View Full Version : Defining division by zero?
Blue_Jaunte
Jul30-09, 01:03 AM
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?
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
Jul30-09, 10:04 AM
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
Jul30-09, 10:09 AM
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 cant 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
Jul30-09, 03:44 PM
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
Jul30-09, 03:57 PM
Interesting thread. Tac-tics response was great. I wish I'd seen it explained like that a long time ago.
g_edgar
Jul30-09, 04:16 PM
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
Jul30-09, 05:13 PM
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.
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 (http://en.wikipedia.org/wiki/Wheel_theory) 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
Jul31-09, 12:18 AM
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 lets you walk along Stern's diatomic series.
:approve:
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
Sep27-09, 11:37 PM
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
Sep28-09, 11:13 AM
You can use induction on the rationals, as long as your inductive step lets 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
Sep28-09, 11:32 AM
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.
:devil:
CRGreathouse
Sep28-09, 03:56 PM
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
Oct11-09, 06:52 AM
is the idea of Blue Jaunte to define a ring of 'infinite numbers?
I(R)=(a,b =Real |a+b\infty = I )
Mentallic
Oct12-09, 09:29 AM
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
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
Oct12-09, 10:05 PM
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 :smile:
demoivre
Oct14-09, 08:58 AM
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
Oct14-09, 05:20 PM
And logic will tell us that any integer divided by itself is equal to one.
Not if it is zero
Since I've started learning division, it is learnt 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'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
Oct27-09, 10:48 AM
Division by zero? NO! Wes Hughes
CRGreathouse
Oct27-09, 06:51 PM
Since I've started learning division, it is learnt 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.
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