Why would 0 to the zero power equal 1?

Question

Can someone please explain...

Answer 1

Check MathForum.org:

Question:
What is the quantity 0^0 (zero to the zeroth power)? Would it be
0, since 0 times anything is 0, or would it be 1, since anything to the 0 power is 1? Or is this undefined, too?

Answer:
This, too, is an indeterminate form. Its logarithm is 0 * infinity.

More detail from the website:
Say you wish to introduce a new element (or maybe two elements) infinity which you wish to append to the real number system. That is not prohibited. After all, that is how we got from natural numbers to integers (appending negative integers and zero), and from integers to rationals (appending ratios of integers), and from rationals to reals (appending limits of convergent sequences), and from reals to complexes (appending the square root of -1). What you end up with is not the real number system, however. Furthermore, if you wish to define the four operations + - * and / for this new system, you probably want them to be the same on real numbers, and just add on the definitions of things like infinity + r and r/infinity, for any real number r. Some of these work fine. It makes sense to define:

infinity + r = r + infinity = infinity
(-infinity) + r = r + (-infinity) = -infinity
infinity + infinity = infinity
(-infinity) + (-infinity) = -infinity
infinity - r = infinity
(-infinity) - r = -infinity
r - infinity = -infinity
r - (-infinity) = infinity
infinity - (-infinity) = infinity
(-infinity) - infinity = -infinity
infinity * r = r * infinity = infinity for r > 0
(-infinity) * r = r * (-infinity) = -infinity for r > 0
infinity * r = r * infinity = -infinity for r < 0
(-infinity) * r = r * (-infinity) = infinity for r < 0
infinity * infinity = (-infinity) * (-infinity) = infinity
infinity * (-infinity) = (-infinity) * infinity = -infinity
infinity / r = infinity for r > 0
(-infinity) / r = -infinity for r > 0
infinity / r = -infinity for r < 0
(-infinity) / r = infinity for r < 0
r / infinity = 0
r / (-infinity) = 0

Where we get into trouble is with defining the following:

infinity + (-infinity)
(-infinity) + infinity
infinity - infinity
(-infinity) - (-infinity)
0 * infinity
infinity * 0
0 * (-infinity)
(-infinity) * 0
infinity / infinity
infinity / (-infinity)
(-infinity) / infinity
(-infinity) / (-infinity)
infinity / 0 = infinity
(-infinity) / 0 = -infinity

These expressions are called "indeterminate forms." These can all have a large range of different values, depending on exactly where the "infinity" parts came from.

As a result, the system you construct is not closed under addition, subtraction, multiplication, or division.

Other indeterminate forms are 0^0 and 1^infinity.

So the answer, is 0^0 is undefined or indeterminate.

Answer 2

Well the general rule is that anything to the zero power is one... I'm trying to think how zero to the zero would work...

the thing is that with most numbers (say the number is n)

(n^1) / (n^1) = n^(1-1) = 1

but (0^1)/(0^1) is undefined.

Try graphing x^x power and see what happens at zero. That should help. :)

UMRmathmajor

Answer 3

Yes, previous poster, 0^0 is undefined.

You might also consider the following two functions:

f(x) = x^0. This function is the same as f(x) = 1, except we don't know about x=0. However, this function has a limit at x=0, namely 1.

f(x) = 0^x. This function is the same as f(x) = 0, except again we don't know about x=0. However, this function has a limit at x=0, namely 0.

In short, both equations look the same at x=0 and both have a limit, but the limits are different - there's a good example why 0^0 as undefined.

Answer 4

It has been a long time since I was in school, but my first inclination is to say that this is undefined. After all, zero to any other power is zero, but any number to the zero power is one. Hmm... Hopefully, someone who is more current on this subject will respond. :-)

Answer 5

you need to use Rule of Exponents to figure this one out.

Rule of Exponents

(x^a)/(x^b) = x^(a - b)

now lets say "a" = "b"

(x^a)/(x^a) = x^(a - a) = x^0

but if you notice, you have x^a on top and bottom, and what happens when you have identical values in both the numerator and denominator. You get 1 as your value, therefore

x^0 = 1

Answer 6

0^0 is NOT equal to 1

Answer 7

It doesnt. Anything to the zero power implies dividing by that thing, a forbidden operation with 0.

Answer 8

It is DEFINED as such.

As I recall. Except, I think, for negative numbers, which, when taken to the power of zero, equal -1.

************

Oooops. Nope. They go to positive one also. Sorry!

Answer 9

it is???