Exploring the Undefined: Why is 0/0 Undefined and Why Does 0^0 = 1?

[ibcnunabit]

I think the simplest explanation for 0/0 being undefined is:

X*0=0 for all x

If 1*0=0, and 2*0=0, and 3*0=0, and 1,000,000*0=0,

Then which value of x does 0/0 belong to? If all the above are true, then

0/0=1, 0/0=2, 0/0=3, 0/0=1,000,000...(etc.) or any other number. But if THAT were true, then 1=2=3=1,000,000, etc. because of the transitive property. But this would be contradictory, therefore 0/0 must undefined.

--Mike from Shreveport

P.S. the answer to 0^0 depends on context and how it is interpreted, but my HP 50G (the finest calculator made) gives the answer 1. It can also be interpreted as undefined, but in the cases where it is, if epsilon (the very smallest possible number) is added to it, the answer becomes 1.

 

[slider142]

P.S. the answer to 0^0 depends on context and how it is interpreted, but my HP 50G (the finest calculator made) gives the answer 1. It can also be interpreted as undefined, but in the cases where it is, if epsilon (the very smallest possible number) is added to it, the answer becomes 1.

If epsilon is added to which 0? If epsilon is added to the base, the answer becomes 1, if added to the exponent the answer becomes 0, if added to both, the answer becomes $$\epsilon^\epsilon$$, which is neither.
Also, in the system of real numbers, there isn't really a definitive "the very smallest possible number", since that number divided by 2 is also a real number and is strictly less than the original number (assuming this number is positive) unless that number is 0.

 

[3.14159]

0! = 1 because 3! = 3*2! and 2! = 2 * 1! and 1! = 1 * 0!
so 0! must be 1

 

[Jake1802]

0!= 1 because mathematicians defined it as such to make calculations involving it easier. Thats what I read at least.

 

[D H]

0! is defined to be one because no other value makes sense and because this particular value does make sense from several different perspectives. Here are two; there are others.

• The recursive definition of n! is that (n+1)! = (n+1)*n!, starting with 1!=1. Using this recursive definition in reverse, (n-1)! = n!/n. Setting n=1 and using 1!=1 yields 0!=1.
• The extension of the factorial to the reals (and the complex numbers as well) is the gamma function. For all positive integers, n!=Γ(n+1). Γ(1)=1, so in this sense too it makes sense to say 0!=1.

Compare this with the problem of defining 0/0 and 0^0. Halls addressed this issue five years ago in this very thread:

By the way, we often say that terms such as 0/0 or 0⁰ are "undetermined" rather than "undefined" because the problem is not that there is no way to define it but, rather, too many ways.

 

[Klockan3]

I agree with Hurkyl. It is better to say 0^0 is undefined. Why undefined? First and foremost, because saying that 0^0 is one leads to contradictions, just as defining 0/0 as 1, or any other number, leads to contradictions.

Secondly, look at it in terms of limits. For any non-zero number a and positive real ε, there exists a neighborhood of a such that ||a^z-1||<ε for all z in that neighborhood. In other words, saying a⁰=1 is consistent with the concept of limits. This isn't true for 0⁰. Consider the function f(x,y)=x^y where x is a positive real number and y is real. This function takes on all real values in any half-neighborhood of 0 (x>0, x²+y²<δ²). Extending to complex numbers, u^v takes on all complex numbers in the neighborhood of 0. The limit of u^v as u,v approach zero does not exist.Defining 0^0 as one is an abuse of notation -- but a very convenient abuse of notation.

Just because 0^x isn't continuous do not mean that it can't have a defined value.

0^0=1 is true in every case were you are not evaluating its surroundings, and in its surroundings then the value at 0^0 doesn't matter any way since you aren't evaluating its value at the point.