable physically.


The "biggest possible number" is in question can it be realized hypothetically whether there is a "biggest possible number", it seems more that one can always add 1 to a number no matter how big a number is hypothetically.
Which doesn't address the question of numbers larger than infinity.

But numbers aren't required to be usable physically.


Far from requiring I am merely curious if x/0 is undefined is hypothetically realized to be physically 'usable', it's like being curious about the truth.


Which doesn't address the question of numbers larger than infinity.


For now is it hypothetically realized whether there is a "biggest possible number", knowing that should better the understanding of infinity.

If infinity is the biggest "possible number" you can have how is it possible (or even conceivable) to have something larger than infinity?
Yet we do and it's a solid part of mathematics...There is no single 'infinity', there's actually infinitely many infinities, many of which can be proven to be larger than others. For example, \aleph_{k+1} \equiv 2^{\aleph_{k}} > \aleph_{k}. Another way of saying that is that given a non-empty set it's power set is always strictly larger than it. Things get worse when you get into the realms of \aleph_{\aleph_{0}} and ordinals. They have issues like \omega + 1 \neq 1 + \omega.

Shh! I was over-simplifying it. ;)

By 'usable' I mean it's conceivable. i is clear in that it doesn't beg any question from one but 0/0 "makes" one ask questions about 'nothing'.Not to a mathematician. The only people who really struggle with 1/0 and the like are people who think their high school education told them all they need to know about this stuff. Often there's a significant intersection with the people who don't grasp that 1 = 0.9999.... is true.

As I said earlier in the thread, formally there is no division in arithmetic, it's multiplication by more advanced concepts. In the integers there is no division concept because there's no whole number X which satisfies 2X=1. You build the rationals by noting you can write such an expression using integers and thus the rationals are those X which satisfy things of the form aX=b for a,b integers and a not equal to 0. Then you can build the algebraics using higher order polynomials, ie \mathbb{Z}[x] is the ring of polynomials with integer coefficients and the algebraic numbers are those which satisfy equations of the form f(x) = 0 for f(x) \in \mathbb{Z}[x]. You can define the reals by the Cauchy completion of the rationals. No where have I needed to mention division, only multiplication and addition. Heck, even subtraction is just defined by addition, in the same way. I define a number X by saying it satisfies X+5=0. Only used addition there. Now what is 7+X? Well I know 7 = 5+2 so 7+X = 2+5+X and I know that 5+X=0 so 7+X = 2+0 = 2. So 7+X=2. Only done using addition, no need to actually say "X is equal to minus 5" or "Adding X is like subtracting 5". That's how you're told about it in school but it's a short hand, it gets you through the day but it's not fundamental.

It might seem a bit pointless and I thought that when I first sat lectures on this stuff. It seemed to be proving obvious things with stupidly lengthy methods but it's how you do things right and it's important when you get onto more complex stuff like algebraic geometry and Galois theory.

0/0 shouldn't make you think about "What is infinity?" or "Is maths broken?", it should make you think "What does 1/3 or 1/7 mean? Therefore what does 1/0 mean? Oh, it is no more a number than 'elephant' or 'blue'! Problem solved".

Does this happen with 1/0? Suppose you said 0 multiplies 1/0 to give 1. What multiplies 2/0? Well 2/0 = 1/(0/2) = 1/0 so it's partner is 0 again. But then that's not a unique pairing.

Below is my faulty prove of the bolded statement using just multiplication and the knowledge learnt from the pms, red are the problematic steps. (Put this here cause it seems generally relevant, unlike the other items in the pms)

Define
a0=1 ---(1)
b0=2 ---(2)
ac=1 ---(3)
bd=1 ---(4)

(3)*0
ac0=1*0
1c=0
c=0 ---(5)

(4)*0
bd0=1*0
2d=0 ---(6)

Put (5) into (2)
bc=2 ---(7)

Put (6) into (1)
2ad=1 ---(8)

Put (7) into (8)
abcd=1 ---(9)

(9)*2
2abcd=2 ---(10)

Put (5), (6) into (10)
abc(2d)=2
ab0=2 ---(11)

Put (2) into (11)
ab0=b0
(a-1)(b0)=0

Using cancellation property
Since b0=2=/=0 (given)
(prove of 1a=/=0?)
Therefore a-1=0
a=1 ???!

BUT using another approach
(1)*2
2a0=2
a0=2---(12)

(2)=(12)
a0=b0
a0-b0=0
(a-b)0=0

Using cancellation property (if pr=qr, then p-q=0 => p=q or r=0 if r,p,q=/=0 and pq=/=0) Since r=0 in this case, therefore p=/=q
Therefore
a=/=b

P.S. Alpha's quote here does not mean this post is a response to the comment, rather it is a reference tag on what this prove is about

Secret:

In your post above, is o supposed to be zero, or is it just another variable denoted by the letter 'o'?

If it is the former, then your definitions of variables appear to be inconsistent. If the latter, then your post is very confusing at best and probably wrong at worst. I haven't checked through the whole thing yet.

The 0 is supposed to be 'zero' but sciforums tend to flatten it into the letter 'o'

And the a,b,c,ds are actually 'numbers'. I'm trying to apply the 'multiplication only' way learnt here to seeing how the 'numbers' move around to form the statement that 2/0=1/(0/2)=1/0 (I understand this is valid but I want to see how it can be proved in the fundamental level mentioned by alpha, so it will be convincing enough to myself to stop trying to mess with the n/0 s)

The definition should be read as follows:
E.g a0=1 means 'a unique number which multiplies by 0 equals 1'

If my prove can yield the aforementioned statement, it means I have proved by contradiction how the n/0 s break down the unique pairing of multiplication and it's inverse. In theory that will settle anything regarding the n/0 s (with the possible exception of 0/0 since it cannot be expressed in terms of the n/0 s and thus 'a bit different')

To put it simply:

Aim of the prove: prove that 2/0=1/(0/2)=1/0
using symbols according to the definitions a=b(=1/d)

Since a,b,c,d are defined to be unique numbers, the a=b result will contracdict the definitions, thus it can then be proved it is impossible to have unique pairing involving n/0 s

The problem is however, I get bizarre result and I've no idea where have I got it wrong.

Ok. Let's take a look...


Below is my faulty prove of the bolded statement using just multiplication and the knowledge learnt from the pms, red are the problematic steps. (Put this here cause it seems generally relevant, unlike the other items in the pms)

Define
a0=1 ---(1)
b0=2 ---(2)

Before we go any further, please explain to me how a finite number times zero can equal 1 or 2.

By definition x \times 0 = 0 \times x = 0 for all finite x.

Ok. Let's take a look...

Before we go any further, please explain to me how a finite number times zero can equal 1 or 2.

By definition x \times 0 = 0 \times x = 0 for all finite x.
Yup

To explain your question, have a look at alpha's attempt in explaining to me why and how the n/0 s failed to be defined



Remember, a number like 1/3 is not "1 divided by 3", it is "The UNIQUE number which when multiplied by 3 gives 1", 1/3 is just a nice notation. Likewise with any rational number a/b, it is the UNIQUE number which when multiplied by the integer b gives the integer a. For instance, 1/3 and 3 form a pair, where one is the unique number which multiplied by the other gives 1. Does this happen with 1/0? Suppose you said 0 multiplies 1/0 to give 1. What multiplies 2/0? Well 2/0 = 1/(0/2) = 1/0 so it's partner is 0 again. But then that's not a unique pairing. So allowing there to be a partner to 0 makes for all sorts of issues. That is why there is no multiplicative inverse of an additive identify in such types of fields.

My interpretation of the above is as follows:

1. A number like 1/3 is not "1 divided by 3", it is "The UNIQUE number which when multiplied by 3 gives 1

Can be written as
Define
d be the UNIQUE number where 3d=1

2. Likewise with any rational number a/b, it is the UNIQUE number which when multiplied by the integer b gives the integer a

Can be written as
Define
p be the UNIQUE number where pb=a

3. Suppose you said 0 multiplies 1/0 to give 1. What multiplies 2/0? Well 2/0 = 1/(0/2) = 1/0 so it's partner is 0 again.

Can be written as
Define
r be the UNIQUE number where 0r=1

Using similar logic, it should be reasonable to define 2/0 as follows:
Define s be the UNIQUE number where 0s=2

Thus alpha's entire post can be reduced to:

Define d be the UNIQUE number where 3d=1
Likewise, define p be the UNIQUE number where pb=a
For instance: 3 and d forms a unique pair
where 1 is the UNIQUE number where 1*1=1

Does this happens with 1/0?
SUPPOSE we define r be the UNIQUE number where 0r=1
What multiplies s, where 0s=2?
Well s=1/(0/2)=r so 0s=1
Therefore both s and r pair with 0 and that is not a UNIQUE pairing
"So allowing there to be a partner to 0 makes for all sorts of issues. That is why there is no multiplicative inverse of an additive identify in such types of fields."


The keyword here is "SUPPOSE"
From all the maths I've learnt so far, alpha is probably using "proof by contradiction" to illustrate how the n/0 s be invalid



Suppose A is true
(A series of steps)
As there is a contradiction, A must be false

Therefore despite both me and alpha know that 0x=x0=0 for all x in R, it is likely alpha assume the existence of a multiplicative inverse of zero (i.e. a number w where w0=0w=1) and then proof by contradiction to show that they cannot exist (at lest in R, the reals)

My faulty proof is trying to simplify 2/0=1/(0/2)=1/0 into just multiplication, but so far unsuccessful

So I might misinterpret alpha's message somehow, but i'm not sure

Secret:

I think the gist of Alphanumeric's post is that if you assume that there is a UNIQUE number x such that x0 = 1, so that x is "paired" with 1/0, then if you consider y0 = 2, you should be able to prove that y = x. But x is then also paired with 2/0, meaning that it is not a UNIQUE inverse.

I think that's what he's saying, anyway.

So the question is how to prove y=x step by step and only with multiplication

Clarification: What I mean is how to use multiplication only to prove y=x? (where y and x follows the conditions stated in post #71)

Therefore despite both me and alpha know that 0x=x0=0 for all x in R, it is likely alpha assume the existence of a multiplicative inverse of zero (i.e. a number w where w0=0w=1) and then proof by contradiction to show that they cannot exist (at lest in R, the reals)No, they explicitly do not exist by construction. We want a number system with the following properties (among others)

Let S be some set of objects and # and @ be two binary operations.

- If x and y are in S then x@y and x#y are too
- If z is in S then x@(y@z) = (x@y)@z and likewise for #, so we don't need to write the brackets, x@y@z is well defined etc
- There is some e such that e#x = x = x#e for all x in S and likewise there is some f such that f@x = x = x@f for all x in S
- Commutativity in #, x#y = y#x
- Distributivity of @ over #, x@(y#z) = (x@y)#(x@z)

To give an example, just set # to + and @ to *. Then e=0 and f=1, 0+x=x=x+0 and 1*f = f = f*1 and x*(y+z) = x*y + x*z.

No mention of division yet, these are valid in set ups which don't have division like rings (ie such as integers or polynomials). We define division and subtract in similar ways to one another but we need to be careful. Thus far we've said a few things about what S must contain so we're kind of building it as we go.

Given an x suppose we just say that there's an X such that x#X is the associated identity, which for # is e, so x#X=e defines X from x. Likewise y@Y=f defines Y from y. Could we just say that S must contain all such defined X and Y? Seems plausible enough but in fact it isn't because though we haven't stated it explicitly we have actually defined a system where anything multiplied by 0 gives 0, is 0*x=0 for all x in S, while we might think we can say there's an y in S such that x*y=1.

I'll do this formally, so I want to show e@x = e (ie 0*x=0) for all x in S, while we might have thought there's a y such that y@x=f (oe y*x=1). We start with the tautology x=x and go from there

x = x@f - By definition of f
x = x@(f#e) - By definition of e I've used f = f#e.
x = x@(f#f#F) - Here F is the # opposite of f (ie if f=1 and #=+ then F=-1)
x = (x@f)#(x@(f#F)) - Distributivity
x = x#(x@(f#F)) - Definition of f under @
x#X = (x#X)#(x@(f#F)) - #X on both sides and using commuting of # terms
e = e#(x@(f#F)) - Definition of X under #
e = (x@(f#F)) - Definition of e# identity
e = x@e - Definition of f#F

Similarly you can get e = e@x. Thus e@x cannot be anything other than e, regardless of which x you pick from S, so unless e=f (which it only does in the trivial ring) there is no x such that x@e=f or more commonly, there's no x such that 0*x=1. I haven't had to assume it exists and constructed a contradiction, I've shown it cannot exist within those axioms.

To allow for such an element in S you have to throw away one of the axioms I used in the step by step derivation. For example, what if there is no X such that x#X=e (ie x+X=0)?

Suppose we do throw away an axiom to let us include this magical number, let's call it Q. We don't want to break things too much so let's say Q is the only number such that 0*Q=1 (or rather Q@e = f). Well then we are led to the annoying property that Q+x=Q for any x in S. We know by the axioms Q+x is in S so which element is it? We note e@(x#Q) = (e@x)#(e@Q) using distributivity. If x is not Q then e@x=e so using Q@e = f, e@(x#Q) = (e@x)#(e@Q) = e#f = f. So now we have both x#Q and Q satisfying e@y = f but Q is unique, so Q=Q#x for any x not equal to Q. So adding anything to Q doesn't change things so now you have a system where x+1=x can be true, despite it being very akin to 1=0.

As a result of this the 'nice' mathematical number systems people work with are those which allow for x#X=e and y@Y=f to define X and Y in all cases except when y=e. Hence all numbers in the Reals have a negative such that x+X=0 and all numbers except the additive identity (ie 0) have inverses such that y*Y=1.

I'm very interested in the number 0
And I was not very satisfied when divison of zero is undefined (except in projective geometry, but the "division sign" does not mean the division in a simple sense)

One day I came up with this
Given
0N=0 ---(*) (where N is any real number)
(But we will consider N=0 separately below unless specified as it seemed to does not work)

If we define a new relation (0-1) which act as the inverse of "multiplied by 0" (Note 0-1 =/= 1/0), then the above becomes
N=0-1 ---(1)

Now given
0/N=0 (where N is any real number) ---(2*)
(However unless specified, we only consider N=/=0)

(0)(1/N)=0
Since 1/N=/=0, it is just a constant
Using (1), it now becomes
1/N=0-1 ---(2a)

Furthermore using (1)

Therefore (1)=(2a)
and ()

Therefore I get two bizarre results
1/N=N
1/0-1)=0-1)
(N=/=0)

Now consider
1/0
By (*) we get
1/0=1/0N
=(1/0)(1/N)
Using (3), it now becomes
(1/0)(N)
=N/0
Therefore another bizarre result
1/0=N/0---(4)
Since N is any real number therefore
1/0=2/0=3/0=1.5/0=root(2)/0 ... EXCEPT 0/0

Now consider when N=0 in (*), i.e.
(0)(0)=0
Using the relation invented in (1)
0=0-1 (N=0) ---(5)

Unfortunately, I still get nowhere when dealing with 0/0
0/0
=0(1/0)
Using (4)
=0(N/0)
When N=0
Using (5)
=0-1(N/0-1)


For Tl;dr
Red means some possibly flawed steps and green means steps that are proved true (NOTE if one step is flawed that means the subsequent steps are also flawed, however I only highlighted the steps that seemed to assume something)

Please tell me what's wrong in this "proof" and debunk if necessary


all you have here is a bunch of numbers

It's a faulty proof
I'll deal with it later after digesting the new knowledge

@AN
---------
I 'm glad you finally reply:D

Thanks, that fully explained the origin of the axiom 0n=0 for all n in reals and how 1/0 does not exist in the reals.
However, I'm still digesting the details and doing some maths in the background to apply it to similar cases.

IMO This solved the crucial portion of the proof that 0/0 and 1/0 are not real numbers (or even not a number at all). Right now I'm using the knowledge you present to work out the other n/0s, hoping to minimize my disturbance to you as much as possible by avoid asking questions in the process.

Technically, the division by zero problem is not solved yet but it's core is dealt with formally and successfully.

-----------

Consider this thread done, if you like. I'll fix my OP later (even disregard it as false if necessary) and wrote a concluding post for future reference. then this thread will be truly finished

Conclusion
Why division by zero (\frac N0) is undefined in the reals
Short answer
Because our maths teacher said so/Because maths said so

Undefined is the answer.

Medium answer (Credit to Prometheus)
0 \times N=0 for any N
Therefore \frac N0 is undefined for any N

Longer answer (Credit to James R)
Using the axiom (actually it's partially correct, this is explained in the fundamental answer section)
0 \times N=0 (where N is any arbitrary number)

Assume zero has a multiplicative inverse (For newbies, assume we can divide by zero)
\frac 10
Multiply both sides by the inverse you get
0 \times N \times \frac 10=0 \times \frac 10
Now we run into the problem of \frac 10 \times 0
If we said
\frac 10 \times 0=1
Then the result is
N=1
This contradict the given condition that N is an arbitrary number
Therefore zero has no multiplicative inverse (i.e. We cannot divide by zero)

Bit longer answer (Credit to Alphanumeric)(Some portions inspired from Alphanumeric)
Fundamentally, there are only addition and multiplication.

Subtraction can be defined as follows:
E.g. Let x be a UNIQUE number such that
x+3=0 (Additive inverse of 3)
Now find
5+x
We know
5=2+3
Therefore
5+x=2+3+x
Since 3+x=0, therefore
5+x=2+3+x=2
In the above process, notice there is not even a single instance of -3

Division can be defined as follows:
E.g. Let y be a UNIQUE number such that
y \times 4=1 (Multiplicative inverse of 4)
Now evaluate
92 \times y
We know
92=23 \times 4
Therefore
92 \times y=23 \times 4 \times y
Since 4y=1, therefore
92 \times y=23 \times 4 \times y=23 \times 1=23
In the above process, notice there is not even a single instance of \frac 14

Now for the case of the n/0s:
E.g. \frac 10
Let q be a UNIQUE number such that
q \times 0=1
E.g. \frac 20
Let r be a UNIQUE number such that
r \times 0=2
Begin proof:

Let q,r,s and h be four UNIQUE numbers such that
q \times 0=1 ---(1)
r \times 0=2 ---(2)
s \times r=1 ---(3)
h \times 2=1 ---(4)


2/0 = 1/(0/2)

Multiply (3) by 0 and h and using the axiom 0 \times N=0 (except when N=q or r as defined)
s \times r \times 0 \times h=1 \times 0 \times h
s \times 2 \times h=0 or s \times r \times 0=0
s \times 1=0 or s \times 2=0
s=0
For the other case, multiply by h
s \times 2 \times h=0 \times h
s \times 1=0
s=0
Therefore both case are equivalent
s=0 ---(5)

1/(0/2) = 1/0


Put (5) into (3)
s \times r=1
0 \times r=1
r \times 0=1---(*)

Now for the case of \frac 00:
Let d be a UNIQUE number such that
d \times 0=0
Note this is identical to the axiom
0 \times N=0
Therefore any d except q and r satisfy the above equation
As there are at least two distinct d which can give 0 when multiplied by 0
Therefore there is no UNIQUE d
Thus d does not exist in the reals

For (*), there are two interpretations:
1. Since
r \times 0=1 and r \times 0=2
It can be said that r is a number which when multiply by 0 equals 1 and 2 (or by modifying the definition of r, equal to any arbitrary number except 0). Therefore r is not a multiplicative inverse of 0 as it does not give a UNIQUE answer.
Since
q \times 0=1
Therefore there are at least two numbers which give 1 when multiplied by 0.
Therefore 0 has no multiplicative inverse

2. r \times 0=1
Using the axiom 0 \times N=0
r cannot be anything other than q (As other numbers will give zero when multiplied by zero)
Therefore
q=r
(By modifying the definition of r, we can show \frac n0 are all equivalent to \frac 10 thus are all equivalent to each other for any n except when n=0. Therefore there are only two distinct types of division by zeros \frac n0 where n=/=0 and \frac 00)
Since we have shown
q=r
Therefore there are two numbers that can give 1 when multiplied by zero
Therefore neither q and r are multiplicative inverse of zero
Therefore 0 has no multiplicative inverse


Regardless of which interpretation you use, we still showed

2/0 = 1/(0/2) = 1/0
Therefore there is no UNIQUE number n such that n \times 0=1
Therefore \frac n0 does not exist in the reals and remain undefined

Fundamental/TRUE answer(Credit to Alphanumeric)


No, they explicitly do not exist by construction. We want a number system with the following properties (among others)

Let S be some set of objects and # and @ be two binary operations.

- If x and y are in S then x@y and x#y are too
- If z is in S then x@(y@z) = (x@y)@z and likewise for #, so we don't need to write the brackets, x@y@z is well defined etc
- There is some e such that e#x = x = x#e for all x in S and likewise there is some f such that f@x = x = x@f for all x in S
- Commutativity in #, x#y = y#x
- Distributivity of @ over #, x@(y#z) = (x@y)#(x@z)

To give an example, just set # to + and @ to *. Then e=0 and f=1, 0+x=x=x+0 and 1*f = f = f*1 and x*(y+z) = x*y + x*z.

No mention of division yet, these are valid in set ups which don't have division like rings (ie such as integers or polynomials). We define division and subtract in similar ways to one another but we need to be careful. Thus far we've said a few things about what S must contain so we're kind of building it as we go.

Given an x suppose we just say that there's an X such that x#X is the associated identity, which for # is e, so x#X=e defines X from x. Likewise y@Y=f defines Y from y. Could we just say that S must contain all such defined X and Y? Seems plausible enough but in fact it isn't because though we haven't stated it explicitly we have actually defined a system where anything multiplied by 0 gives 0, is 0*x=0 for all x in S, while we might think we can say there's an y in S such that x*y=1.

I'll do this formally, so I want to show e@x = e (ie 0*x=0) for all x in S, while we might have thought there's a y such that y@x=f (oe y*x=1). We start with the tautology x=x and go from there

x = x@f - By definition of f
x = x@(f#e) - By definition of e I've used f = f#e.
x = x@(f#f#F) - Here F is the # opposite of f (ie if f=1 and #=+ then F=-1)
x = (x@f)#(x@(f#F)) - Distributivity
x = x#(x@(f#F)) - Definition of f under @
x#X = (x#X)#(x@(f#F)) - #X on both sides and using commuting of # terms
e = e#(x@(f#F)) - Definition of X under #
e = (x@(f#F)) - Definition of e# identity
e = x@e - Definition of f#F

Similarly you can get e = e@x. Thus e@x cannot be anything other than e, regardless of which x you pick from S, so unless e=f (which it only does in the trivial ring) there is no x such that x@e=f or more commonly, there's no x such that 0*x=1. I haven't had to assume it exists and constructed a contradiction, I've shown it cannot exist within those axioms.

To allow for such an element in S you have to throw away one of the axioms I used in the step by step derivation. For example, what if there is no X such that x#X=e (ie x+X=0)?

Suppose we do throw away an axiom to let us include this magical number, let's call it Q. We don't want to break things too much so let's say Q is the only number such that 0*Q=1 (or rather Q@e = f). Well then we are led to the annoying property that Q+x=Q for any x in S. We know by the axioms Q+x is in S so which element is it? We note e@(x#Q) = (e@x)#(e@Q) using distributivity. If x is not Q then e@x=e so using Q@e = f, e@(x#Q) = (e@x)#(e@Q) = e#f = f. So now we have both x#Q and Q satisfying e@y = f but Q is unique, so Q=Q#x for any x not equal to Q. So adding anything to Q doesn't change things so now you have a system where x+1=x can be true, despite it being very akin to 1=0.

As a result of this the 'nice' mathematical number systems people work with are those which allow for x#X=e and y@Y=f to define X and Y in all cases except when y=e. Hence all numbers in the Reals have a negative such that x+X=0 and all numbers except the additive identity (ie 0) have inverses such that y*Y=1.
More simply:
Let R be some set of objects and # and @ be two binary operations
Let x, y and z be the elements (the objects) of R

R has the following axioms:
1. Given x,y are in R, x#y and x@y are also in R
2. Commutativity in @ i.e. x@y=y@x
3. Commutativity in # i.e. x#y=y#x
4. Distributivity of @ over # i.e. x@(y#z)=(x@y)#(x@z)
5. # identity: a (a#x=x#a=x)
6. @ identity: m (m@x=x@m=x)
9. Associativity in @ i.e. (x@y)@z=x@(y@z)=x@y@z
10. Associativity in # i.e. (x#y)#z=x#(y#z)=x#y#z

Using these axioms we then define:
7. Negative numbers (additive inverses)
Let X be some elements in R such that x#X=a
8. Reciprocals (multiplicative inverses)
Let Y be some elements in R such that y@Y=m

It seems plausible that all X and Y are defined in R
However based on the above axioms of R, we actually implicitly prove that
0 \times x=0 for any x

Formal prove:
x=x
x=x@m=m@x (using 6.)
x=x@(m#a)=(m#a)@x (using 5.)
x=(x@m)#(x@a)=(m@x)#(a@x)(using 4.)
x=x#(x@a)=x#(a@x)
x#X=x#(x@a)#X=x#(a@x)#X (#X on both sides)
a=a#(x@a)=a#(a@x) (using 3. and 7.)
a=x@a=a@x (using 7.)
Replacing a by 0 and @ by * you will get
0 \times x=x \times 0=0 for any x

Since from 6. m@x=x@m=x while from the above we proved that a@x=x@a=a
Therefore
a=/=m
Therefore
a@x=x@a=m
Or present in a more familiar way
0 \times x=x \times 0=1
Is impossible in R

Therefore the multiplicative inverse of 0 (i.e. \frac n0) DOES NOT EXIST in R (which happened to be the reals) due to the consequence of the axioms of R

External resources (Which also deal with other messy zeros such as 0! and 0^{0})
http://www.physicsforums.com/showthread.php?t=530207

As for what happens if the axioms were modified in another set S to allow the \frac n0 to exist is beyond the scope of this thread

Division means , dividing something with some other thing . Here both the things are existing and have some values .

Zero means nothing , empty or non-existence .

So, how can something be divided with nothing ?

Instead 0+ or 0- exists ; which is a very small value .

So, division with 0+ or 0- , should be considered ; instead of 0 .

@hansda

For 1-3rd lines
This is how some illustrate n/0 (except 0/0) is undefined using daily perspective and examples

For 4-5th lines
0 is neither positive nor negative

Assume n is a positive number
If you divide n by 0+ you get a very large positive value
If you divide n by 0- you get a very large negative value

The magnitude of the value tends to infinity as your 0- and 0+ approaches to 0, however the signs remains the same

In mathematical terms, what you are saying is
\lim_{x \to 0} \frac nx = \lim_{x \to 0+} \frac nx = \lim_{x \to 0-} \frac nx

(For layman, \lim_{x \to a} L means "The value L you approach to as x approaches as close as you want/possible to the value a from both sides (i.e. from smaller than a and from larger than a)")

However
\lim_{x \to 0+} \frac nx = +oo
\lim_{x \to 0-} \frac nx = -oo

Both limits does not exist (as infinity is not a number)
Plus
\lim_{x \to 0+} \frac nx =/= \lim_{x \to 0-} \frac nx (As one is positive and one is negative)

As the left hand limit does not agree with the right hand limit, by the definition of limits
\lim_{x \to 0}\frac nx DOES NOT EXIST

For n= negative number, the final result are the same but with the result of the left and right hand limits inverted

However this is not the fundamental reason why n/0 is undefined
(As there exist some types of piecemeal functions that \lim_{x \to a+} f(x) = \lim_{x \to a-} f(x) =/= f(a). Thus you can argue my demonstration above cannot determine whether f(x)=\frac nx where x=0, actually exist)

@hansda

For 1-3rd lines
This is how some illustrate n/0 (except 0/0) is undefined using daily perspective and examples

For 4-5th lines
0 is neither positive nor negative

Assume n is a positive number
If you divide n by 0+ you get a very large positive value
If you divide n by 0- you get a very large negative value

The magnitude of the value tends to infinity as your 0- and 0+ approaches to 0, however the signs remains the same

In mathematical terms, what you are saying is
\lim_{x \to 0} \frac nx = \lim_{x \to 0+} \frac nx = \lim_{x \to 0-} \frac nx

(For layman, \lim_{x \to a} L means "The value L you approach to as x approaches as close as you want/possible to the value a from both sides (i.e. from smaller than a and from larger than a)")

However
\lim_{x \to 0+} \frac nx = +oo
\lim_{x \to 0-} \frac nx = -oo

Both limits does not exist (as infinity is not a number)
Plus
\lim_{x \to 0+} \frac nx =/= \lim_{x \to 0-} \frac nx (As one is positive and one is negative)

As the left hand limit does not agree with the right hand limit, by the definition of limits
\lim_{x \to 0}\frac nx DOES NOT EXIST

For n= negative number, the final result are the same but with the result of the left and right hand limits inverted

However this is not the fundamental reason why n/0 is undefined
(As there exist some types of piecemeal functions that \lim_{x \to a+} f(x) = \lim_{x \to a-} f(x) =/= f(a). Thus you can argue my demonstration above cannot determine whether f(x)=\frac nx where x=0, actually exist)

What i mean to say is that , in reality '0' does not exist ; because '0' means absence or non-existence .

Instead 0+ or 0- exists in reality . However small they may be but they have some value of their existence .

Dividing a number with 0+ or 0- is possible ; though the result will be infinity plus or minus . That is a different issue .

But dividing a number with 0 is just not possible . This will not make any sense also .

If it is considered that , l 0+ l = l 0- l ; then N/( l 0+ l ) = N/( l 0- l ) .

Here l x l means mod x and N is any number .

What i mean to say is that , in reality '0' does not exist ; because '0' means absence or non-existence .

Instead 0+ or 0- exists in reality . However small they may be but they have some value of their existence .

Dividing a number with 0+ or 0- is possible ; though the result will be infinity plus or minus . That is a different issue .

But dividing a number with 0 is just not possible . This will not make any sense also .

If it is considered that , l 0+ l = l 0- l ; then N/( l 0+ l ) = N/( l 0- l ) .

Here l x l means mod x and N is any number .

Even in such scenario, |0+|=|0-|=/=|0|=/=0
Thus N/|0+|=|N/0-| does not imply N/|0|

Thus N/0 is still undefined

Your application suggested above seemed to have something to do with surreal numbers, however I do not understand the specifics, thus I cannot comment any further on your suggestion without getting things wrong

http://en.wikipedia.org/wiki/Surreal_number

@Alphanumeric:
The hypothetical number Q you mentioned in the aside of your proof seemed to have something more pathological than just 0=1. I'll post the details of my investigation later when I have time

Note that I don't mean your proof is wrong, but that your proof is right and the "What if" scenario for the properties of the number Q is much nastier than what is presented in the post

Of course, once you open that door even a tiny bit then all manner of pathological behaviour comes rushing in. Once you can 'prove' something contradictory then you can prove anything. It's known as the principle of explosion (http://en.wikipedia.org/wiki/Principle_of_explosion).

(This is the promised analysis that have been delayed from posting due to busy uni life)


Pathological properties introduced if n/0 is allowed
It has been previously shown and proved that the reals (and complex numbers) does not fundamentally allow division by zero

For the purpose of the illustration of the pathological "can of worms" that division by zero can introduce, however, lets give the following definition:

Let q be the only element ∉ C such that
q x 0 = 1

Using the definition of multiplicative inverse and Def 1

For any number, a is a multiplicative inverse of b iff axb=bxa=1 and a is unique
q should be an inverse of 0
Therefore q-1=0

Other definitions and results that would be used in the illustration:

A number a is an multiplicative identity iff axn=n and nxn-1=a where n is any number

Property of 0 in the set C: 0xn=0 for any n∈C

Property of 0 in the set C: 0xn=0 for any n∈C

Commutative law for complex numbers: axb=bxa, a+b=b+a

Property 1: Multiplication of q is non associative
Assume we are given the following expression and were asked to evaluate it
qx0xn

If we consider
(qx0)xn
=1xn (Def 1)
=n (Def 3)

However if we consider this instead:
qx(0xn)
=qx0 (result 1)
=1 (Def 1)

Therefore multiplication involving q is non associative

we observed that
(qx0)xn=n but qx(0xn)=1, therefore we can apply a trick, inspired from the property of cross products

One property of the cross product: axb=-bxa
and have the following definition

Property of the multiplication of q:
(qx0)xn=nx(qx(oxn))

Property 2: Multiplication of q is non distributive
Consider
qx(0+0)
=qx(0x2)
=qx0 (Result 1)
=1 (Def 1)

Now consider
qx(0+0)
=qx0+qx0
=2
=/=qx0

Therefore multiplication of q is non distributive

Property 2 also highlighted the most disturbing property of division by zero (which compound with the observations in the the context of limits, result in it being undefined)

Property 3: "Spontaneous generation/Trivial Ring property"
Consider the following case again
qx0
By the property of the addition of real numbers and Result 1
qx0=qx(0+0+0+0+0+...)

Property 2 showed that
qx0=/=qx(0+0+0+0+0+...)
In addition
qx(0+0+0+0+0=...) = qx0+qx0+qx0+qx0+qx0+... = ???

Thus if the results of both sides are regarded as the same, then this implies the following disturbing fact
0=1=2=3=4=5=6=7=...
I.e. The entire complex numbers collapsed into the trivial ring (discussed earlier in another thread)

(This also explained why division by zero memes were often associated with black holes)

Conclusion: In order to define a sensible number system for division by zero, the following hurdles must be overcome:
1. Evaluate qx1, q+0, qxq, q+1
2. Fix property 3
3. Address and solve property 2 in detail
4. Proof or disprove that qxn=nxq (n∈C)

Again I must emphasize it is undefined anywhere in the reals, complex or possibly the quarternions (however I don't really have get into much detail with that yet)

Math is useful for many things but some of the basic procedures can lead to conceptual problems, when trying to find an analogy within reality. For example, 1/(1/2) =2. This violates the conservation of energy and implies perpetual motion. For example, if we start with one battery and divide by a half, aI will now have two batteries. The world's energy crisis is solved by math.

Division by zero also allows one to play the same conceptual trick in reality. If we divided the battery by zero we get infinite batteries. If we assume the conservation of energy, the dividing by zero would require infinite energy and therefore may not be possible within the physical reality of a finite universe.

Math is useful for many things but some of the basic procedures can lead to conceptual problems, when trying to find an analogy within reality. For example, 1/(1/2) =2. This violates the conservation of energy and implies perpetual motion. For example, if we start with one battery and divide by a half, aI will now have two batteries. The world's energy crisis is solved by math.

Division by zero also allows one to play the same conceptual trick in reality. If we divided the battery by zero we get infinite batteries. If we assume the conservation of energy, the dividing by zero would require infinite energy and therefore may not be possible within the physical reality of a finite universe.

@1st block
This explained why when doing maths on physical systems, one must be careful on the conditions they have, as some of the operations are simply forbidden by the physical conditions (i.e. there is no amount<0, at best you can only say the change in amount <0 (i.e. the amount decreses))

@2nd block
Technically you cannot say division by zero = infinity as if you apply such reasoning, you can say e.g. 2/0=1/0=3/0=infinity
Now assume if you can treat them as numbers then you will end up with infinityx0=1,2,3..... (because the definition of division/multiplicative inverse is a UNIQUE number a such that axb=1) which not only the rule of multiplicative inverse is violated, but also that 0xn=0 for any n in complex numbers is also violated. (Even when you say infinity is not a number, the multiplicative inverse rule is still violated)

Physically, as many of mentioned before near the beginning of this thread, divide some objects by zero means you are are not actually carrying out the division, thus the result is undefined (e.g. how can a basket of apples be distributed evenly among no people? there's not even a person to receive an apple! thus it makes no sense to say how many apples do each people get)

Even in such scenario, |0+|=|0-|=/=|0|=/=0
Thus N/|0+|=|N/0-| does not imply N/|0|

Thus N/0 is still undefined

Your application suggested above seemed to have something to do with surreal numbers, however I do not understand the specifics, thus I cannot comment any further on your suggestion without getting things wrong

http://en.wikipedia.org/wiki/Surreal_number

I think in this case , we should first understand 0 (zero) . Then only N/0 can be understood .

So , what is zero (0) ?

I think any new math that explains division by zero would have to show that 0/0=1. I read once a long time ago that division by zero is the biggest problem with theoretical physics today, so when my math teacher derived the tangent line to a circle something really caught my eye that I won't forget. They started out saying that h was the slope of the curve from two different points. Then they said that as the limit of h approuches zero that would give the tangent line, the two points on the curve become the same point. But, when this happens h is zero, and the h is canceled out from the top and bottom of the equation because before they said it was the limit. But then the tangent line is when h is zero, so then how did they just cancel out? Does writting the word limit in front of an equation change operation by zero? I didn't see how it affected division by zero other than just saying that it was okay in that instance. So then 0/0 would have to be able to be canceled out just like any other a/a, so then 0/0 would be one unless it just happened to be some kind of fluke where it just so happens in that instance. But if writing limit doesn't change rules of operations besides allowing canceling of zero's then why couldn't a/a just equal one for all reals? It seems to work in that instance but not in others. It would be like saying, well if I have no apples then I don't divide them one time...

I think in this case , we should first understand 0 (zero) . Then only N/0 can be understood .

So , what is zero (0) ?

I think I'll better google it cause I don't really know much about the specific properties of zero, other then it can be used to represent the magnitude of a number (e.g. 0.00001 vs 100000), an additive identity (i.e. 0+a=a) and that 0xn=0 for all complex numbers n


I think any new math that explains division by zero would have to show that 0/0=1. I read once a long time ago that division by zero is the biggest problem with theoretical physics today, so when my math teacher derived the tangent line to a circle something really caught my eye that I won't forget. They started out saying that h was the slope of the curve from two different points. Then they said that as the limit of h approuches zero that would give the tangent line, the two points on the curve become the same point. But, when this happens h is zero, and the h is canceled out from the top and bottom of the equation because before they said it was the limit. But then the tangent line is when h is zero, so then how did they just cancel out? Does writting the word limit in front of an equation change operation by zero? I didn't see how it affected division by zero other than just saying that it was okay in that instance. So then 0/0 would have to be able to be canceled out just like any other a/a, so then 0/0 would be one unless it just happened to be some kind of fluke where it just so happens in that instance. But if writing limit doesn't change rules of operations besides allowing canceling of zero's then why couldn't a/a just equal one for all reals? It seems to work in that instance but not in others. It would be like saying, well if I have no apples then I don't divide them one time...

When using limits

\lim_{h \to a} f(h)=L
e.g. \lim_{h \to 0} \frac hh=1

It means when h approaches arbitrary close to a/as close to a as you like, the value equals L. Note that h does not actually get to a, just as close as you want, therefore using the example above, h=/=0, thus you can cancel them out

f(a) has no relationship with \lim_{h \to a} f(h), i.e. f(a) does not need to be = \lim_{h \to a} f(h) (graphically it means as you approach a from both sides of the function f(x), you will eventually reach a certain value. However the actual value f(a) may be a dot above or below the point where the lines joined, therefore creating a gap), if they are equal, it means the function is continuous at a (i.e. there is no gap, jump or hole in f(x) at x=a)

0/0 cannot be =1 because 0xn=0, thus any complex number n can satisfy this equation. Division is an inverse of multiplication, thus it should undo what multiplication does. Therefore there should be only one possible number n that can satisfy nxm=1. We then say n is the multiplicative inverse of m. Clearly this is violated in 0/0 as 0xn=0 for any complex number n

But by definition isn't the tangent line the line that only intersects a curve at one point? How does it do this if the lim of h approuching 0 is then said to be when h is actually zero? Wouldn't this mean that the current definition of the tangent line would prove to be impossilbe and that it wouldn't be able to only intersect at a single point?

I don't see how zero times any number equalling zero proves anything. You could still put 0/0 in for n then say it is one and then get zero as a result. But if you had 1/0 as n then you would get 1 as a result, the zero's would cancel and give you 1 times 1. So then the statement 0xn=0 would not be true, but 1/0 does not satisfy any currently defined number and does not quallify for being a n since it would be undefined.

I think a system that involded a new way to deal with zero's effectivly would have to define n/0, and in this system we couldn't say that anything times zero is equal to zero. That could be part of the reason it doesn't work, because somehow it would have to explain how you could get the definition of tangent without using limits. In other words it would have to put out something that has been shown or verified to work everytime. If you didn't get the same result of the tangent line it would have to be wrong.

Scarry thing is that anytime you are dealing with unknowns they could be equal to zero at any time, and then you would just treat them as though they are any other number even though they could have been equal to zero at one time or another or the whole time you worked the equations...

I think I'll better google it cause I don't really know much about the specific properties of zero, other then it can be used to represent the magnitude of a number (e.g. 0.00001 vs 100000), an additive identity (i.e. 0+a=a) and that 0xn=0 for all complex numbers n

Following websites are interesting for 'division by zero' , 'zero' and 'division' .


http://en.wikipedia.org/wiki/Division_by_zero

http://en.wikipedia.org/wiki/0_(number)

http://en.wikipedia.org/wiki/Division_(mathematics)


Let us consider the operation of 'division' . What does division mean ? Division means dividing something into equal parts . So , N/m is less than N as long as m is greater than 1 .

When m is 1 , N is equal to N/m.

When m is less than 1 ; N/m is greater than N . Here actually , physically division becomes multiplication ; thats why N/m is greater than N . Here multiplication factor is 1/m , which is greater than 1 .

N/p is greater than N/m ; where p , m both are less than 1 and p is less than m . Going by this logic N/0 should be something greater than N/0+ . As N/0+ is infinity . So, N/0 also should be considered as infinity ; as something greater than infinity is infinity only .

Even if we did assume that n/0 was infinity, what better off would we be? We would have the same problem have having numbers that we could no longer deal with. So then it seems we would have to have multiplicitive rules for infinity in order to also have multiplicitive rules of zero.

I have come across one theory that can add/subtract infinities. It was called infinite sums. Basically you just write out the series of the reals involved, then you find a patern to those numbers. So if you said all whole numbers, then you would write out 1, 2, 3, 4, 5... Then you could write out all even numbers 2, 4, 6, 8, 10... Then you could add them down to get 3, 6, 9, 12, 15... Then by looking at the patern of the numbers you got you figure that line is every mulitple of 3. So then all whole numbers plus all even numbers would be all multiples of 3. For some reason this didn't become a valid mathmatically theory, but I am not sure exactly why. Some higher mathmaticas to me can seem even more arbitruary than this one. I think it may be because there was nothing to test it on, and the only thing I know of where it would be useful is string theory. But then, string theory can not be tested.

Even if we did assume that n/0 was infinity, what better off would we be? We would have the same problem have having numbers that we could no longer deal with. So then it seems we would have to have multiplicitive rules for infinity in order to also have multiplicitive rules of zero.

I have come across one theory that can add/subtract infinities. It was called infinite sums. Basically you just write out the series of the reals involved, then you find a patern to those numbers. So if you said all whole numbers, then you would write out 1, 2, 3, 4, 5... Then you could write out all even numbers 2, 4, 6, 8, 10... Then you could add them down to get 3, 6, 9, 12, 15... Then by looking at the patern of the numbers you got you figure that line is every mulitple of 3. So then all whole numbers plus all even numbers would be all multiples of 3. For some reason this didn't become a valid mathmatically theory, but I am not sure exactly why. Some higher mathmaticas to me can seem even more arbitruary than this one. I think it may be because there was nothing to test it on, and the only thing I know of where it would be useful is string theory. But then, string theory can not be tested.

Let us compare N/0 with N/0+ .

N/0+ = infinity . There is no doubt about it . So , we can write the following :

N = 0+ + 0+ + 0+ + ... upto infinity number of times 0+ .

But if we write, 0 + 0 + 0 + ... upto infinity number of times 0 ; the result will still remain 0 and the result will not add upto N , like the earlier case of N/0+.

So, we can write N in terms of 0 as , N = 0 * infinity + N . Here N remains as remainder , though N is greater than 0 .

Thus , N/0 is something greater than N/0+ .

Kind of reminds me of the 0.999... = 1 debates. But then came the question of can you denote the next number down from 1 as being 0.999...8. An infinite amount of numbers with a different value on the end of it. They all seemed to agree that it wasn't a valid form of notation, but they "prove" 0.999... = 1 by multiplying it by 10 and then subtracting all the 0.999... from it and then ruducing it to one.

It is kind of like what I did by adding different values of infinity. I was reading Guth's paper on inflation the other day and he says the reason why this is not valid mathmatics is because you could write all the values in any order you wanted to. If you write them in a different order than you get a different relation to those numbers. So then it would seem like the mathmatics that proves 0.999...=1 would work because we know that it is just going to be all nines or you could say that any number that changes the value of each digit would always change that value of the digits in the same order.

Mathmatics seems plagued with a problem with dealing with certain types of remainders, so I think you may be on to something. Say you divide 10 by 3. You get 3.333... Now multiply that number by 3 you get 9.999... You just did multiplication and division by the same number but you ended up getting a slightly different answer. So by saying that 9.999... = 10 you are simply correcting the muliplication/division errors brought about by the way remainders deal with numbers that do not divide evenly.

In modular arithmetic, the integers mod 4 has four elements: {0, 1, 2, 3}.

Under multiplication, we have: 0x0 = 0 (mod 4), 1x1 = 1 (mod 4), 2x2 = 0 (mod 4), 3x3 = 1 (mod 4).

So therefore: 0 ≡ 2 ≡ 0x2-1 (mod 4). So 2 is a 'zero divisor' in Z4.
But 0 doesn't have an inverse (i.e. 1/0 is not defined, so nor is 0-1), and since 2 ≡ -2 (mod 4), it also says -2 ≡ 0x2-1 (mod 4).
Hence 2 is its own multiplicative and additive inverse in Z4.

0 and 0+ are ideal terms of mathematics. If we consider practical physics, 0+ can be considered as smallest unit for any variable. For example if we consider distance as variable, Planck's distance (LP) is the smallest distance and there is no smaller distance possible. So for distance, LP can be considered as 0+ . Here if we consider N as a finite distance , N/0+ = N/LP = some finite value(x) . So N = x*LP .

N also can be writen as N = 0*x + N or N = 0*infinity + N .

N/0 , if we write this as a*b + c ( the equation form of division ,where a=0; b=N/0 and remainder c=N )

Here we can see that N/0+ is a finite value but N/0 is infinity .

I think N/0 is incomplete division.

Let us consider the operation of division in the form of subtraction.

So, if we write N = a*b + c . Then (N-c) - b - b - b - ... (a times - b) = 0 .

But in the case of N/0 ; N - 0 - 0 - 0 - ... ( infinity times - 0 ) = N ( and not zero ) .

So, perhaps i think , N/0 is incomplete .

Despite the attempt in avoiding infinities and 1=2=3=4=... by simply left out the RHS of the non distributive property of D (which I still yet to find a solution to it), a new paradox had arose when attempt in division (despite applying the unique properties of the non associative multiplications in D)

Result derived from the division section
q^-m=0 , m>0 (sensible according to the definition)

1=q^(m-1) is not despite it might still make sense saying positive powers of q=1 and negative powers of q=0

BUT

Let m=2
you will end up with

q=1

Which contradict with the definition near the beginning (q=/=1)

I doubt this can be resolved

Given all these 3 weeks of failed attempt...

Is there a mathematical proof that for any conceivable number system, division by zero (including both n/0 and 0/0) MUST result in paradox???
If the above is false, then at least which axioms need to be broken for a number system to be divisible for all members within it
Is it impossible to have any arbitrary number system where every element has both an additive inverse and multiplicative inverse? If yes how to prove it?

So far in mathmatics the closest example is Riemann sphere, with 1/0=complex infinity. However 0/0 is still undefined even in such number system

There is also wheel algebra, but the division has two types (left and right) and is defined very differently from the usual one

5684 (summary of my attempt in building D)

Other than that, it seems D obey nearly all the index laws (including that q^0=1)
The none associative property can be summarized using those three rules, which allow multiplication to be handled. Of those properties, (q.0).n=(q.n).0 is a surprise to me
I'm still at lost of dealing with the nasty non distributive property of D, thus I cannot define what is e.g. 1/(q+1)
My approach on the division is inspired by the division in complex numbers, which seemed to work quite well until q^-m=0 pops up
NB D is the following set (C,+,*,q) which I am trying to build
NB2 Note that distributive law breaks down only when said distribution will produce a q.0 pair (although this is precisely how division by zero collapse a number system into the trivial ring if distributive law is assumed to hold) and for the associative law of multiplication, breaks down partially only when (q.n).0 and q.(n.0) with n=/=q (anything within the brackets and the () as one object . term outside are commutative, thus can be rearrange in any order without affecting the result) and can be rationalized by (q.n).0=n.((q.(n.0))) for n=/=q

Of course, once you open that door even a tiny bit then all manner of pathological behaviour comes rushing in. Once you can 'prove' something contradictory then you can prove anything. It's known as the principle of explosion (http://en.wikipedia.org/wiki/Principle_of_explosion).

Someone could argue that this very statement in itself could fall into a catagory that is a result of the same underlying principle, but I won't, or have I already?

5727

Sorry you are wrong. This is exactly what I did for 3 weeks in an attempt to construct a division by zero number system to no avail.

It seems in order to divide by zero, one must kill some of the field axioms which causes 0n=0 to result.
But then you end up something that is not a field, as the physics forum mentioned


5728
Interestingly, (at least according to my 3 week analysis) the non associative property of a number system including division by zero with trying to preserve the field axioms is not that difficult to fix, with only q as the special case (Only 3 multiplication rules is required when manipulating 3 terms, with (q.0).n=(q.n).0 being the most surprising one which can make the multiplication less difficult to deal with than expected), until you start messing with division defined in a way similar to division in complex numbers, which end up with the killer q=1 contradiction. Still it is mush better than the non distributive properties of such hypothetical number system which I'm still at loss of fixing it


Someone could argue that this very statement in itself could fall into a catagory that is a result of the same underlying principle, but I won't, or have I already?
What principle? I don't think AN's statement is itself a result of the "Principle of Explosion", if that's what you mean


In modular arithmetic, the integers mod 4 has four elements: {0, 1, 2, 3}.

Under multiplication, we have: 0x0 = 0 (mod 4), 1x1 = 1 (mod 4), 2x2 = 0 (mod 4), 3x3 = 1 (mod 4).

So therefore: 0 ≡ 2 ≡ 0x2-1 (mod 4). So 2 is a 'zero divisor' in Z4.
But 0 doesn't have an inverse (i.e. 1/0 is not defined, so nor is 0-1), and since 2 ≡ -2 (mod 4), it also says -2 ≡ 0x2-1 (mod 4).
Hence 2 is its own multiplicative and additive inverse in Z4.

Thanks! It intuitively explains what a "zero divisor" means, which confuse me since the first day I came across it some months ago



0 and 0+ are ideal terms of mathematics. If we consider practical physics, 0+ can be considered as smallest unit for any variable. For example if we consider distance as variable, Planck's distance (LP) is the smallest distance and there is no smaller distance possible. So for distance, LP can be considered as 0+ . Here if we consider N as a finite distance , N/0+ = N/LP = some finite value(x) . So N = x*LP .

N also can be writen as N = 0*x + N or N = 0*infinity + N .

N/0 , if we write this as a*b + c ( the equation form of division ,where a=0; b=N/0 and remainder c=N )

Here we can see that N/0+ is a finite value but N/0 is infinity .

I think N/0 is incomplete division.

Let us consider the operation of division in the form of subtraction.

So, if we write N = a*b + c . Then (N-c) - b - b - b - ... (a times - b) = 0 .

But in the case of N/0 ; N - 0 - 0 - 0 - ... ( infinity times - 0 ) = N ( and not zero ) .

So, perhaps i think , N/0 is incomplete .

0*infinity is undefined. I'll let the others to explain the details to you since I'm not good at tackling infinities, thus you cannot say such thing
Infinity is not a complex number. In the context of surreal and hyperreal numbers, arithmetic can be done on infinity as ordinal numbers, which I only briefly read about and don't really understand the details except the arithmetic is completely different from the more familiar complex numbers, and require careful treatment

You can still say N=0*x+N since for any x, 0x=0 (Even in your context that x=N/LP where N is an arbitrary complex number, x is still a complex number thus basically the equation reduces to N=N, which does not tell us anything useful)

I've also apply the second approach when I was still a 7th grader, but N/0=n (n is arbitrary) with reminder N does not do anything useful at all
After learning that division is multiplication by a multiplicative inverse, which requires the inverse to be unique, I understand that a/b cannot be any arbitrary n, thus the above approach does little help in the division by zero issue

P.S. Note that the two biochem.jpg are not identical, I'm just too lazy to use another name