# [SOLVED] Power of Zero

• Jun 12th 2006, 08:04 PM
Gini
[SOLVED] Power of Zero
Something I should remember but can't! What do numbers to the power of Zero equal? To the power of one they are themselves and squared, cubed etc is obvious but zero?? Anyone remember
• Jun 12th 2006, 08:20 PM
malaygoel
Quote:

Originally Posted by Gini
Something I should remember but can't! What do numbers to the power of Zero equal? To the power of one they are themselves and squared, cubed etc is obvious but zero?? Anyone remember

it is equal to 1
• Jun 12th 2006, 08:21 PM
CaptainBlack
Quote:

Originally Posted by malaygoel
it is equal to 1

Except for $\displaystyle 0^0$, which is traditionaly undefined.

RonL
• Jun 13th 2006, 10:23 AM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlack
Except for $\displaystyle 0^0$, which is traditionaly undefined.

RonL

Sometimes it is useful for Infinite Series to define that to be one.
In England that is not the style though.
• Jun 13th 2006, 04:34 PM
Quick
Quote:

Originally Posted by ThePerfectHacker
Sometimes it is useful for Infinite Series to define that to be one.
In England that is not the style though.

I don't see how that could ever be the style because $\displaystyle 0^0$ isn't undefined because of tradition, it's because of the rules of powers.

$\displaystyle a^3 \cdot a^{-3} = a^{(3-3)} = a^0 = 1$
or it could be written as...
$\displaystyle a^3 \cdot a^{-3} = \frac{a^3}{a^3} = 1$

so $\displaystyle 0^1 \cdot 0^{-1} = 0^{(1-1)} = 0^0 =$undefined
or it could be written as...
so $\displaystyle 0^1 \cdot 0^{-1} = \frac{0^1}{0^1} = \frac{0}{0} =$ undefined
• Jun 13th 2006, 04:51 PM
ThePerfectHacker
Quote:

Originally Posted by Quick

The tradition is that the bases are always positive numbers.

This is my 14:):)th Post!!!
• Jun 13th 2006, 09:07 PM
JakeD
Quote:

Originally Posted by Quick
I don't see how that could ever be the style because $\displaystyle 0^0$ isn't undefined because of tradition, it's because of the rules of powers.

$\displaystyle a^3 \cdot a^{-3} = a^{(3-3)} = a^0 = 1$
or it could be written as...
$\displaystyle a^3 \cdot a^{-3} = \frac{a^3}{a^3} = 1$

so $\displaystyle 0^1 \cdot 0^{-1} = 0^{(1-1)} = 0^0 =$undefined
or it could be written as...
so $\displaystyle 0^1 \cdot 0^{-1} = \frac{0^1}{0^1} = \frac{0}{0} =$ undefined

How to define to $\displaystyle 0^0$ is an old controversy. This page has a good discussion and claims "Consensus has recently been built around setting the value of $\displaystyle 0^0 = 1.$"
• Jun 14th 2006, 02:34 AM
Quick
Quote:

Originally Posted by JakeD
How to define to $\displaystyle 0^0$ is an old controversy. This page has a good discussion and claims "Consensus has recently been built around setting the value of $\displaystyle 0^0 = 1.$"

It seems like they want it to be $\displaystyle 0^0=1$ for a lot of reasons, but they give no proof that $\displaystyle 0^0=1$, although it might be convenient, I don't think it's very mathmatical.
• Jun 14th 2006, 04:37 AM
CaptainBlack
Quote:

Originally Posted by Quick
It seems like they want it to be $\displaystyle 0^0=1$ for a lot of reasons, but they give no proof that $\displaystyle 0^0=1$, although it might be convenient, I don't think it's very mathmatical.

There can be no proof, it will be defined to have a particular value
if/when the convention of it being undefined is changed.

RonL
• Jun 14th 2006, 04:45 AM
CaptainBlack
Quote:

Originally Posted by JakeD
How to define to $\displaystyle 0^0$ is an old controversy.

It appears to be still controversial, which means that unless an
author defines what they want it to mean, it is ambiguous and
hence undefined :D

Quote:

This page has a good discussion and claims "Consensus has recently been built around setting the value of $\displaystyle 0^0 = 1.$"
• Jun 14th 2006, 11:57 AM
topsquark
Quote:

Originally Posted by CaptainBlack
It appears to be still controversial, which means that unless an
author defines what they want it to mean, it is ambiguous and
hence undefined :D

I can't see why it could EVER be defined as one or the other since:
$\displaystyle \lim_{x \to 0}0^x \to 0$

and

$\displaystyle \lim_{y \to 0}y^0 \to 1$

As the limits do not agree we can't say that $\displaystyle 0^0$ is defined. What confuses me is why there is even discussion about defining it to be one or the other?? :confused:

-Dan
• Jun 14th 2006, 12:06 PM
TD!
Have you thought of letting x approach 0 as base and power symmetrically?

$\displaystyle \mathop {\lim }\limits_{x \to 0} x^x = 1$

This should make the "convention" at least more plausible.
• Jun 14th 2006, 12:06 PM
Quick
Quote:

Originally Posted by topsquark
I can't see why it could EVER be defined as one or the other since:
$\displaystyle \lim_{x \to 0}0^x \to 0$

and

$\displaystyle \lim_{y \to 0}y^0 \to 1$

As the limits do not agree we can't say that $\displaystyle 0^0$ is defined. What confuses me is why there is even discussion about defining it to be one or the other?? :confused:

-Dan

I would like to point out that when something can't be defined it becomes UNDEFINED

I also think this post should be moved to chat room or miscellaneous.