Why is 0^0 considered 1 while 0^1 equals 0?

[kaskus]

why why why why ??

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

 

[Dragonfall]

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

 

[g_edgar]

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.

 

[kaskus]

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

Calculator’ shows that 0^0 = 1

 

[Dragonfall]

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.

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.

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

 

[JCVD]

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.

 

[CRGreathouse]

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

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 ({}, {}, {}).

 

[kaskus]

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.

i don't believe that calculator is wrong

can you explain me

 

[Preno]

For example:

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

 

[Santa1]

i don't believe that calculator is wrong

can you explain me

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

 

[D H]

in discrete math we like to say 0^0=1 ...

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 let's 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 x^y approaches a as x approaches zero.

 

[kaskus]

ok!


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

sorry, I'm confused about this case

 

[D H]

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) = x^y 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.)

 

[Whatever123]

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...

 

[CRGreathouse]

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...

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:

> 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.

 

[Whatever123]

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.

 

[D H]

Here's Pari's opinion:

> 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.

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