Division by zero

[ScandinavianWarrior]

Why do math teachers say you cannot divide by zero. Dividing by zero would result in an infinite number. Zero divided by zero therefore would have to be any real number due to the modulus operator. So why can't teachers just come out and say that?

[Nintendofreak]

People usually refer to division by zero as being undefined, which makes sense, because there is no definite value to the expression.

And many people say that anything divided by itself is going to be equal to one. I disagree with that. I have to agree with you as it being all numbers, rather than just one.

You can split zero up into as many groups of zero as you want, and the remainder is still going to be zero, meaning that you are done dividing.

But to answer your question, dividing by zero is mathematically impossible. Theoretically, it does equal infinity, but the best way to express it is by just simply saying that it's undefined, and the math cannot be done. This is the reason why when you try to do so with a calculator you get an error. Computers don't like infinity.

[ScandinavianWarrior]

People usually refer to division by zero as being undefined, which makes sense, because there is no definite value to the expression.

And many people say that anything divided by itself is going to be equal to one. I disagree with that. I have to agree with you as it being all numbers, rather than just one.

You can split zero up into as many groups of zero as you want, and the remainder is still going to be zero, meaning that you are done dividing.

But to answer your question, dividing by zero is mathematically impossible. Theoretically, it does equal infinity, but the best way to express it is by just simply saying that it's undefined, and the math cannot be done. This is the reason why when you try to do so with a calculator you get an error. Computers don't like infinity.

So we are agreed that division by zero results in an infinite number. Well math equations have the variable "i" as being the square root of negative one if my memory serves me correctly. Shouldn't math have a defined variable for an infinite value?

[Nintendofreak]

A defined variable? Sounds a bit oxy-moronic, but I see what you're saying.

And rather than saying "an infinite number", just say "infinity". You are implying that there are multiple infinite numbers. And infinity is expressed as that sideways "8" thing, which you already know. It is also commonly referred to as "~" when typed out, because using the charmap to find the symbol for infinity (or alt codes) is just a waste of time.

[ScandinavianWarrior]

A defined variable? Sounds a bit oxy-moronic, but I see what you're saying.

And rather than saying "an infinite number", just say "infinity". You are implying that there are multiple infinite numbers. And infinity is expressed as that sideways "8" thing, which you already know. It is also commonly referred to as "~" when typed out, because using the charmap to find the symbol for infinity (or alt codes) is just a waste of time.

Okay. So then zero divided by zero is all real numbers. How would that differ from the rule that all numbers divided by themself equals 1?

[Nintendofreak]

Working with zero can always cause a lot of problems, and may result in an undefined number. As you know 1/1 is equal to 1, and 2/2 is also equal to 1. Everything divided by itself is equal to 1, except in the instance of zero.

0/0 is simply all real numbers.

Let's look back at 2/2. If you had two stones, and you were going to organize them in groups of two, you would only be able to make one group.

If you had zero stones, and you wanted to organize them in groups of zero, you could give yourself one group of zero, and have a remainder of zero (going back to what you said about 0%0), or two groups of zero, and STILL have a remainder of zero. If you have a remainder of zero, that usually means that you have found the answer.

I guess it all depends on what parameters you make when dividing. I guess some people may agree that 0/0 is equal to infinity as well, because if you have a remainder equal to the denominator, you can divide it again. In the instance of zero, however, you will always have the remainder of zero.

[ScandinavianWarrior]

I suppose no matter how you arange the equation and define the parameters you still cannot get a definite answer?

[Nintendofreak]

I don't quite understand the question, there.

Give me another example of how to express 0/0? O.o

[ScandinavianWarrior]

Is there any possible way to get a defined answer when dividing by zero?

[Nintendofreak]

I really doubt it. If you have zero in the denominator, then no matter what, you will never have an exact, precise answer that you can express using only numeric characters (0-9).

[ScandinavianWarrior]

I really doubt it. If you have zero in the denominator, then no matter what, you will never have an exact, precise answer that you can express using only numeric characters (0-9).

What about variables?

[Nintendofreak]

I don't know if calling it a variable is appropriate, but there are symbols for numbers like infinity and root(-1) as you already mentioned.

And those symbols are used *because* of the fact that the number is undefined.

I'm sticking with my answer of no, it can't be defined.

[ScandinavianWarrior]

I don't know if calling it a variable is appropriate, but there are symbols for numbers like infinity and root(-1) as you already mentioned.

And those symbols are used *because* of the fact that the number is undefined.

I'm sticking with my answer of no, it can't be defined.

I want other opinions on the matter. See what other people have to say. However, I agree with you.

[wendy]

For various values of zero, one can indeed do division by zero, but this must never be handled straight on.

For example, the limit of 1/u is 0, as u goes large. If one treats this accordingly, one can indeed cancel out 1/u for large values, and pick up supplemental terms. It is accurate and precise enough for there to be based a geometry on it (euclidean geometry corresponds to a sphere of diam 1/u, and we are dealing with u+m vs u+n giving a real difference of m-n.

On the other hand, 1/0 is not a place but a direction. That is, it's not like "36", but like "zillions and zillions". None the same, not all mathematical operators work, still, some do.

Note that 0 ^0 = 1. This is because the product of zero terms leaves the product undisterbed, regardless of what size the terms are.

W

[bo198214]

Assume x/0 would be defined (as possibly something outside the rational numbers) but the laws of the rational numbers are kept for all those new values. Then would
0/0 + 0/0 = (0+0)/0 = 0/0
and after subtracting 0/0 from both sides
0/0 = 0

On the other hand would
0/0 * 0/0 = (0*0) / (0*0) = 0/0
and after dividing by 0/0 on both sides
0/0 = 1

So dividing by 0 leads to contradictions, thatswhy it is prohibited.

Note that 0 ^0 = 1

Thats no equality but a definition. For the operation ^ to be defined for certain real numbers x and y, for every rational approximation of x, say by the sequence (x_n), and for every rational approximation of y, say by the sequence (y_n), the sequences (x_n^ y _n) must all approximate the same real number. Thats not the case for x=0 and y=0. Though the definition seems to lead to no contradictions.

[moonlord]

I always thought 0/0, 0^0,~-~ and other things like this must be avoided when calculating limits. I don't see how else they would come it...

EDIT: Now I see why 0^0 is ok to be 1. I just remembered a joke saying the Big Bang happened because God divided by zero...

[thigle]

no answers from me but here is some background...

there are certain obvious dualities between Geometries & Algebras.
let's consider how the Real axis was slowly built and then found to correspond to Reals algebra. (at least in western tradition)

old greeks (not knowing they are old) at first worked with only positive integers, without zero, and not 'too big'.

then, as the ratios between integers were explored, the free space between the few points of 'small positive integers' on yet-to-be-found Real axis started to fill up.

however, even when 0 & negative numbers (integers as well as rational) entered the discussion, there were still hidden 'points' or free slots on this line.

these were filled up by irrational numbers, which allowed the FULLNESS of the future Real axis, even though its ends (extremes, '- & + infinity') were still invisible in the in the fog of infinite distance.

then with the advent of calculus & limits, the 'infinity' started to be consistently used. however, it was still 'infinite' infinity (or 'open', 'potential' infinity, as it is sometimes called)

& then Cantor stepped out of the finite into the transfinite & closed the infinity, actualising it. (anyway that's another ramification of this story we don't have to follow.)

though, synthetic, projective geometry did the same.

another important step was realisation of 1 to 1 mapping between the geometry of Complex Plane (CP2) and the algebra of the complex numbers by Riemann(?) by stereographic projection from the top of the sphere touching the complex plane at its origin by its nadir.

as a multiplication by complex number equals to rotation in complex plane, by analogy, rules for multiplication of triplets was desperately seeked by Hamilton. knowing about the anomaly of the root i of the complex algebra, he became so obsessed with finding new rules for multiplying triplets (with 2 imaginaries) to represent completely the rotations in 3-space (check out the explanation with the images of hand&book), that after years of seemingly-futile trying, even his small children would ask him at the breakfast table: "well papa, can you multiply triplets ?"
and then there is the famous story of Hamilton, how after struggling for years to find rules for 2 'imaginary' roots of his algebra of triplets supposedly corresponding to the way rotations work in 3d, he realised (walking on a bridge!) that the solution to this 'problem' is that it has NO solution ! there exist no algebra of 'triplets of numbers' corresponding naturally to the structure of 3-space ! the minimal system for handling the nature of 3-rotations consists of quadruples of numbers - quaternions. and so he carved his discovery into the laying-by stone: ijk=i²=j²=k²=-1

thus quaternions were found to be the algebra of 3-space, and soon octonions were found to be even more inclusive.

*

now how is this related to the issue of dividing by zero ?

Cayley-Dickson construction lead to the understanding of the generalised process of constructing algebras. from Reals to Complex to Quaternions to Octonions ( these were called Division Algebras) and beyond.

this beyond is where there starts the possibility of Zero divisors (or check wikipedia for a bit less technical presentation)...
...for it was found, that the further this process of constructing algebras goes, the more algebraic properties are lost. division is the last one that is discarded.

so 1-dim real numbers are have all the commonly used algebraic properties
2-dim complex numbers, which are not ordered - this can be seen from the nature of ordered rotations in 3-space.
4-dim quaternions, are not commutative;
and 8-dim octonions, are not associative.

tonySmith states:

As Okubo has noted in his book Introduction to Octonion and Other Non-Associative Algebras in Physics (Cambridge 1995), the theorem that real division algebras must have dimension 1,2,4,8 "...has been derived on the basis of topological reasoning on a seven-dimensional sphere. A pure algebraic proof of the theorem is still unknown."

to sum up, defined ways to divide by zero are found from algebras up from octonions, in the so called Zero-divisor algebras, where one can have ab=0 for non-zero a & b.

[houserichichi]

Somewhere on this forum there is another discussion I had with RQ on the matter. To break it down to its most fundamentals (assuming you're talking about what I'm talking about)

A field (you can look up the definition yourself) F is defined to have no zero divisors. Real numbers form a field. Thus, any real number divided by zero is not in the field so it cannot happen.

Also, 1/0 does not equal infinity, it just tends to it. Infinity is not a number so a quotient cannot "equal" it, the quotient can only approach infinity as one of the numbers (in this case the bottom one) approaches some other number (in this case 0). That is, 1/a approaches infinity as a approaches zero. Again, problem with this is which side do you come from? If you approach zero from the left then 1/a approaches negative infinity. If you approach zero from the right then 1/a approaches positive infinity. Do we just say that "infinity" is BOTH negative and positive infinity?

I found the discussion there. Hope it helps.

[bo198214]

division is the last one that is discarded
...
defined ways to divide by zero are found from algebras up from octonions, in the so called Zero-divisor algebras, where one can have ab=0 for non-zero a & b.

If division is no more possible, what sense does it make to speak of division by zero?

[houserichichi]

He's just talking about division algebras, a special case of an algebra. You can still divide, but now you can divide by zero whereas before you couldn't. There are other systems where you can't divide by other numbers...for instance, the Wiki on division by zero shows an example where 2/2 is undefined in Z/6Z since 2x=2 has two, independent solutions in that ring (notice that I said ring and not field, the kicker).

[bo198214]

He's just talking about division algebras, a special case of an algebra. You can still divide, but now you can divide by zero whereas before you couldn't.

He said already that the only real division algebras are real numbers, complex numbers, quaternions and octonions. He talked about zero divisors in higher dimensional real algebras, which are then necessarily no division algebras. So there is no division and so it makes no sense to talk about division by zero.

[bo198214]

Wiki on division by zero

There is an interesting mention of wheels with the statement "The real numbers can be extended to a wheel, as can any commutative ring."

But as one already could imagine there is some drawback. In the case of wheels the subtraction is no more possible. As I already showed for the rational numbers defining division by zero leads to a contradiction. The proof easily extends to fields (i.e. addition and multiplication abelian group).

[houserichichi]

Some elements will still have mutliplicative inverses (ie: they can be "divided"), but not all.

Infinite dimensional division R-algebras are not isomorphic to R, C, H, or O. As I mentioned in the previous point he was talking about (finite dimensional...should have included that bit) division R-algebras, a special case that follows the 1, 2, 4, 8 Cayley Dickson process.

I already showed for the rational numbers defining division by zero leads to a contradiction. The proof easily extends to fields

The rationals are a field.

[wendy]

The notion that 0^0 is something that requires a separate definition is wrong. This particular notion depends on the only way of arriving at zero is to do 1-1, which is similar to saying that you can't proove there are no elephants in your lounge unless you put one in there, and then remove it.

Put simply, 0 is a counting number (meaning none found), and one can show that 0^0 is then the same as the identity (1), with affixed zero copies of the string "*0", ie 1[*0]_0 = 1.

A separate definition about 0^0 is much the same notion that the ancient greeks were hung up about whether 1 was a number (since it is not plural). One might throw stones at the greeks for this, with counting to one, and then still not be able to count to zero.

W

[bo198214]

The rationals are a field.

Yes, thatswhy I mentioned that the contradiction is not rational specific, but is valid in every field.

The notion that 0^0 is something that requires a separate definition is wrong

No, Wendy, you may surrender to an illusion.
When regarding x^y as an operation on the reals, we must exclude x<=0 as I will explain later.
Of course for real x and natural y we would define 0^0=1. That poses no problem when we start the induction at 0: x^0=1 and x^(n+1)=x*x^n. But already if we extend to integer y by x^(-n) = 1/x^n, this poses a slight problem for x=0 (and n=0), but of course we can stick to our definition and only for y<0 we exclude x=0.

Then let us extend this definition to the reals. I know two ways how to do that:
1. We define e^x by an infinite sum (of which we show that it always converges). Then we show that e^x is a bijection from R to R_>0 so we can define the inverse function ln(x) for x>0 and then define x^y=e^(y*ln(x)). So in this definition x=0 must already be excluded.
2. We define x^y first for rational y and then extend it to the reals by converging/Cauchy sequences. When extending to rational y via roots, we must exclude x<0, but we can still keep 0^0=1. When we now extend to real y then each Cauchy sequence (y_n) of rationals must yield a converging sequence x^(y_n). If we regard now x=0 and sequence y_n=1/n then x^(y_n) converges to 0. If regard on the other hand the constant sequence y_n=0. Then x^(y_n)=1. And when we mix both sequences we can make x^(y_n) even not converging. So x^y has no proper definition for x=0 and y=0. (I dont need to mention that x^y is well defined for x=0, y>0 and all x>0, arbitrary y in this extension to the reals)

So from the last consideration we could conclude that 0^0=0 and that 0^0=1 on the reals. This is kind of similar contradiction as we get when assuming that 0/0 is defined on the rationals. On a non-strict level maybe that 0^0=0^(1-1)=0/0 is convincing.

This particular notion depends on the only way of arriving at zero is to do 1-1

No, its exactly the opposite, i.e. that there are different ways to arrive at zero and each gives another result to 0^0. As I already said in my previous post.

[thigle]

recently, i heard about a parrot from England (forgot his name so no credits to this bird) who exhibits an unusual feature for his kind: recognition of an empty set, of nothing, of zero.

consequently, groups of scientists gathered to 'study' this 'unusual' bird, to find out whether this is a 'real' thing.

maybe they should just take a sip of tea and relax.

[PWrong]

So from the last consideration we could conclude that 0^0=0 and that 0^0=1 on the reals. This is kind of similar contradiction as we get when assuming that 0/0 is defined on the rationals. On a non-strict level maybe that 0^0=0^(1-1)=0/0 is convincing.

There are easier ways to do this. Just try to take the limit of x^y as (x,y) approaches (0,0). You get different answers on different paths. This proves, by definition, that the limit does not exist.

Or you could look at a 3D graph of z=x^y. It's easy to see that the graph is discontinuous.

Now that I think about it, z=x^y is an absurdly complicated object. If all the variables are complex, it has 6 dimensions, no symmetries to speak of, and contains just about every simple curve we know of as a subset. It's no wonder Jinydu and I never worked out that continuous tetration thing.

[bo198214]

But even if we let out the reals and convergence, but only consider natural numbered exponent:
0^0 =: a
a=0^0=0^(2*0)=(0^0)^2=a^2

Then we have only the both possibilites a=0 or a=1. And I see no way to decide algebraicly which value is the "right".

@PWrong
I anyway avoid to use exponentiation for complex numbers because, its not uniquely defined.

[wendy]

Let 0^0 = n

Now, we note that "a" = "a * m^0", since there are sero occurances of m in the LHS

We then note that this is true regardless of the value of m.

We note this is true for all values of a.

Therefore m^0 = identity(*) = 1, for all values of m

Therefore 0^0 = 1:

QED

[bo198214]

Wendy! I thought you were a mathematician! This "proof" would not even satisfy a physician!

We know that 0^m = 0 and we notice that this is true for all m, hence 0^0=0, QED.
Would *you* accept such a "proof"?