View Full Version : 0 ^ 0 = 1

Yarin

05-09-2009, 10:04 PM

I read that zero to the power of zero is suppose to be 1...

Does anyone know why this is the case? It seems silly to me. Wouldn't the logical answer be zero? To my understanding, exponentation is simply recursive multiplication, 3 ^ 4 is 3 * 3 * 3 * 3, 8 ^ 2 is 8 * 8. This would make zero to the power of x nothing. The way I see it, this should give zero, or, at the very least, an 'undefined' result. (much like dividing by zero).

?

Exponentiation - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Exponentiation#Zero_to_the_zero_power)

also

Computer programming languages that evaluate 0^0 to 1[16] include bc, Haskell, J, Java, LISP,MATLAB, ML, Perl, PHP, Python, R, Ruby, Scheme, and SQL. In the .NET Framework, the method System.Math.Pow treats 0^0 to be 1.

Among spreadsheet applications, Microsoft Excel issues an error when it evaluates 0^0, while OpenOffice.org 3 returns 1.

Microsoft Windows' Calculator and the calculator in Google search[17] evaluate 0^0 to 1.

Maple simplifies a^0 to 1 and 0^a to 0, even if no constraints are placed on a, and evaluates 0^0 to 1.

Mathematica simplifies a^0 to 1, even if no constraints are placed on a. It does not simplify 0^a, and it takes 0^0 to be an indeterminate form.

The TI-84 returns a Domain Error when given 0^0 to solve, but the TI-89 returns 1. The TI-89 Titanium returns undef.

Salem

05-09-2009, 10:19 PM

Division can be done by subtracting powers.

Any number can be written as some base, raised by some exponent (floating point does this all the time)

x = n ^ a

y = n ^ b

so

x / y == n ^ ( a - b )

Dividing a number by itself gives 1, which in the above identity would mean a and b are the same, that is a-b is 0

laserlight

05-09-2009, 10:44 PM

Based on the same reasoning that n^0 is 1 for some n not equal to 0 (which is what Salem just outlined), I reason that 0^0 = 0/0, and hence it is indeterminate. However, I have never asked a professional mathematician for his opinion.

bithub

05-10-2009, 01:40 AM

0^0=1 is true simply for convenience, yet it is not something that is universally agreed upon. Read this (http://en.wikipedia.org/wiki/Exponentiation#Zero_to_the_zero_power) for more information.

twomers

05-10-2009, 10:08 AM

>> I reason that 0^0 = 0/0, and hence it is indeterminate.

0^0 is perfectly defined.

laserlight

05-10-2009, 10:16 AM

0^0 is perfectly defined.

Please elaborate.

twomers

05-10-2009, 11:37 AM

Whoops. Though I wrote it as 0^0 I meant x^0.

laserlight

05-10-2009, 12:06 PM

Though I wrote it as 0^0 I meant x^0.

Right, in which case you completely missed or ignored the "based on the same reasoning that n^0 is 1 for some n not equal to 0" part of my sentence. (Though in retrospect it should be any n rather than some n, but whatever.)

twomers

05-10-2009, 12:19 PM

Thanks, laserlight. I'd love an omelette.

to calculate 0^0 one can take lim(x^y) when x,y -> 0

Depending on the relative speed of approximation to zero the result will be different

Snafuist

05-11-2009, 02:37 AM

The recursive definition of the product is:

prod() = 1

prod(x,y,z, ...) = x * prod(y,z, ...)

Thus, prod(5,3) = 5*prod(3) = 5*3*prod() = 5*3*1 = 15.

1 has been chosen because it is the neutral element of multiplication. Hence, a^0, 0^0, 0! are all equal to 1.

Check your favorite math book for "empty product", "empty sum", "empty function"...

EDIT: and it makes sense, too: how many functions are there from X->Y? Exactly |Y|^|X| functions. What if X=Y=<empty set>? Then |Y|^|X| = 0^0 = 1, and there's exactly one function from the empty set to the empty set: the empty function.

Greets,

Philip

Magos

05-11-2009, 03:34 AM

x^0 (where x -> 0) = 1

0^x (where x -> 0) = 0

So 0^0 will get different values depending on how you approach it. The actual value is thus undefined.

laserlight

05-11-2009, 03:42 AM

So 0^0 will get different values depending on how you approach it. The actual value is thus undefined.

You can always define it out of convenience (and possibly elegance), which is what the cited Wikipedia article notes, as do bithub and Snafuist. If that Wikipedia article is to be believed, mathematicians do concur with my analysis as well, to some extent.

Yarin

05-11-2009, 06:38 PM

I've looked around more (already seen Wiki's say on it), it seems that most agree that the value is (in reality) undefined.

Glad that's cleared up. :) My mind often has a mind of it's own and goes off pondering things that I really don't care about.

>> Thanks, laserlight. I'd love an omelette.

Forgive me for not understanding this, but... what?

brewbuck

05-11-2009, 07:21 PM

to calculate 0^0 one can take lim(x^y) when x,y -> 0

Depending on the relative speed of approximation to zero the result will be different

But that assumes that x and y can both be varied. If you treat y as a constant 0, then you are left with lim (x ^ 0) x->0 which is clearly 1.

In practice these undefined points can be defined in whatever way is most convenient for the problem at hand (usually, equal to the limit when approaching from the appropriate direction)

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