Zero to the Zero

[JaKiri]

Isn't it [math]0^0[/math'], which means "how many 0s are there in 0" = 1.

 

That's 0/0, and in any case your reasoning is hideously flawed.

 

One 0 makes 0.

 

Two 0's (0*0 or 2*0) make 0.

 

Three 0's (0*0*0 or 3*0) make 0.

 

And so on to infinity.

 

Which one of those is 'right'?

[lqg]

They most definitely are not.

 

However' date=' 0^0 should be strictly speaking undefined, as the proof by notation that works for x^0 where x =! 0 doesn't hold if x = 0.[/quote']

why not?

 

0/0 is like 0^a/0^a=0^(a-a)=0^0

 

now why this is wrong?

 

 

wait a minute perhaps it's not the same because zero can be represented in infinite state of powers:

0^a/0^2a=0^(a-2a)

[Guest eyedea]

0^0 is an "indeterminate", which is different than an undefined value. It means that it has been proven to be unexplainable. There are other indeterminates such as infinity^0, infinity/infinity, infinity/0, and some others.

[Primarygun]

Sorry, I am too lazy to have a look at this passage.

I support what lgg say, 0^0=undefined=0/0, right?

[paganinio]

hmm you're not right

1, In High School Math, 0^0 and 0/0 are both undefined.

2, But in University, I am not sure.

3, Even if 0^0 is undefined, you cannot say 0^0 'EQUALS' undefined. It's a serious mistake.

[Dapthar]

hmm you're not right

1' date=' In High School Math, 0^0 and 0/0 are both undefined.

2, But in University, I am not sure.

3, Even if 0^0 is undefined, you cannot say 0^0 [b']'EQUALS'[/b] undefined. It's a serious mistake.

1. True.

2. They still are undefined at the University level, and beyond.

3. True.

[Aeschylus]

Hmm I've never been clear on this: is indeterminate value (such as 0/0) also undefined? Often 0^0 is simply defined as 1.Any way read this:

 

http://mathworld.wolfram.com/Indeterminate.html

 

and this:

 

http://mathforum.org/dr.math/faq/faq.0.to.0.power.html

[123rock]

0^0 is the same as 0/0, that's why x^0, where x is not 0 is x/x=1

[Aeschylus]

0^0 is the same as 0/0, that's why x^0, where x is not 0 is x/x=1

 

What you're arguing from is the fact that that x^n/x^m generally equals x^(n-m), but it dods not neccessarily follow that x^0 = 0/0. You should be very, very careful about equating any intdeterminate or undwefined values.

[MolecularMan14]

its like dividing anything by zero, it must be undifined...but mathmatically it comes out to 1...but thats probably just off of a theorem...Im sure there's a proof behind it.

[ydoaPs]

my TI83+ said it is undefined. it said "ERR:DOMAIN"

[Dave]

its like dividing anything by zero, it must be undifined...but mathmatically it comes out to 1...but thats probably just off of a theorem...Im sure there's a proof behind it.

 

Say what now?

[MolecularMan14]

is there a proof behind dividing something by 0...or a proof for having 0 to the 0??

 

My palm calculator probably just made it =1 because it was raised to the 0th power, and anything to the 0th is 1...even tho its undefined

[Dave]

Dividing anything by zero is an undefined operation. It's not really linked to 0⁰ as far as I know.

[MolecularMan14]

yea i know, but its the same outcome...thats all i was saying

[Dave]

Well, not really. Read post #5: generally 0⁰ is regarded to be 1.

[MolecularMan14]

strange that so many calculators would come out with undefined...I have only tried it on one calc and found it to be 1

[Primarygun]

Well, not really. Read post #5: generally 0^{0[/sup'] is regarded to be 1.}

^{So far, my book told me x^0=1 when x is not equal to 0.}

[pulkit]

0^0 is an "indeterminate", which is different than an undefined value. It means that it has been proven to be unexplainable. There are other indeterminates such as infinity^0, infinity/infinity, infinity/0, and some others

 

0^0 is just undefined.

 

You end up with indeterminate quantities when while taking a limit both numerator and denominator approach 0. But then too there is the easy way out of applying L'Hospital's rule and getting a value. Indeterminate merely implies that it may have different values in different situations.

[Dave]

So far, my book told me x^0=1 when x is not equal to 0.

 

Read the link in my first post on the subject.

[123rock]

0^0 is not 1.

0/0 does not equal one, because 0 can be contained in 0 in more times than just 1. It's perhaps all real values, kinda like if x=a^2; then x has two solutions, a and -a, but that doesn't mean a=-a. In order for modern mathematicians to get this problem out of their way instead of solving like the good ol' times, they went ahead and said, "It's undefined."

[Dapthar]

0^0 is not 1.

Correct.

 

0/0 does not equal one,

Correct.

 

because 0 can be contained in 0 in more times than just 1.

Nope. It's due to the fact that [math]a/b = c \iff a = b \cdot c[/math], where [math]c[/math] is unique for a given [math]a[/math] and [math]b[/math]. Note that if [math]b=0[/math] and [math]a[/math] is fixed, c can be any value, and therefore, it is not unique. Thus, division of any number by [math]0[/math] is undefined.

 

It's perhaps all real values,

Nope. Division of two real numbers is not a multiple valued operation. An example of a multiple valued operation is [math]\pm[/math]. An example is the quadratic formula (i.e. if [math]f(x)=ax^2+bx+c, x=\frac{-b\pm \sqrt {b^2-4ac}}{2a}[/math]).

 

In order for modern mathematicians to get this problem out of their way instead of solving like the good ol' times, they went ahead and said, "It's undefined."

Assigning a numerical value to these concepts that does not conflict with the definitions of division is currently, and may never be, possible. Thus, the "undefined" moniker.

[123rock]

0/0=R and all imaginary numbers.

if a/b=c, and a is 0, then c is 0 and can't be any other number but 0.

+or-, thanks for the reminder. For the sake of logical thought and processing, since there can be two values for a quadratic equation, and 3 for a cubic, and so on, for x/0, or rather 0/0 there are in infinite number of solutions.

 

0/0=? 1

0/0=1/1*0/0

0/0=0/0 correct

All real numbers and imaginary numbers follow the same pattern if 1 is replaced by x or any variable that you choose.

[ydoaPs]

0/0=R and all imaginary numbers.

if a/b=c' date=' and a is 0, then c is 0 and can't be any other number but 0.

+or-, thanks for the reminder. For the sake of logical thought and processing, since there can be two values for a quadratic equation, and 3 for a cubic, and so on, for x/0, or rather 0/0 there are in infinite number of solutions.

 

0/0=? 1

0/0=1/1*0/0

0/0=0/0 correct

All real numbers and imaginary numbers follow the same pattern if 1 is replaced by x or any variable that you choose.[/quote']

 

isn't 0/0 undefined?

[123rock]

It's not only undefined, it's infinity.