Why, exactly, is any number to the 0th power equal to one?

Question

I know it is to complete the sequence of powers, i.e. in binary, or any other base, n^0 has to be one so you can have a ones place. But is that a reasonable explanation for this?

Answer 1

Yes, it comes from the rules for exponnents
when you divide, you subtract exponents. That is, (a^m)/(a^n) = (a^(m-n))
For example,
(2^5)/(2^2) = 2^(5-3) = 2^3 = 32 / 4 = 8
So (a^m)/(a^m) = a^(m-m) = a^0 = ?
So it seems that anything divided by itself is something^0
So what does this mean?
Well, anything divided by itself is 1.
For example,
(2^3)/(2^3) = 8/8 = 1
(the reason for this is because 1x8 = 8 and 1xN = N for any N)
So it follows that anything^0 must be 1, except 0^0 which is indeterminate.

Answer 2

This proof uses the laws of exponents. One of the laws of exponents is:

n^x
--- = n^(x-y)
n^y

for all n, x, and y. So for example,

3^4
--- = 3^(4-2) = 3^2
3^2


3^4
--- = 3^(4-3) = 3^1
3^3


Now suppose we have the fraction:

3^4
---
3^4

This fraction equals 1, because the numerator and the denominator are the same. If we apply the law of exponents, we get:

3^4
--- = 1 = 3^(4-4) = 3^0
3^4

So 3^0 = 1.

Answer 3

Because it simply implies that the number is being divided by itself, and any number divided by itself is equal to one.

From the laws of exponents we can see how this works.

Given a^m and a^n, a^m / a^n = a^(m-n), and when n = m, a^(m-n) = a^ (m-m) = a^(0)

It's the last case which is of interest to us.

Note that when n = m, it is implied that a^m = a^n, because a given base raised to equal powers yields the same number. It also implies that when n = m, a^m / a^n = 1.

Now we can make the following assertion:

When n = m, then a^m = a^n, and a^m / a^n = 1 = a^(m-m) = a^(0).

Therefore, any number or base, not equal to 0 and raised to the 0 power is equal to 1. The special case of 0^0 produces what's called an indeterminate form, which must be dealt with using calculus techniques.

Answer 4

Take any number x. Take the n power of x: x^n
The way to get from x^n to x^(n+1) is always to multiply by x.

So...
x ( x^n ) = x^(n+1)

Now, x^1 = x.
So, x^0 has to be the number that will give x when you multiply it by x.
That number is 1. (and it doesn't matter what x is)

Here's an example...
Let x = 3, then...
3^0 = 1
3^1 = 3
3^2 = 9
3^3 = 27

To go from any row of this "table" to the next row below it, you have to multiply by 3. That rule applies when going from 1 to 3, just the same as it applies when going from 3 to 9.

Answer 5

One way to think of this is as follows: For exponentiation involving real numbers in general, one could define a^b as:

a^b = e^(b*ln a).

Now as b tends to 0, the expression on the right tends to e^0. HOWEVER, remember that e^x has an independent definition given as follows:

e^x = 1+x+x^2/2 + x^3 /3! + x^4/4!+...

Clearly, then, e^0 = 1 + 0+0+... = 1. Therefore, a^b tends to 1 as b tends to 0.

For more information on this, see any introductory book on real analysis. I recommend Rudin, Principles of Mathematical Analysis.

Answer 6

Remember dividing exponentials means subtracting the exponents itself. So n^1/n^1=1. So if you wanted to subtract the exponents, you would get n^1/n^1=n^0. Using the transitive property, we can conclude that since n^1/n^1=n^0 and n^1/n^1=1, n^0=1.

Answer 7

Um, well yeah... IMO that explanation is good enough.

I guess you could also mention the rule by adding exponents in multiplication, i.e. x^a * x^b = x^(a+b). Now consider that 1 = x^a/x^a = x^a * x^(-a) = x^(a-a) = x^0. So, for this rule to stay consistent, 1 = x^0

Answer 8

Because 1 is the multiplicative identity and is what you get whenever you multiply zero things together.

But it is wrong to say that x^0 = x^(n-n) = x^n / x^n, because 0^0 = 1, but 0/0 is undefined!

Answer 9

Yes, any number you take to the power of 0 will equal one.

Answer 10

n^2 / n^1 = n^(2 - 1) = n^1

So... to have a number to a zeroeth power
n^2 / n^2 = n^(2 - 2) = n^0
n^2 divided by itself makes 1
so n^0 = 1