Question

What is `0^0`? (0 to the power of 0)

Answer 1

In elementary mathematics, this is often left undefined.

Explanation:

But with the notion of limits available, we can show that `lim_(xrarr0^+)x^x = 1`. So some give the definition `0^0=1`

As a cardinal number, `0` is the cardinality of the empty set.

And for Cardinal numbers `A` and `B`, `A^B` is defined to be the cardinality of the set of all functions from `B` into `A`.
With this definition and the observation that there is exactly one function from `O/` to `O/` (namely the empty function), we have a proof that `0^0 = 1`.

Answer 2

It depends...

Explanation:

A case for `0^0 = 1`

Suppose we define:

`a^n = overbrace(a * a * ... * a)^"n terms"`

for any non-negative integer `n`.

Then `a^0` is an empty product, so should have the value `1` (similar to the way that an empty sum has the value `0`), regardless of the value of `a`.

This definition is useful in contexts where you are dealing purely with non-negative integer exponents.

A case for indeterminacy

Note that:

`lim_(x->0) x^0 = 1`

`lim_(x->0^+) 0^x = 0`

More generally, let `a > 1` be a real number.

For `n = 1, 2, 3,...` let:

`{ (x_n = a^(-n)), (y_n = -1/n) :}`

Then as `n->oo` we find:

`x_n -> 0`

`y_n -> 0`

and:

`x_n^(y_n) = (a^(-n))^(-1/n) = a`

If instead, we have `0 < a < 1` then define:

`{ (x_n = a^n), (y_n = 1/n) :}`

to get similar results.

If `a < 0` then you have to be more careful and explicit as to what you mean. For example, if `a^(1/n)` denotes the real `n`th root of `a`, then we might use:

`{ (x_n = a^(1-2n)), (y_n = 1/(1-2n)) :}" "` if `a < -1`

or

`{ (x_n = a^(2n-1)), (y_n = 1/(2n-1)) :}" "` if `-1 < a < 0`

So according to how we approach the point `(x, y) = (0, 0)`, the function `f(x, y) = x^y` can take any real value you like.

This all points towards considering `0^0` to be indeterminate in contexts where non-integral exponents are being used.

Answer 3

`1`

Explanation:

Anything to the `0` power equals to `1`. Ex:

`2^0 = 1`

`100000000^0 = 1`

Therefore, `0^0 = 1`

Hope this helps!