But numbers aren't required to be usable physically.
Which doesn't address the question of numbers larger than infinity.
Division by zero,seems flawed
- Thread starter Secret
- Start date
Big Chiller
Registered Senior Member
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.
For now is it hypothetically realized whether there is a "biggest possible number", knowing that should better the understanding of infinity.
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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 $$\omega + 1 \neq 1 + \omega$$.
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".
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
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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.
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.
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.
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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 $$x$$.
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
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:
secret said:
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
Reference of proof by contradiction said:
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
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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.
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.
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.
It's a faulty proof
I'll deal with it later after digesting the new knowledge
@AN
---------
I 'm glad you finally reply
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
I'll deal with it later after digesting the new knowledge
@AN
---------
I 'm glad you finally reply
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 answerWhy division by zero ($$\frac N0$$) is undefined in the reals
Because our maths teacher said so/Because maths said so
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:
secret said:
secret said:
secret said:
Regardless of which interpretation you use, we still showed
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)
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
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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 .
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)
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)
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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 .