Can Division by Zero Be Defined in a More General Number Group?

[Blue_Jaunte]

I assume everyone who saw this thread first rolled their eyes. Sorry for that. My question is an innocent one, as I am completely uneducated in number theory.

My understanding is that numbers are grouped in a sort of Russian doll fashion, with each successive group encompassing all previous ones. I suppose there is an infinite number of ways to define these groups, but the way I order them goes like this:
-Natural numbers
-Rational numbers
-Real numbers
-Complex numbers

Now, since complex number require the special definition of √(-1), could you just define division by zero (arbitrarily, say, as "1/0 = m") and make an even more general number group?

Blog Insight: Division by zero

 

[JCVD]

Division is the inverse process of multiplication, so if one wants to allow division by zero one would have to accept 0/0 = 1, as 1 is the multiplicative identity. Then, assuming division by zero is valid, one would have to accept the following argument:

1*0 = 0, 2*0 = 0, 0=0 --> 1*0 = 2*0 --> 1*0/0 = 2*0/0 --> 1 = 1*1 = 1*0/0 = 2*0/0 = 2*1 = 2.

 

[Tac-Tics]

could you just define division by zero (arbitrarily, say, as "1/0 = m") and make an even more general number group?

What's in a name?

In math, you are free to use whatever definitions you want. But once you choose one, you have to stick with it! You can't change your mind, mid-proof.

Division is a binary operation. An operation is just a function with fancy syntax. So instead of writing x/y, let's write it out in familiar function notation as d(x, y). It's defined on all points in the (x,y) plane except for the line made up of all points where y = 0.

If we want to allow division by zero, all we need to do is "fill in" this line. We can just make up whatever values we want for d(x, 0) for each x. They can be real numbers, or we can create numbers and fit them into the rest of the numeric framework we are creating.

Let's start off with a naive attempt. Let's just say that d(x, 0) = 0 for all x. So division by zero is now defined. Have we succeeded? Well, we can now say that 1/0 = 0. And 2/0 = 0. But by doing this, we implicitly invalidate some theorems we're familiar with. Namely, division is no longer the inverse of multiplication. It's still true that (2 * 3) / 3 = 2, but it's NOT true for all x and y that (x * y) / y = x. Why not? Because that rule would imply (2 * 0) / 0 = 2, when in fact, by our definition, (2 * 0) / 0 = 0. Oops.

And that is the crux of the problem. The minute we define a value for x/0, division is no longer the inverse of multiplication. It can't be, regardless of the value we choose for it. There is a proof of this that isn't that hard to understand, but I'll leave it to you to convince yourself.

 

[ramsey2879]

As JCVD pointed out division must have its inverse operation to get back to original thing that was divided. Thus n/4 * 4 = n* 4/4 = n * 1 = n. This can be done with zero since zero times anything is zero. Division by a is thus like dividing something into a equal parts that add up to the thing divided. You can't divide by zero since zero equal parts of anything is always zero.
A number group that allows division by zero would be completely useless. On the otherhand, complex numbers have many practical applications, especially in electronics and wave theory.

 

[SW VandeCarr]

A number group that allows division by zero would be completely useless. On the otherhand, complex numbers have many practical applications, especially in electronics and wave theory.

It's hard to tell what might be completely useless in the future. Grassmann numbers were invented in the mid 19th century, but didn't find a use until the 20th century (in physics.)

 

[chem123]

Interesting thread. Tac-tics response was great. I wish I'd seen it explained like that a long time ago.

 

[g_edgar]

There is a system known as the "Riemann sphere" where operations + and \cdot are defined on a sphere. Except a few combinations are undefined. This does have a 0 and division by 0 is defined (with an exception 0/0 is undefined). It also has \infty. For example a/0 = \infty if a \ne 0. And 0 \cdot \infty is undefined. Often thought of as an extension of the complex numbers by adding one extra point \infty.

If this seems interesting to you, see if you can look it up.

 

[SW VandeCarr]

This does have a 0 and division by 0 is defined (with an exception 0/0 is undefined).

x/x=1 for all real non zero x. Could 0/0 be defined as zero? Would that have any major consequences for Riemann sphere arithmetic or even standard arithmetic? For example we have the Dirac delta "function" which is zero everywhere except when x=0 where the "function" is infinite. The Dirac delta is defined for all real numbers and need not be affected by defining 0/0 as zero.

 

[Hurkyl]

My understanding is that numbers are grouped in a sort of Russian doll fashion, with each successive group encompassing all previous ones.

This is a reasonable description of the number systems a layperson typically encounters , but it doesn't work when you start studying more interesting things such as: modular arithmetic, angular position, polynomial rings, function fields, tensor fields, cardinal numbers, extended real numbers.

And some other interesting things would fit in your hierarchy, but it would cause branching. For example, one interesting ring is the Gaussian integers are the set of all complex numbers of the form m+ni where m and n are integers.


Even within the hierarchy you mentioned, you generally lose properties as you go up. The natural numbers has the property that mathematical induction works for them, but it doesn't work with the rational numbers. The real numbers have an ordering, but the complex numbers do not.

Of course, you gain properties too: the complex numbers are algebraically closed -- every complex polynomial has a root -- but the real numbers do not have that property.

Now, since complex number require the special definition of √(-1), could you just define division by zero (arbitrarily, say, as "1/0 = m") and make an even more general number group?

You can define things where division by zero is sometimes or even always defined. But this is where "losing properties" rears its ugly head: if we insist on keeping all of the algebraic identities we know and love, you can prove that all numbers are equal, and so this is a very boring mathematical structure.

Probably the most useful structure that allows division by zero is the projective numbers -- but in this system, 0/0 is still undefined, as is 1/0 + 1/0. Wheels are closely related structure where the four arithmetic operations are always defined, but arithmetic identities become more complicated, such as

x \cdot z + y \cdot z = (x + y) \cdot z + 0 \cdot z​

(Pay careful attention to the fact that in a wheel, it is possible to have 0 \cdot z \neq 0)

 

[CRGreathouse]

Even within the hierarchy you mentioned, you generally lose properties as you go up. The natural numbers has the property that mathematical induction works for them, but it doesn't work with the rational numbers.

You can use induction on the rationals, as long as your inductive step let's you walk along Stern's diatomic series.



OK, so I'm feeling snarky today. Hurkyl's point about losing properties as you move 'up' the hierarchy is actually a very useful concept.

 

[Labyrinth]

Now, since complex number require the special definition of √(-1), could you just define division by zero (arbitrarily, say, as "1/0 = m") and make an even more general number group?

No, because division by zero is not division. The operation of division simply doesn't apply when the divisor is 0. It means that no operation is to take place, so it cannot be given a number or variable. You'd have to use some other (new?) operation that is not division.

I do wish there was an accepted symbol for undefined though.

 

[Ignea_unda]

You can use induction on the rationals, as long as your inductive step let's you walk along Stern's diatomic series.

That's an interesting concept. Is there a particularly text that would provide me with a good explanation of Stern's diatomic series?

 

[DaveC426913]

Hurkyl's point about losing properties as you move 'up' the hierarchy is actually a very useful concept.

Bah. Mathematicians are thieves. The concept was stolen from the object-oriented programming concepts of 'prototypes' and 'inheritance', which preceded the invention of math by many centons.

 

[CRGreathouse]

That's an interesting concept. Is there a particularly text that would provide me with a good explanation of Stern's diatomic series?

I don't know of one. Here are some web pages:
http://www.research.att.com/~njas/sequences/A002487
http://mathworld.wolfram.com/SternsDiatomicSeries.html
http://en.wikipedia.org/wiki/Calkin–Wilf_tree

 

[zetafunction]

is the idea of Blue Jaunte to define a ring of 'infinite numbers?

I(R)=(a,b =Real |a+b\infty = I )

 

[Mentallic]

I do wish there was an accepted symbol for undefined though.

Oh as do I. It just doesn't seem as elegant when being able to express every defined solution concisely in a mathematical sense, then for the undefined solutions, I need to write in words "undefined". It just... suks.

I've come to understand for myself that even though 1/0 is undefined, it still has useful properties such as being able to imagine it as being \infty and thus being able to solve equations with this little issue. e.g. tan\theta=1/0 would give \theta=\pi/2\pm n\pi.
Also for 0/0, I see this as (from my personal experience) that this is going to be a finite, non-zero value. The only thing holding you back from finding this value is to find the limit of such equations that use this. e.g. finding the value of f(0) in f(x)=xcotx

 

[Hurkyl]

I've come to understand for myself that even though 1/0 is undefined, it still has useful properties such as being able to imagine it as being \infty

We don't imagine -- we construct a new number system (e.g. the projective numbers) in which 1/0 is defined, and use that one instead of the real numbers.

By the way, both of these limits are of the form 0/0:

\lim_{x \rightarrow +\infty} \frac{x}{x^2}

\lim_{x \rightarrow +\infty} \frac{x^2}{x}​

 

[Mentallic]

We don't imagine -- we construct a new number system (e.g. the projective numbers) in which 1/0 is defined, and use that one instead of the real numbers.

Ahh ok I didn't know that, thanks.

By the way, both of these limits are of the form 0/0:

\lim_{x \rightarrow +\infty} \frac{x}{x^2}

\lim_{x \rightarrow +\infty} \frac{x^2}{x}​

Did you mean for the limits to be approaching 0?
In that case, yes it's true, but for all cases that I can think of, they can always be transformed into multiple limits such as
\lim_{x \rightarrow 0} \frac{x}{x^2}=\lim_{x \rightarrow 0} \frac{x}{x}* \lim_{x \rightarrow 0} \frac{1}{x}=1*\frac{1}{0}=\infty on those projective numbers you've mentioned

 

[demoivre]

Hey
I'm a first year math student at university and I've been trying to define this whole division by zero as well.
Everybody keeps going on about the division being the inverse of multiplication and all of that. What I think you guys missed is that zero is neither positive nor negative. Zero is where everything starts. In the system of numbers zero i defined as a integer and a whole number. And logic will tell us that any integer divided by itself is equal to one.

Now what I want to know is can one define 0 as a integer or a whole number...
Please respond to this and give me some ideas. I am not a very intelligent guy

 

[ramsey2879]

And logic will tell us that any integer divided by itself is equal to one.

Not if it is zero

 

[Cogent]

Since I've started learning division, it is learned that division by zero is undifined. But in the book "Introduction to Analytic Number Theory" by Tom M. Apostol, I've found that "Zero divides only zero" [Theorem 1.1 (h)]. What does this statement mean?

 

[Hurkyl]

I've found that "Zero divides only zero" [Theorem 1.1 (h)]. What does this statement mean?

The set of things which zero divides is {0}.

 

[Wesley Hughes]

Division by zero? NO! Wes Hughes

 

[CRGreathouse]

Since I've started learning division, it is learned that division by zero is undifined. But in the book "Introduction to Analytic Number Theory" by Tom M. Apostol, I've found that "Zero divides only zero" [Theorem 1.1 (h)]. What does this statement mean?

k | n is defined as "there exists an integer m such that mk = n". So 0 | 0 means that there exists some integer m such that 0m = 0.

 

[AButterfield]

Oh as do I. It just doesn't seem as elegant when being able to express every defined solution concisely in a mathematical sense, then for the undefined solutions, I need to write in words "undefined". It just... suks.

In theoretical computer science there is a generally accepted convention
that uses the symbol \bot to denote undefined. This is because undefinedness in programming language semantics (e.g. non-termination) is usually characterised as the least element in
a partial ordering where all (interesting) sets of values have a least element, as does the whole ordered space, hence the use of "bottom".

If integer division is implemented as repeated subtraction, then a
naive treatment of division by zero will result in an infinite loop,
modelled theoretically in programming semantics as \bot.

 

[HallsofIvy]

Since I've started learning division, it is learned that division by zero is undifined. But in the book "Introduction to Analytic Number Theory" by Tom M. Apostol, I've found that "Zero divides only zero" [Theorem 1.1 (h)]. What does this statement mean?

I haven't read that book (or any book on "Analytic Number Theory") but this might be relevant.

\frac{a}{0}, for a non-zero is not defined, in the real number system, because if we set a/0= b we get a= b(0)= 0 which is not true for any b. But \frac{0}{0} which is also not defined is a separate situation- if we set 0/0= b we get 0= b(0)= 0 which is true for all b. I rather suspect that this theorem is connected to that distinction.

 

[JSuarez]

Since I've started learning division, it is learned that division by zero is undifined. But in the book "Introduction to Analytic Number Theory" by Tom M. Apostol, I've found that "Zero divides only zero" [Theorem 1.1 (h)]. What does this statement mean?

There is a small imprecision here: Apostol is not talking about division; he is talking about divisibility, which is a diffrent concept in number theory.

Divisibility is a binary relation, defined on pairs of integers (or elements of a more general ring), defined by:

a|b iff there is an integer k, such that b = ka

With this definition, it's easy to see that 0|0 and, if x|0, then x = 0. In particular, there is no mention of invertible elements.

Division, on the other hand, is (usually) a function defined on a ring by:

a/b \equiv a \times b^{-1}

Here, b must be invertible, otherwise, as was pointed out by Hurkyl, if I'm not mistaken, if we allow 1/0 to be an element of the ring and keep the other operations unchanged, we end up with 1 = 0, and the entire eing collapses to {0}.

 

[CRGreathouse]

In theoretical computer science there is a generally accepted convention
that uses the symbol \bot to denote undefined.

Be cautious, though; that's also used in logic to denote falsity.

 

[okkvlt]

i think 0^-1=x.
x is every point on the complex plane, including the infinite and finite in magnitude.
and
0/0=0*any number on the complex plane
0/0=0*infinite=finite
0/0=0*infinetsimal=0
0/0=0*finite=0

so 0/0 is every point on the complex plane with a noninfinite magnitude, including zero.

 

[Mentallic]

i think 0^-1=x.
x is every point on the complex plane, including the infinite and finite in magnitude.
and
0/0=0*any number on the complex plane
0/0=0*infinite=finite
0/0=0*infinetsimal=0
0/0=0*finite=0

so 0/0 is every point on the complex plane with a noninfinite magnitude, including zero.

and what about lim_{x -> \infty} \frac{x^2}{x}=\frac{0}{0}=lim_{x -> \infty}x=\infty

 

[zetafunction]

and how about the function f(s)=0^{-s} many textbook take f(s) to be alwyays 0 by analytic regularization

 

[TurkishVanCat]

I do wish there was an accepted symbol for undefined though.

There is, depending on context of course. In recursive function theory, undefined is denoted by ↑ , to indicate that the function diverges (meaning its program/algorithm runs in an infinite loop and never stops) on a particular value, and thus is undefined on that input. I don't see why a similar notation couldn't be adopted for general math.

As to the original question, as those before me have stated, if you work with the more common systems of numbers, in order to define division by zero and maintain the 'nice' algebraic properties of these monoids/rings/fields, you wind up with the trivial ring, of which there is only one element, namely 0 (which is fairly boring).

Along with your question, you cited the fact that we give a definition to √-1, which seems like a rather arbitrary, silly thing to define. I'd just like to point out that this definition is actually very natural and makes a lot of sense within the context of other areas of math and science. As far as I'm concerned, i is no more 'imaginary' than √2 is. :)

 

[Gib Z]

Along with your question, you cited the fact that we give a definition to √-1, which seems like a rather arbitrary, silly thing to define. I'd just like to point out that this definition is actually very natural and makes a lot of sense within the context of other areas of math and science. As far as I'm concerned, i is no more 'imaginary' than √2 is. :)

No really so much a definition as it is assigning it a symbol.

 

[zetafunction]

Anyway could we make a 'o ring' i mean the ring with elements

a+ b.(1/0) with a and b considered real numbers , so this set have RING properties

 

[Hurkyl]

What is the sum and product you propose?

 

[zetafunction]

the sum would be the usual

a+b.(1/0)+c+d.(1/0)= (a+b)+(c+d).(1/0)

the problem with the product is that you make it worse , in the product a term (1/0).(1/0) would appear

 

[Hurkyl]

Well, then, with those operations, the set of {a + b (1/0)} is not a ring.

Of course, the set of all polynomials in one variable is a ring, and we can name that variable 1/0 if we choose. But does that have anything to do with division by zero?

 

[jcender]

Now, since complex numbers require the special definition of √(-1), could you just define division by zero (arbitrarily, say, as "1/0 = m") and make an even more general number group?

Consider defining a number m and also redefining zero with it as a way of extending the number system and thus defining division by zero. By defining both zero and "m" through division instead of subtraction, they can be multiplicative inverses. How might this be carried out?

Let's follow your example of the special definition of the √(-1) where i indicates an imaginary number when written together with a Real number, and choose a symbol that will indicate nothingness when written with a Real number. In other words, instead of the single number 0, there will be a set of absent numbers represented by Real numbers together with an absence symbol.

So let m be n\infty where n is a Real number, and let the absence symbol be \overline{\infty}. These numbers of nothing would then be n\overline{\infty}. In general, division by zero numbers would be \frac{x}{n\overline{\infty}} = (x/n) \infty. As with a reciprocal under a fraction bar, the inverse inverts. A zero number divided by another zero number would equal the quotient of their Real parts.

Note that the example of the imaginary numbers can only be followed so far. Normally at least, the additive identity property would preclude forming numbers made up of Real numbers plus absent numbers in a way analogous to the complex numbers.

The bar in the absence symbol is an inverse bar. It indicates absence and functions somewhat like a fraction bar except there is never a numerator. (This prohibition entails an exception for the division rule of equality although one much less onerous than the lack of definition of division by 0.) The infinity symbol here represents an array of the Real numbers. (Other definitions are possible.) To see some of the advantages of this notation, look to pp. 379-382, 580-2, 897-907, 916-25 in the book The Road to Reality by the physicist and mathematician Roger Penrose. He uses it for n-real dimensional space, without, however, a multiplicative inverse. Also note the compatibility of the new numbers with the complex extended plane or Riemann sphere referred to on this thread by g_edgar on Jul30-09. Indeed they may be seen as a generalization of it.

A number of issues not addressed here such as implications for physics, defining the absence symbol more fully, defining subtraction, an unsigned zero, arithmetic with the inverses of zero numbers, multiplication of zero numbers, a geometric interpretation, the empty sets unsuitability as a rigorous basis for the absent numbers, as well as other issues, are dealt with in the attached paper. Also not addressed here is the question "Does this number system form a field or not?"

 

[yuiop]

We don't imagine -- we construct a new number system (e.g. the projective numbers) in which 1/0 is defined, and use that one instead of the real numbers.

By the way, both of these limits are of the form 0/0:

\lim_{x \rightarrow +\infty} \frac{x}{x^2}

\lim_{x \rightarrow +\infty} \frac{x^2}{x}​

Assuming you meant in the limit as x goes to zero (as Mentallic mentioned) then can 0/0 is undefined even in the projective system, so can we not apply L'Hopital's rule and get:

<br /> \lim_{x \rightarrow 0} \frac{x}{x^2} = \lim_{x \rightarrow 0} \frac{1}{2x}=\infty \qquad ?<br />

For the case that the limit of x goes to infinity (in the projective system):

<br /> \lim_{x \rightarrow \infty} \frac{x}{x^2}=\lim_{x \rightarrow \infty} \frac{1}{2x}=0<br />

In fact it seems we can note that x/x² is equivalent to 1/x and obtain the same limits as above even without applying L'Hopital's rule in the projective system.

 

[Max™]

Be careful using up arrows for stuff like this.

http://en.wikipedia.org/wiki/Knuth's_up-arrow_notation