# Why why why why ?

1. Sep 13, 2009

### kaskus

why why why why ??

Can you explain me why 0^0=1 but 0^1=0 ??

2. Sep 13, 2009

### Dragonfall

Re: why why why why ??

0^0 is undefined. It's not equal to 1.

3. Sep 13, 2009

### g_edgar

Re: why why why why ??

Depends on what you mean. For real numbers, say 0.0, we may call 0.0^0.0 "undefined". It is a classic "indeterminate form" which means in any case you cannot define 0.0^0.0 by limits.

On the other hand, for cardinal numbers, say 0, we get 0^0 = 1 since there is exactly one function from the empty set to itself.

For other kinds of numbers, you will have to look at the situation and see what happens.

Finally, x^1 = x for any x, and in fact 0^1 is not indeterminate, so this works even for limits.

4. Sep 13, 2009

### kaskus

Re: why why why why ??

Calculator’ shows that 0^0 = 1

5. Sep 13, 2009

### Dragonfall

Re: why why why why ??

Calculator is wrong. Well, it's not completely right. 0^0 is undefined in general, and in very specific cases which people have pointed out, it may be 0 or 1 or whatever.

There is no function from the empty set to itself, if we don't allow the empty set to be a function.

Last edited: Sep 13, 2009
6. Sep 14, 2009

### JCVD

Re: why why why why ??

in discrete math we like to say 0^0=1 because for any x, x^0 is an instance of the empty product, which logically should be the multiplicative identity (since any number is itself multiplied by the empty product), which is 1. Thus once we see the 0 in the exponent we don't even look to see what is being raised to the zeroth power because it is inconsequential.

7. Sep 14, 2009

### CRGreathouse

Re: why why why why ??

There is exactly one function from the empty set to itself. You can call it f(x) = x^2 if you like; it's the unique function with domain and codomain equal to the empty set.

The 'usual' formalism for functions is an ordered triple (D, C, S) where D is the domain, C is the codomain, and S is a set of ordered pairs of the form (d, c) where d is in D and c is in C. There is exactly one such ordered pair for each member of D.

So in that sense, the empty set is not a function. The function itself is ({}, {}, {}).

8. Sep 14, 2009

### kaskus

Re: why why why why ??

i don't believe that calculator is wrong :uhh:

can you explain me

9. Sep 14, 2009

### Preno

Re: why why why why ??

For example:

$\lim_{x\to 0}0^x=0$, but $\lim_{x\to 0} x^0 = 1$.

10. Sep 14, 2009

### Santa1

Re: why why why why ??

Type in 0/0, 1e308, 1/0 and tell me what it says :) .

11. Sep 14, 2009

### D H

Staff Emeritus
Re: why why why why ??

In lots of places in math we say 0^0=1 as abuse of notation. It is a very convenient abuse of notation, but abuse nonetheless. For example, this abuse of notation lets us write infinite series in the form

$$f(x)=\sum_{n=0}^{\infty}a_nx^n$$

rather than the slightly more verbose

$$f(x)=a_0+\sum_{n=1}^{\infty}a_nx^n$$

However, 0^0 must be undefined. Given any arbitrary number a it is trivial to come up with a function y(x) such that y approaches zero as x approaches zero and such that xy approaches a as x approaches zero.

12. Sep 14, 2009

### kaskus

Re: why why why why ??

ok!!

How can you prove that 0^0=1 wrong?

sorry, i'm confused about this case :uhh:

13. Sep 14, 2009

### D H

Staff Emeritus
Re: why why why why ??

1. Let f(x) = 0^x, and examine how this behaves as x approaches zero. This function is identically zero except at x=0, where it is undefined. So maybe we should define 0^0 to be zero?

2. Let f(x,y) = xy and examine how this behaves as x and y simultaneously approach zero. The limiting value of f(x,y) as x and y approach zero, and even whether a limit even exists, depends entirely on the path taken. One example: Let y(x) = ln(2)/(ln(|x|)-|x|). Note that as x approaches zero, y also approaches zero, and in this case, the limit is 2. So maybe we should define 0^0 to be two? Or 2.71828...? Or 100? (Just replace the ln(2) with 1 and ln(100) to get 2.71828... and 100.)

14. Sep 14, 2009

### Whatever123

Re: why why why why ??

You can think of it as a division by zero, which is why it doesn't work... x^n/x^m = x^(n-m). Therefore, 0^0 = 0^x/0^x, which is, of course, 0/0 and in the indeterminate form...

15. Sep 14, 2009

### CRGreathouse

Re: why why why why ??

That explanation would also forbid 0^1 = 0.

The usual way of dealing with this is that in integer contexts, 0^0 = 1 and in real contexts, 0^0 is undefined. Here's Pari's opinion:
Code (Text):
> 0^0
%1 = 1
> 0.0^0
%2 = 2
> 0^0.0
*** _^_: gpow: 0 to a non positive exponent.
> 0.0^0.0
*** _^_: gpow: 0 to a non positive exponent.

16. Sep 15, 2009

### Whatever123

Re: why why why why ??

Ahhhh... Yeah, you're correct. Maybe I should have spent more than 30 seconds thinking up my "proof." I should have seen that it would say the same for 0^(x+n)/0^x. Thanks for correcting me. I thought that my explanation was a nice simple way to show that 0^0 is in the indeterminate for real contexts.

17. Sep 15, 2009

### D H

Staff Emeritus
Re: why why why why ??

CR, can you explain that second result, %2=2?? Not the %2, I understand that. The right hand side, =2.

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