Why does any number to the power of zero = one?

Answer 1

by definition \

x*(a+b) = x^a * x^b
but b = 0

x^(a) = x^a * x ^ 0
now devide by x^a on both sides(x cannot be zero)

1 = x^0

proved

Answer 2

Note that even 0^0 = 1. The rule holds even for 0. This isn't an arbitrary definition. It's necessary for exponentiation to work.

The reason this is true is that there is always a 1 implied when you raise a number to another number. The 1 is there to serve as a multiplicative identity. This is known as the "empty product." See the sources below for more details.

That is,

x^3 = 1*x*x*x
x^2 = 1*x*x
x^1 = 1*x
x^0 = 1

And thus,

0^0 = 1

because 1 is the only thing that's left. It's vacuously true. This is important because it lets:

x^y = x^(y+0) = x^y * x^0 = x^y

Without x^0=1, then there would be a contradiction. As you can see, this will hold with x=0 as well, so 0^0 certainly shouldn't be undefined. Thus, 0^0 is defined to be 1. You might argue that 0^0 could be defined to be 0 here, but that definition is not convenient and causes many problems. Many very important identities in mathematics depend upon 0^0=1. (see the quote below for one example) In fact, in order for the series expansion of e^0 to converge to 1 (as it should) then 0^0 must equal 1. As an analysis teacher of mine once said, if 0^0 = 0, the whole universe would collapse to a point (0, in fact).

To quote, "Some textbooks leave the quantity 0^0 undefined, because the functions x^0 and 0^x have different limiting values when x decreases to 0. But this is a mistake. We must define x^0 = 1 for all x, if the binomial theorem is to be valid when x = 0, y = 0, and/or x = −y. The theorem is too important to be arbitrarily restricted! By contrast, the function 0^x is quite unimportant." –– Concrete Mathematics, by Ronald Graham, Donald Knuth, and Oren Patashnik, Addison-Wesley, IBSN 0-21-14236-8, page 162 in the first edition, the chapter on binomial coefficients.

Answer 3

Ok, you got a lot of mathematical solutions. Let me give you a graphical one:

Imagine the graph of the function y = a^x. (a>1)

At x=1, y=a. The graph rises steeply on the right of x=1. But on the left, the graph drops, and as x approaches 0, y approaches 1.

To make it more clear:
a^0.5 = square root of a
a^0.25 = fourth root of a .... and so on.
Since a is a positive no. greater than 1, any root of a should be greater than 1. And as x gets closer and closer to 0, a^x gets smaller and smaller, but not smaller than 1.

So clearly, a^0 = 1.

On the negative side of x-axis, the graph drops gradually from (0,1), reaching y=0 at negative infinity.

For 0
You could check it out using a graphing program such as f(x)-viewer, or advanced grapher,etc. But I'm sure you're bright enough to imagine (or draw) the graph yourself!

Answer 4

All excepted zero. Zero to power of zero is undefined number.

Let's take N, N not null. N ^ 0 = N ^ (1 - 1) = (N ^ 1) / (N ^ 1) = N/N = 1.

Zero is a problem, because at the same time 0 power N (with N not null) is zero. And actually 0 divided by zero is undefined.

Answer 5

The value of X^0, while X is not equal to zero can be evaluated based on simple division and law of indices, that says X^a/X^a=X^(a-a)=X^0=1.

Answer 6

Embliri is correct. But a little complex to read

Y=X^2/X=X
which can be rewritten as
Y=X^(2-1)=X
when both numerator and denominator are raised to same powers,
Y=X/X =1
when rewritten
Y=X^(1-1)=1
Y=X^0=1
Something related:
You know of square roots and cubic roots. For higher we call them as fourth root, fift root etc.

0th root of any number excluding 0 is infinite.

square root can be written as x^(1/2)
Cubic root can be written as x^(1/3)

0th root can be written as x^(1/0)
which is x^(infinite) and is
infinite for all x natural numbers
0 for 0
+or - for all x non zero negative integers.

Answer 7

x/x=1 => x*(x^-1)=1 => x^(1-1)=1 => x^0=1

Answer 8

let a^0=b
transferring to logarithimic form,
0=logb to base a
from rules of logarithims,we know that only log1to any base=o
therefore,we obtain that b=1
substituting,we have that a^0=1

Answer 9

x^(n+1) / x = x^n
when n = 0
x / x = x^0
therefore x^0 = 1

Answer 10

well, I think it is just a rule that was defined years ago. However, there are ways around it.

Take for example this:

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

Pretty cool right?