WHY Does A^0=1 ?

Question

Everyone knows that any number raised to the 0 power will equal one. The first person to answer why this is in the form of a mathmatical proof, gets the best answer..
Its fun

Answer 1

Hi,

Really there's not much to prove regarding why any number raised to the zero power is 1. This is more or less a known definition in mathematics and we appreciate it for what it's known for. If you wanted to test this out you could pick A = 5 and show the following:

5 ^ 0 = ?

5 ^ 1 = 5

5 ^ 2 = 25

5 ^ 3 = 125

Notice that 5 ^ 0 needs to be less than 5 and we aren't raising it to any power, hence the zero. Then if we aren't raising a number to any power, then that must mean that it's 1! Also, think about A = 1, if that was the case, it wouldn't matter what the exponent would be because it would ALWAYS amount out to being 1.

That is probably where the definition comes from!

I hope that helps you out! Please let me know if you have any other questions!

Sincerely,

Andrew

Answer 2

This simple example would say why it is 1

The rule is: x^b / x^c = x^(b-c).

Ex.1: 5^3 / 5^2 = 125 / 25 = 5
5^3 / 5^2 = 5 ^(3-2) = 5^1 = 5

Ex.2: 5^3 / 5^3 = 125 / 125 = 1
5^3 / 5^3 = 5 ^(3-3) = 5^0 =
... and the result must be 1. So 5^0 =1.

The below link gives a detailed description. But the above statements are the summary of it.

Answer 3

It is not so much a proof as a definition.

For "normal" exponents like A^7 and A^4, it's easy to see that :

A^x / A^y = A^(x-y)

For example: A^7 / A^4 = A^(7-4) = A^3. This is easy to see because 4 of the A's in the denominator cancel out 4 of the A's in the numerator.

If we accept that A^x / A^y = A^(x-y) for any x and y, then it should be true that:

A^z / A^z = A^(z-z) = A^0

But we know that (A^z / A^z) must be 1. So if we want the rule to work for all exponents, we must define that A^0 = 1.

Answer 4

any thing raised to the power 0 is 1.

because A^0 = A^(1-1) = A^1 * A^(-1) = A^1/A^1 = A/A = 1

Answer 5

Aplying the exponencial rules:

A^n / A^n = A^(n-n) = A^0 = 1

Because also,

A^n / A^n = 1 (qty divided by itself is always 1)

Then A^n / A^n = A^0 = 1

Answer 6

x^y is defined as
e^(y*log(x))
If y is 0, we have
x^y=e^y
By the definition of the exponential function,
e^y=1+y/1+y^2/2!+y^3/3!+...
Since y is zero,
e^y=1
Thus,
x^y=x^0=1

Answer 7

A^1 is A times itself 1 time.
A^2 is A times itself 2 times.
so, A^0 is A times itself 0 times which is 0 because you are not multiplying by anything. that's just what I think.

Answer 8

0^0 is also defined as one, by the way.
A^n is defined as 1 times A, n times.

Answer 9

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

Similarly
A^(x-y) = (A^x)/(A^y)

Assuming A≠0:
A^0
= A^(x-x)
= (A^x)/(A^x)
= 1