Let's first look at an example. Let's look at the list of numbers
3^1, 3^2, 3^3, 3^4, ....
Finding the actual values, we get 3, 9, 27, 81, ....
So what is the pattern in the bottom sequence? Well, every time you move to the right in the list you multiply by 3, and every time you move to the left in the list you divide by 3. So we could take the bottom sequence and keep going to the left and dividing by 3, and we'd have the sequence that looks like this:
..., 3^-3, 3^-2, 3^-1, 3^0, 3^1, 3^2, 3^3, 3^4, ....
..., 1/27, 1/9, 1/3, 1, 3, 9, 27, 81, ....
So now we know what all the powers of 3 are! Actually, we just did the integer powers of 3. But that's probably enough for now.
While the above argument might help convince your intuitive side that any number to the zero power is 1, the following argument is a little more rigorous.
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.
We can plug in any in number in the place of three, and that number raised to the zero power will still be 1. In fact, the whole proof works if we just plug in x for 3:
x^4
x^0 = x^(4-4) = --- = 1
x^4
Here's another explanation from Aldo Daniel Completa, a contributor from Argentina:
In the FAQ there is an explanation about " n^0 (any number to zero power) ." I think another is this:
Let's begin with examples.
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 rule is: x^b / x^c = x^(b-c).
In order to generalize this rule for the case b=c, it must be defined that x^0 = 1 (x is any number different from 0). In mathematics, usefulness and consistency are very important. This convention allows us to extend definitions of power that would otherwise require treating 0 as a special case.
This method can also explain the definition: x^(-b) = 1 / x^b.
Ex. 5^2 / 5^4 = 25 / 625 = 1 / 5^2
5^2 / 5^4 = 5 ^(2-4) = 5^(-2)
2006-08-31 03:32:26
·
answer #1
·
answered by Truth Seeker 3
·
3⤊
0⤋
You should award vijay the points. He went to the most trouble explaining the answer. Maybe I can elaborate a little more on what Vijay wrote. See, we define a^n as a*a*a*a....*a (n times). Simly, we define a^m as a*a*a...*a (m times). Then we can define products, quotients, etc. If we define the product, then we have: (a^n)*(a^m) = a^(m+n) Yes? If we define the quotient, then we have (a^n)/(a^m)=a^(n-m). But we know that if n=m, then the quotient is always 1, yes? Check this! Well, if the quotient is always 1 when n=m, then what does n-m equal to? Hmmm, must be zero eh? So anything to the power of 0 must be 1 by definition. Get the idea? Post another question if you are still confused.
2006-08-31 03:43:26
·
answer #2
·
answered by Anonymous
·
0⤊
0⤋
try dividing 234^7 by 234^7 . it is 234^(7-7)=234^0.
we divided something by the same number. so answer is 1.
so anything divided by itself = the thing ^0=1
provided you dont try 0^n/0^n , n any number. this division (by 0 ) is not defined. so
0^0 is also not defined.
2006-08-31 03:33:56
·
answer #3
·
answered by jaco 3
·
0⤊
0⤋
It's part of the series of powers. For example, 3^3=27, 3^2=9, 3^1=3, it's dividing by 3 each time you subtract one from the exponent, so 3^0=(3^1)/3=3/3=1.
And I'm not a math genius.
2006-08-31 03:22:04
·
answer #4
·
answered by Anonymous
·
3⤊
1⤋
234^0=234^(1-1)=234^1/234^1=1
2006-08-31 03:33:19
·
answer #5
·
answered by raj 7
·
1⤊
0⤋
Not a math genius but I'll take 2 points anyway. By definition, multiplication is just serial addition. If you don't take any and then you don't add anything more to it, you still don't have any. The same logic applies to negative numbers. If you don't take a number and you don't subtract anything from it you still don't have anything.
2006-08-31 03:25:15
·
answer #6
·
answered by Kuji 7
·
0⤊
0⤋
I usually could help you, yet seeing which you're soliciting for each answer to thirty issues, that in simple terms tells me you published your finished homework for somebody else to do for you. by making use of the time you finished typing those up, you ought to've been performed with it. provide up being so lazy. those issues are so elementary it is not even humorous. wager that makes all people a genius.
2016-12-14 15:26:45
·
answer #7
·
answered by ? 4
·
0⤊
0⤋
By definition.
2006-08-31 03:32:42
·
answer #8
·
answered by Anonymous
·
0⤊
0⤋
Exactly what He ^ he said .......... And I AM a Math Genuis So I guess you should Eliminate him :-) lol
2006-08-31 03:24:11
·
answer #9
·
answered by D B 4
·
1⤊
2⤋