Dividing by zero

[houserichichi]

If you look up a paper by Baez called Octonions the first paragraph briefly breaks down the four normed division algebras...

There are exactly four normed division algebras: the real numbers (R), complex numbers (C), quaternions (H), and octonions (O). The real numbers are the dependable breadwinner of the family, the complete ordered field we all rely on. The complex numbers are a slightly flashier but still respectable younger brother: not ordered, but algebraically complete. The quaternions, being noncommutative, are the eccentric cousin who is shunned at important family gatherings. But the octonions are the crazy old uncle nobody lets out of the attic: they are nonassociative.

Doesn't tell you much other than yes, you're right - the quaternions are noncommutative. Both i,j,k are roots of negative 1 and they have an algebra of their own.

As for preferring algebra over analysis or vice-versa ( god forbid ) it happens, usually in the second or third year (depending on how your degree is structured). You'll most likely have to take analysis close to the end, and if you intend on taking a masters program afterward I don't think there's much choice, but don't quote me there. Algebra fits the same bill, but in my own experience, folks tend to one side more than the other...either because they prefer the challenge or the topics come more naturally. (I find analysis more difficult, but that's just me.) If you're lucky you'll love both, but it doesn't mean it will make things any easier.

Anyways, this 0/0 discussion fits into both categories...algebraically it can't be done in the cases we're looking at, but analytically it can. So who's right? Depends on the context I suppose.

[RQ]

Anyways, this 0/0 discussion fits into both categories...algebraically it can't be done in the cases we're looking at, but analytically it can. So who's right? Depends on the context I suppose.

Actually I think it's the other way around. We showed with arithmetic that 0/0 cannot equal all numbers, but 0/0 can't be graphed in any function, and limits do not determine its actual value, thus indeterminate. The confusion with 0/0 arises when you think that just because 0=0 then 0/0 can equal anything because 1*0=0, 2*0=0, 1.2321*0=0, but the problem is that x=0/0 and x*0=0 are two different equations where the value of 0/0 is not evaluated but is trying to be found, with no original assumptions.

No, the complex numbers are a field as well, and in fields multiplication is always commutative. If, however, we used matrices instead, it's normally the case that AB does not equal BA...they don't form a field, but again you can't "divide" by zero (the zero matrix, that is). So if you want to solve for A in AB = 0, for instance, you have to multiply the RIGHT by the inverse of B whereas for BA = 0 you'd have to multiply on the LEFT by the inverse of B. That's the simplest example I could think of off the top of my head. (Let's assume we were using square matrices too - to make the discussion easier.)

We werent talking about matrices nor quaternions, but about numbers.

[houserichichi]

We werent talking about matrices nor quaternions, but about numbers.

The two main differences to those that haven't studied algebra before between a "number" and anything else, be it matrices, mappings, vectors, or anything from the gambit of choices, is 1. the way they look (a matrix doesn't look like a number does for example) and 2. their algebra (obviously matrix arithmetic is much different than "number" arithmetic). In the end, however, they turn out to be examples of the same things, though the reals have properties the matrices don't. The point in introducing them was to show the universality of the 0/0 argument - it applies to a lot more than "numbers", whereby I assume you mean the reals. If you want to move into the complex number field then you might as well continue on and take the quaternions while you're at it - just to see if it works.

The reason 0/0 can't be determined algebraically is because it isn't allowed...it's undefined. (I'm talking field theory here wherein "0" has no multiplicative inverse by its own existence, not classical algebra). Analytically there are tricks (ie. l'Hôpital's rule, for instance) to show where the limits approach different values depending on how we come at it. You're right that they don't determine the actual value of a function (say x/x as x->0) but that's like arguing that the sum of (1/2)^n from n=0 increasing as n->infinity doesn't have an answer...may be true, but it still tends to 2 and that's half the fun of first year calculus - dealing with tendencies. In the analytic world there are a plethora of solutions to equations of indeterminate form - in fact I believe there are uncountably many of them.

[RQ]

...but that's like arguing that the sum of (1/2)^n from n=0 increasing as n->infinity doesn't have an answer...

You learn this in Precalculus and it's called an infinite geometric series for which the formula is: 1/(1-r) where r is the fraction by which each subsequent number is multiplied and is less than 1. This has absolutely nothing to do with 0/0 and (1/2)^infinity is infinity. The answer to 1/2+1/4+1/8+... is 1.

0=0/0 because

0=0^1
0=0^(2-1)
0=0^2/0^1
0=0/0

QED

P.S.: What does QED stand for? Quite Excellently Done?

[houserichichi]

Ok, I chose a bad example for the infinite sum , but nonetheless an infinite sum can be done using limits and produces strange results. It solves paradoxes, but seems counterintuitive at first.

(1/2) ^ (infinity) can't be done from straight arithmetic...you have to use limits...and the answer is 0, not infinity, because 2^n increases faster than 1^n (which doesn't increase at all).

QED stands for Quod Erat Demonstratum and is Latin for "That which was to be demonstrated, was..."

(1) 0=0^1
(2) 0=0^(2-1)
(3) 0=0^2/0^1
(4) 0=0/0

Again, in your proof that 0 = 0/0 you're correct in lines 1 and 2, and I know the steps you took in line 3 but it's not right. Let's pretend you threw in line 2.5 which reads

0 = (0^2)(0^-1)

(obviously the step you didn't write that would get you to line 3) which, if you go on to line 3, you've assumed that 0^-1 even exists, which it doesn't. The additive identity in a field doesn't have a multiplicative inverse, and that's exactly what zero is, so you've used valid arguments in your proof, but you missed a crucial existence...so the proof is false.

Hope that helps a little!!

[RQ]

(obviously the step you didn't write that would get you to line 3) which, if you go on to line 3, you've assumed that 0^-1 even exists, which it doesn't. The additive identity in a field doesn't have a multiplicative inverse, and that's exactly what zero is, so you've used valid arguments in your proof, but you missed a crucial existence...so the proof is false.

Hope that helps a little!!

Well if you mean that 0^-1 doesn't exist, the definition is 0^0/0, since each number to the power of an integer, indicates how many times that number is multiplied by itself, and a number to a negative integer would mean how many times this number is divided by itself or how many times you have to multiply it to get its reciprocal, and in this case 0 doesn't have a reciprocal since 1/0 is 0. The very fact that you accept that zero has no inverse shows that you believe that x/0 equals 0, but besides that the definition of 0^0 is 0^(1-1) and thus 0^1/0^1 and 0/0, thus 0^0 is undefined as textbooks say in Algebra I.

[houserichichi]

the very fact that you accept that zero has no inverse shows that you believe that x/0 equals 0

The very fact that I accept that zero has no inverse (which is a consequence of definition, no need for acceptance) shows that things like x/0 have no value. It's not that they "equal infinity" or "zero", it's that they just don't exist. Sometimes it's more convenient to see where they tend to, but the actual value of them isn't defined in a field since x/0 I assume means x*(0^-1) and * is the field multiplication. It's a direct consequence of the axioms that 0^-1 isn't in a field and so neither is x*0^-1.

As I've said before, in an algebra course they're not valid "numbers" in a field (by the axioms), but in an analysis course you can see where they tend to, and thus are indeterminate forms since from different questions they tend to different values. It depends on the context in which you're looking at them.

I'm starting to hate 0/0 more than ever

[Keiji]

the definition of 0^0 is 0^(1-1)

The definition of n^0 is 1, so 0^0 = 1.

[houserichichi]

Yar...0^0 is undefined because there's a contradiction with

lim x^0 = 1
x->0

AND

lim 0^x = 0
x->0

Sometimes it's more convenient to define 0^0 to be 1, for instance, we need that for the binomial theorem to work, but there's a discontinuity of x^y at (x,y) = (0,0) so it is undefined (it's actually indeterminate, but I'm willing to give that word up just to get things straightened out). :wink:

[RQ]

Sometimes it's more convenient to define 0^0 to be 1, for instance, we need that for the binomial theorem to work, but there's a discontinuity of x^y at (x,y) = (0,0) so it is undefined (it's actually indeterminate, but I'm willing to give that word up just to get things straightened out).

That theorem doesn't make you divide by 0 because there is no parameter set for x. Until you find the value of x then you can say you are dividing by 0 otherwise if you divide for example (x^2+2x+1) by x+1 you get x+1 which equals 0, which then shows that x=-1, but you haven't gotten that value before dividing by x+1.

Also when you divide any binomial that equals to zero by any other value that equals to zero, you always get a value that equals to zero, thus you are continuing to emphasize my point.

Example:

x^2+2x+1=0

(x^2+2x+1)/(x+1)=0/(x+1)

x+1=0

x=-1

(-1)^2+2(-1)+1=0

1-2+1/(-1+1)=0/0

(-1+1)=0/0

0=0/0

[houserichichi]

(-1)^2+2(-1)+1=0

1-2+1/(-1+1)=0/0

(-1+1)=0/0

The first line is fine (you subbed -1 in for the x value in your previous polynomial...)

The second line is where I have a problem... on the left hand side you're dividing by zero (can't be done)

(1-2+1)/(-1+1) = 0/0

(1-2+1)/(0) = 0/0

1/0 - 2/0 + 1/0 = 0/0 ??? Can't add undefined terms together

...you can take a limit, but you can't divide by (-1+1 =) 0. Also note, if you break down the numerator a little more so that

1-2+1

= (1-2)+1

= -1+1

and sub THAT into your equation, you get

(-1+1)/(-1+1) = 0/0

1 = 0/0

which contradicts what you wanted to show.

As for the binomial theorem part, I was referring to 0^0, not 0/0, my bad!! :wink:

[RQ]

...and sub THAT into your equation, you get

(-1+1)/(-1+1) = 0/0

1 = 0/0 ...

if x has a solution of -1, then ur above equations (x+1)/(x+1)=0/0 and getting 1=0/0 means that you have assumed that 0/0=1 by dividing the monomial by itself and giving it a value of one. This isn't valid.

...you can take a limit, but you can't divide by (-1+1 =) 0. Also note, if you break down the numerator a little more so that

1-2+1

= (1-2)+1

= -1+1

This is correct, but since if x has a solution of -1, thus the two polynomials are equal, x^2-2x+1=x-1. You show that, but we don't do that to find the other solution for x by dividing 0 by 0 in a polynomial form which also equals 0, and we find a solution.

Using polynomials can be misleading, because although a solution for x that makes the denominator of two polynomials dividing equal to zero.

Dividing by polynomials is not the same as 0/0 because you divide one polynomial that for a solution of x that makes it equal to zero, by another that also has a solution of x equal to zero, you get a third polynomial that equals say 4, since when you multiply 4 times 0 you get 0, and this is basically the inverse. Take this for example:

x^2-1=x-1

x^2-1/x-1=x-1/x-1

(x+1)(x-1)/x-1=x-1/x-1

x+1=0

x=-1 invalid solution, x=1 and 0, thus 0/0=1 and 2

The reason this happens is because in line 3, we have (x+1)times 0/0 and in line 4 we get a solution that assumes 0/0 as 1, thus the right side assuming 0/0=0 gets an invalid solution, but assuming it is 1, gives us one of the two. Also, the solution we get from 0/0 is the solution of the value that x +/- some other number gives us that x substituted in the other polynomial gives us 0, so that we can multiply the two and give us 0, and thus 0/0 would give us the value of that polynomial that we originally multiplied 0 with.

Thus the polynomial version of 0/0 is invalid because you are assuming that 0/0=1 and that they are values of x and if you change one on the nonzero polynomial, you change the zero into something else other than zero.

Hope I didn't rant too much, I hate long posts.

[houserichichi]

See I wasn't dividing (x+1)/(x+1), I was dividing (-1+1)/(-1+1). The first of these two is a polynomial, the second of these two is a division among real numbers - there's a very big difference. While the division of two numbers resulted from a polynomial, it's completely allowable once we "sub in" the "solutions" of x.

Since you're trying to find a value for 0/0 (I'll assume a real number and nothing more technical) you have to make sure that it's the same in all cases.

(-1+1) / (-1+1) = 0/0 (adding then dividing)
(-1+1) / (-1+1) = 1 (dividing first)

The first result contradicts multiplicative unity in a field which says that for all x in the the field F we have x^-1 such that

x*x^-1 = 1

and in this case dividing by (-1+1) is the same as multiplying by the inverse of it. We SHOULD have 1 on the right hand side by the field axioms.

The second result contradicts multiplication by the zero in a field. The axiom states that for all x in field F we have a 0 such that

0*x = 0

(we also require x*0 = 0, but we're dealing with the reals for sake of argument, where multiplication is commutative). Now forgetting that we're multiplying 0 by 1/0, we should have zero on the right. However, now remembering that we're multiplying 0 by 1/0 we notice that we can't even perform the operation at all because 1/0 isn't in a field by definition.

Thus, since the first line contradicts multiplicative unity, and the second contradicts both multiplication by the additive identity and the existence of inverse of said number, the value 0/0 is undefined in the reals (the field in question).

I also hate long posts...sorry if this was.

[Keiji]

(-1+1) / (-1+1) = 0/0 (adding then dividing)
(-1+1) / (-1+1) = 1 (dividing first)

Very true

[RQ]

(-1+1)/(-1+1) does = 0/0 because -1+1=0 by the axiom that x=x

x*x^-1 can be proven by substituting for real values, and in this case 0 doesn't work.

x+1 and -1+1 is the same thing because I stated in the beginning that x has a solution of -1, and for convenience and to avoid confusion, I put x and whether you put -1 or x doesn't matter. In fact putting x shows what you're trying to do more clearly.

-1+1/-1+1 does not equal to 1 because 0/0 does not equal to one, since if you multiply each side by any number other than 0 you get that if 0/0=1 then 0/0= all numbers, and subsequently all numbers equal each other, which is a contradiction as to x=x.

If 0/0 did not exist in the first place as you say, then 0/0 does not equal 0/0 which means there is no way to write 0/0, which there is, and that the axiom x=x is untrue, since 0/0 does not equal itself.

So my question is to you, what are you going to do about the fact that

0=0^(2-1)

0=0/0?

[houserichichi]

since if you multiply each side by any number other than 0 you get that if 0/0=1 then 0/0= all numbers, and subsequently all numbers equal each other

That's what I've been saying all along...it's an indeterminate form because you have no idea what it equals. It doesn't tell you anything until you inspect it analytically, not algebraically.

As for 0^(2-1) = 0/0 = 0, as I've said before, you need to break down

0^(2-1) = 0^2 * 0^-1

where * is the multiplication and ^ is the exponentiation. Since, by definition, 0^-1 is the multiplicative inverse of 0, that is

0*0^-1 = 1 (by definition of *-inverse)

you can't perform the multiplication at all because that term doesn't exist within the field. Hence, 0 does not equal 0/0 because you forgot the crucial detail that your arithmetic fails at the line I singled due to aforementioned definition.

[RQ]

Yes, I agree that for the equation x*0=0 there are an infinite number of solution which would make 0/0 indeterminate, but the problem is, that the equation 0/0=x and x*0=0 are not the same, because to get one from the other, you need to use arithmetics by either multiplying or dividing each side by 0 which would give you the other equation from the first if and only if 0/0=1. This is why:

0/0=x and to get that x*0=0 we have to:

[0/0]*0=x*0

we get the above equation and the only way it equals the equation x*0=0 is if 0/0= to any number that has a real or imaginary value. This means that 0/0 can be assumed to be a number that exists and isn't undefined yet in the end it becomes undefined. A bit bogus, don't you think?

This could make 0/0 indeterminate however 0/0 cannot equal all numbers because:

if 0/0=x where x doesn't equal 0 for convenience and y isn't 1, then:

[0/0]y=xy

0/0=xy

x=xy --> contradiction

The only value that does not impose a contradiction for the equation 0/0=x which suggests that 0/0 has a valid value is 0.

[RQ]

you can't perform the multiplication at all because that term doesn't exist within the field. Hence, 0 does not equal 0/0 because you forgot the crucial detail that your arithmetic fails at the line I singled due to aforementioned definition.

If 0/0 doesn't equal itself, 0/0=0/0, then what's going to be indeterminate? 0/0 is the expression for the value that may or may not be undefined. This fraction, ratio is still subject to laws of arithmetic if it can be written down as a fraction which it obviously can be. What it evaluates to is what may or may not exist, but the fraction itself can be expressed in terms of a numerator and denominator, which when if 1/0 is multiplied by 2 it becomes 2/0. You can't just say 0/0 is undefined and then undefined * 2 is not a valid answer.

[houserichichi]

Try this...define 0/0=1 like you wanted. Then multiply both sides by -1 so that we have

(-1) * (0/0) = (-1) * 1 = -1

so far so good.

However, let's make it look prettier and lose that middle step so we just have

(-1) * (0/0) = -1

so that

(-1*0)/0 = -1

and since (-1 * 0) = 0 we must have

0/0 = -1

which is a contradiction.

The field axioms don't directly forbid 0 to have a multiplicative inverse, but it turns out that they do so indirectly.

If say 0/0 = 0 instead, what we really mean are these two things: first, 0 does have a multiplicative inverse; and secondly, that it satisfies

0*(0^-1) = 0.

But also

0*(0^-1) = 1,

since that's what it means to be a multiplicative inverse. Thus, 1=0 which is a contradiction.

Another way to reach the same conclusion is using the usual rules for adding fractions:

(1/1) + (0/0)
= [(0*1)+(1*0)]/(0*1) (cross multiply)
= [(0+0)/0] (perform multiplications)
= 0/0 (perform addition)
= 0. (by assumption)

On the other hand,

(1/1) + (0/0)
= 1+0 (1/1=1 by definition, and 0/0=0 by assumption)
= 1. (definition of additive identity (zero) in a field)

Once again, we have 0=1 which is a contradiction.

Now, there are number systems where 0 does equal 1, but we're working in a field here which is a ring with unity along with other axioms, thus requiring 1 not be 0 so long as we have more than one element.

So really there is no one set value for 0/0, in fact i just showed that it equals three different integers (-1, 0, and 1). The same can be done for any number in a field which is why 0/0 is indeterminate (better yet, let's drop that word and just say undefined for now because it's a much easier concept to work with).

[houserichichi]

but the fraction itself can be expressed in terms of a numerator and denominator, which when if 1/0 is multiplied by 2 it becomes 2/0

What is a fraction, really? It's a residue class of some equivalence relation. The relation is (a,b)~(c,d) if and only if a*d=b*c. Since (1,0)~(2,0) we must have 1/0 and 2/0 in the same residue class. However, (1,0)~(a,0) where a is any element of the field, so (1,0) is in all residue classes (and thus anything divided by zero is as well) so we say that anything divided by zero is undefined.

You can't just say 0/0 is undefined and then undefined * 2 is not a valid answer.

Sure I can. Undefined means it doesn't even exist...like a pink elephant. If you have one pink elephant and one grey elephant in a room you'd have two elephants (the beginnings of a circus). However, that pink one doesn't exist (otherwise I switch "pink" for "pokadotted") so you really only have one elephant in the room. (contradiction)

However, if you have one grey elephant in the room and no pink ones, you only have one elephant in the room. (that's ok)

See the difference?

[RQ]

Try this...define 0/0=1 like you wanted. Then multiply both sides by -1 so that we have

(-1) * (0/0) = (-1) * 1 = -1

so far so good.

However, let's make it look prettier and lose that middle step so we just have

(-1) * (0/0) = -1

so that

(-1*0)/0 = -1

and since (-1 * 0) = 0 we must have

0/0 = -1

which is a contradiction.

Yes that disproves that 0/0 cannot equal 1 wher x=x which is the original assumption.

If say 0/0 = 0 instead, what we really mean are these two things: first, 0 does have a multiplicative inverse; and secondly, that it satisfies

0*(0^-1) = 0.

But also

0*(0^-1) = 1,

since that's what it means to be a multiplicative inverse. Thus, 1=0 which is a contradiction.

Well those are the rules for numbers other than zero. I say the rule shouldn't apply to 0 because of its special properties such as the fact that any number x times 0 equals 0, or this equation: 5*0=0, 1*0=0, thus 5*0=1*0, but that doesn't mean that 5=1 because if you divide each side by 0 you get 5=1 if 0/0=1 and x=y where x doesn't equal y if 0/0= any number other than 0, which is ridiculous and is a contradiction.

[RQ]

Another way to reach the same conclusion is using the usual rules for adding fractions:

(1/1) + (0/0)
= [(0*1)+(1*0)]/(0*1) (cross multiply)
= [(0+0)/0] (perform multiplications)
= 0/0 (perform addition)
= 0. (by assumption)

On the other hand,

(1/1) + (0/0)
= 1+0 (1/1=1 by definition, and 0/0=0 by assumption)
= 1. (definition of additive identity (zero) in a field)

Once again, we have 0=1 which is a contradiction.

Cross multiplication means you multiply both fractions by 1, yet meaning that you have assumed that 0/0=1. This was however very convincing that 0/0 can't be 0 and believe me I was almost gonna give it up.


I think if we're gonna disprove that 0/0=0 we have to see how 0^-1 is a nonexistent value, but the definition is that n^-1 means n/n

[houserichichi]

Alright...so I'll get back to the reason why 0^-1 is nonexistent. You had a problem with the fact that it is REQUIRED that

0*(0^-1) = 1

being the definition of what 0^-1 means and the fact that we must also have

0*(0^-1) = 0

since we all know 0 times anything is zero. Problem is, the first one is the definition (and I stress that word) of multiplicative inverse. The second one can be deduced from the laws of arithmetic ... it's not a definition, but a consequence of other definitions.

So both of them have to be right in a field, as much as I hate to admit. The only thing is, we define inverses in a field for all nonzero elements. That's a definition, not a choice so right there 0^-1 doesn't exist because we made it so when defining what a field is. Thus, 0/0 is as undefined as 0/0 algebraically.

ONLY when we speak in the language of analysis do we really need to worry about it being indeterminate, but that's because we deal with limits there.

Did that help?[/b]

[RQ]

Yes, that is true, but you have to remember that 0 is not an ordinary "number" thus special rules must apply.

However, I did find a flaw as to the proof of 0/0=0 which makes it insufficient and assumption ridden. It is the fact that 0^1 does not equal 0^(1+0) or anything that adds or subtracts a power equal to or less then 0 to the 0. This is because x^0 is assumed to be an identity, thus x^(1+0) means x^1 since [x^1]*[x/x]; x^1*1. This means that 0^0 is assumed to be 1, which is invalid. I withdraw any proof that I claimed to have for 0/0=0 since it has been found on assumptions that are invalid and misleading contradictions.

I didn't see this at first, but there is no reason not to believe that 1/0=0/0=x/0. This at least holds. Furthermore, this isn't a proof that 0/0 is undefined. It's merely a false one that it is 0.

I wish however that people see that calculations where 0/0 has a value of 0 do not result in any contradictions. The inverse you are talking about only applies to nonzero numbers, since as you said 0 to the power of 0 or less cannot be shown in this way to equal some other power.

I do still say that 0^-1=[0/0]/0 and 0^0=0/0, since 0^1 means 0, and 0^0 would be 0/0.

[houserichichi]

0 is not an ordinary "number" thus special rules must apply.

Well, it's got special properties (as it's the additive identity), but other than that I see no reason to treat it differently. So long as we follow the field axioms we'll come to any and every conclusion necessary...and in those axioms, the zero doesn't have a multiplicative inverse.

As for setting 0/0 = 0 not resulting in contradictions, it may be the case in certain scenarios where that may be the best choice, but it doesn't follow from the axioms - what we'd be doing is making a temporary definition. The rules of algebra say that 0/0 is undefined, plain and simple, so when we give it a value (any value) we're temporarily breaking the rules, or at least choosing to give one priority over another. Same goes for 0^0.

I do still say that 0^-1=[0/0]/0 and 0^0=0/0, since 0^1 means 0, and 0^0 would be 0/0

By definition 0^1 = 0, but why would 0^0 = 0/0? If that's the case then yeah, you've got it bang on...but remember that dividing by zero is the same as multiplying by the inverse...so

0^-1
= [0/0]/0
= [0*0^-1]*0^-1

If we now cancel the 0^-1 from the first line with the 0^-1 on the bottom right, we get

1
= [0*0^-1]
= 0/0

and since you also decided 0^0 would be equal to 0/0 (in the thing I quoted above) then we MUST have

0/0 = 0^0 = 1

which I've proved a couple times isn't the case. Now, I assumed something above...the part where I cancelled out the 0^-1 can't be done. However, algebraically we require that ability, so have HAVE to assume that

0^-1/0^-1
= 0^-1*(0^-1)^-1
= 0^-1*0
= 1

so it's flawed from the beginning!

[RQ]

Well, it's got special properties (as it's the additive identity), but other than that I see no reason to treat it differently. So long as we follow the field axioms we'll come to any and every conclusion necessary...and in those axioms, the zero doesn't have a multiplicative inverse.

As for setting 0/0 = 0 not resulting in contradictions, it may be the case in certain scenarios where that may be the best choice, but it doesn't follow from the axioms - what we'd be doing is making a temporary definition. The rules of algebra say that 0/0 is undefined, plain and simple, so when we give it a value (any value) we're temporarily breaking the rules, or at least choosing to give one priority over another. Same goes for 0^0.

You can't say that the field of axioms apply for 0/0 since you can't show it mathematically given the initial axiom of x=x and the definition of 0. The inverse of 0 may be said to be x/0 which x can be any number essentially. However note that I didn't say that x/0 is the reciprocal of 0, since that would mean that 0/0=1, which is untrue if you multiply both sides by a number y that doesn't equal 1. Inverses do exist, unless they are undefined, then you are talking about the fraction itself which has no evaluation yet can be expressed in a ratio though meaningless in an equation.

By definition 0^1 = 0, but why would 0^0 = 0/0? If that's the case then yeah, you've got it bang on...but remember that dividing by zero is the same as multiplying by the inverse...so

0^-1
= [0/0]/0
= [0*0^-1]*0^-1

If we now cancel the 0^-1 from the first line with the 0^-1 on the bottom right, we get

1
= [0*0^-1]
= 0/0

and since you also decided 0^0 would be equal to 0/0 (in the thing I quoted above) then we MUST have

0/0 = 0^0 = 1

which I've proved a couple times isn't the case. Now, I assumed something above...the part where I cancelled out the 0^-1 can't be done. However, algebraically we require that ability, so have HAVE to assume that

0^-1/0^-1
= 0^-1*(0^-1)^-1
= 0^-1*0
= 1

so it's flawed from the beginning!

Again by dividing each side by 0^-1 you assume that 0^-1/0^-1=1 and only result to get it in the end. This is untrue, but by your equation it arises no contradictions, except the assumption that 0/0 is anything else other than 1.

0^-1
= [0/0]/0
= [0*0^-1]*0^-1

Hmm, this is perhaps the most interesting point brought up in the whole discussion.

If 0^0 represents 0/0 and 0^-1 represents [0/0]/0 and so on, then 0^-1 can be shown to equal where x is a positive nonzero number

{[0^x]*[0^0]}*0^-1 <----- NOte the reason it is 0^-1 and not 0^0 (0/0) is because *0^0 would mean that 0^-1=0/0 which is a false with the given statements assumption.

Thus by this we could say that 0^-1 with the given assumptions under which 0/0 is not undefined as I said equals [(0^x)*(0^0)]*[(0^x)*(0^0)]*0^-1 which would not be invalid if 0/0=0 or 1.

By the reason I gave earlier [0^0]*[0^x] where x is a positive value, does not equal [0^(0+x)]. Thus:

[0^2x]*[0^0]*0^-1=0^-1

Now if we divide this whole process by 0 we get that

{[0^2x]*[0^0]*0^-1}/0=0^-2

{[0^2x*0^-x]*[0^0]*0^-1}=0^-2

and we get that 0^-1=0^-2

or the other version which we don't make /0^x into *0^-x which would be its nonexistent inverse, we get that
0^-2=0^-1/0^x where x can be as many 0's which makes sense since 0^1=0^2=0^3 and so on.

This is extremely circular.

[houserichichi]

You can't say that the field of axioms apply for 0/0 since you can't show it mathematically given the initial axiom of x=x and the definition of 0

Please read this
http://math.la.asu.edu/~kawski/classes/mat371/handouts/realaxioms.pdf
which is the real number axioms. The first 10 are some of the field axioms, the next four are the order axioms, and the last is the supremum axiom. However, look at axiom F7...it's for all numbers in the reals EXCEPT 0.

Again by dividing each side by 0^-1 you assume that 0^-1/0^-1=1

Yes, dividing something by itself SHOULD equal 1...but FIRST the numerator has to be PART of the set in question (in this case the real numbers), and SECOND that particular thing MUST have an inverse. 0^-1 isn't in the reals but if it was, I suggested that its inverse would be 0.

[RQ]

I think due to inconclusive mathematical proof (from either side) we should consider this matter closed until something new and valid comes up with no given assumptions or axioms.

[PWrong]

Ok, it seems the 0/0 argument is over, (presumably RQ is willing to accept the current axioms until he revolutionises algebra).

How do you explain other paradoxes? I think Russell's paradox is the main one:
http://mathworld.wolfram.com/RussellsAntinomy.html
I read somewhere that this underlies all other paradoxes. For instance, here's a simplified version of the same thing:
"This statement is false"

Of course, I don't think mathematics is wrong about set theory or anything, I'm just wondering how you think these paradoxes should be interpreted.

[houserichichi]

Two things...

One, 0/0 isn't a paradox, it's a rule.
Two, Russell's paradox, specifically, is handled in a few different ways, one of them being the introduction of the ZF/ZFC set theory axioms which rids us of such a troublesome issue. There are other methods, if you do a quick search on the net you'll find other methods (one proposed by Russell himself and another by Hilbert, for example.