value of 0 over 0

[sash1]

We're discussing this in math class right now (Geometry Honors - a sophomore class, and I'm only a frosh ). I mean, "anything over itself," as we named it, would show that this is 1. But, something over 0 is undefined. We also went over this last year, but we never got into very much. We finally got into it now, again, in non-Euclidean geometry, dealing with triangles that violate the measure postulate. Basically, making a flat plane a triangle, with an acute angle of 0 or 90 and then finding the sin, cos, tan, cot, sec, and csc of them.

Thats how we got into this, and I want to know what you guys think or know of what the value of 0 over 0 is.

Thanks a lot,

~Aunix

 

[CTho9305]

its undefined. you can take limits as a function approaches 0/0 though... like the limit of sin(x)/x as x approaches zero is defined. there are som calculus-based tests you can do to figure out if there is a limit, and to find that limit.

 

[rimshaker]

Outside of calculus, there's nothing more to say other than 0/0 is undefined, or DNE (does not exist). But in basic calculus, where limits are used in the problem, a 0/0 result is considered an indeterminate form and the function that produced the 0/0 result can be further evaluated using what's called l'Hopital's Rule. Other kinds of indeterminate forms are like infinity/infinity, 0*infinity, infinity-infinity. Anway, you only encounter these when dealing with functions with the same variable in both numerator and denominator. l'Hopital's Rule proves that a real finite answer exists by simply taking the limit of the derivative of the original function. If it's first derivative still produces an indeterminate form, then do it again and take the second derivative... and so on, and so on.

The process stops once you reach a finite number, 0, or infinity. A result of infinity simply means the original function is divergent.

 

[Arschloch]

Actually, I believe that 0 / 0 is technically "indeterminate," not undefined. Undefined is reserved for (x / 0) | (x != 0).

But like the other guys said... there are methods in calculus that can attempt to find what the function -approaches- as it goes toward 0 / 0.

 

[rimshaker]

Let's put it this way. It's undefined for anything below calculus. It's indeterminate for calculus and beyond. This is easier for most students to understand.

 

[Carceri]

It's actually pretty simple. The VALUE x / 0 is undefined for any x no matter if x is value or a function with a limit point.

If we replace 0 by some other function approaching 0 when x also is approaching 0 we have a very different scenario where you can use different methods for finding the value. But in that case it is incorrect to write 0 / 0, since that indicates the value 0.

Instead one would write: f(x) / g(x) where f(x) -> 0 and g(x) -> 0 when x -> a (for some value a). Some books will call this a "0 / 0" case which is fine as long as you dont think of this as the value 0 divided by the value 0.

 

[MadRat]

The rule that anything divided by itself (in this case 0/0) is equal to 1, should supercede the rule that you cannot divide by zero. It comes down to an "order of processes" answer, not necessarily conventional wisdom.

 

[rimshaker]

<< The rule that anything divided by itself (in this case 0/0) is equal to 1, should supercede the rule that you cannot divide by zero. It comes down to an "order of processes" answer, not necessarily conventional wisdom. >>



Typical layman's answer.

 

[MadRat]

You can take that elitist comment and shove it.

 

[rimshaker]

Sorry.. no offense in that way you suggested. It's just that with everyone else's effort to be somewhat technical in the Highly Technical forum, you come out of nowhere and just say...... ONE!! Didn't you bother to read the previous posts?

 

[Carceri]

<< The rule that anything divided by itself (in this case 0/0) is equal to 1... >>


There is no rule stating that. What you are probably referring to is one of the definitions of a field, namely the condition that says:

In a field each element except from the additive identity (0) has a multiplicative inverse.

In the case of the real numbers (which indeed are a field) this condition says:

For any x different from 0 there exists a y such that x * y = 1

You probably ask: Why do we have to assume that x is different from 0?

Well assume that x is 0. Then we need to find y such that 0 * y = 1

This is the same as finding y such that 0 + 0 + ... + 0 (y times) = 1 (by the definition of multiplication)

But 0 is the additive identity in the real numbers, so 0 + 0 = 0, hence we have

0 + 0 + ... + 0 (y-1 times) = 1

We can continue in that fasion until we reach:

0 + 0 = 1, and finally

0 = 1

which is clearly a contradiction ("the number of elements in the empty set is different from the number of elements in the superset of the empty set" if you really want to go back to basics). Hence our assumption that x could be 0 was wrong.

This means that in a field the additive identity can not have an inverse element. For the real numbers this means that any value divided by 0 makes no sense.

Remember that division is not an operation, but just a notation. Dividing by z just means to multiply by the inverse element of z, and since 0 does not have an inverse element, we can't divide by 0

I hope this clears things up a bit

 

[Ryanov]

Carceri, I'm an engineering student who was once a math major and have now completed a math minor instead. Your analysis is indeed what I was pretty much going to provide In fact, that is pretty much along the lines of the argument a mathematician would use to decide that 0/0 is undefined. As for the previous posts about it being indeterminate, this is actually incorrect at a precise level. As a value in and of itself, 0/0 is indeed undefined. "Indeterminate" specifies that it has a value, but that the value cannot be determined. In electrical engineering we deal with many functions of time and use a lot of calculus and differential equations. While it may be true that a *limit* of a function can tend to 0/0, this does not necessarily imply that the value of the actual limit is 0/0. In fact, l'Hopital's Rule in its strongest form guarantees this. For a detailed and precise mathematical proof, do not consult a calculus text but rather a book on Analysis, which is calculus in its rigorous form (completely precise proofs that prove all details of a theorem and guarantee what functions and numbers it will work with). I studied with Walter Rudin's Principles of Mathematical Analysis after having taken three semesters of calculus and differential equations.

Alternatively, one can use axiomatic set theory to prove this. It is more difficult, however, yet it is more fundamental. Set theory is the most fundamental element of mathematics along with logic. Visit metamath.org to find detailed proofs worked out from axioms, but this will take a lot of time to learn.

Good luck

 

[Carceri]

I'm glad that we agree

I am one course away from having a minor in mathemathics, and have had courses in both analysis and calculus.

 

[Uuplaku]

Im going to quote my calculator in answering this question:

"Error"

 

[BlueScreenVW]

Any limit you'll take of 0^0 will be 1, since by definition X^0=1 where X is any number. So it can't be proven that way at least. However, I wonder if this isn't sometimes defined so that 0^0=1?

 

[yomega]

<< Im going to quote my calculator in answering this question:

"Error" >>



I personally prefer the answer my HP-49G gives me:

"?"

 

[Degenerate]

While i see many good explanations on 0/o.. cdan i ask, is there any use use of using 0/0? how can it be used practically?

 

[rimshaker]

<< While i see many good explanations on 0/o.. cdan i ask, is there any use use of using 0/0? how can it be used practically? >>



hehehh, there aren't any useful applications for it! It's simply another mathematical topic introduced in detail in your calculus sequence... namely a small section Calculus II.... i think.

 

[q2261]

my 83p says:

"INVALID DIMM"

 

[Carceri]

My TI-85 is boring, but pretty informative:

ERROR 02 DIV BY ZERO

 

[Shalmanese]

Well, Microsoft calculator gives "Result of function is undefined".

It seems Microsoft Programmers never took calculus

 

[Mday]

you cannot divide by zero. if you divide by zero, the result in IEEE mathematics is NAN, not a number. the IEEE is one standard of treating division by zero.

however, in some cases, you can divide by x, as x approaches zero. the result is either undefined, infinite, or finite as with limit. which is to say, every case depends upon the nature of the function, or series\sequence, which you are taking the limit of.

I will tell you that in most cases which apply to certain aspects of reality, it is safe to assume that a non-zero over zero is considered infinite. the nature of 0/0 is very strange however.

similar ambiguities exist between infinite/infinite. cuz there are varying levels of infinitum.