kitchen table math, the sequel: zero

Monday, April 19, 2010


At least somebody around here is smarter than a 5th grader:
Their answer to question 29, "Any number to the power of zero is equal to what value?" is incorrect. Zero to the power of zero is indeterminate, not 1.

Neils Henrik Abel
Why is this true?


OrangeMath said...

x^a/x^a = x^(a-a) = x^0. However, this last expression has lost the original information of dividing by x^a, if x were zero, this would be undefined. Therefore, 0^0 is undefined.

Anonymous said...

Plus ... in the *limit* any number raised to the power of zero *is* 1. After all, as we approach infinitely many numbers (1, 2, 3, ...) raised to the power of 0, the ratio of wrong answers (1) to correct answers (∞) heads towards zero (1/∞ = 0). If we have 0 incorrect answers, then all our answers are correct.



-Mark Roulo

Allison said...

Actually, the original claim "any number to the power of zero equals 1"

isn't true.

It's DEFINED to BE true by convention, BECAUSE people wanted to be able to do the nice addition and subtraction rules on exponents. But nothing about the reals requires it to be true.

I'll find a reference, hold on.

farmwifetwo said...

Jeez, it's been a LONG time :) But then again I've been getting a refresher course teaching the elder.

Lsquared said...

How about this: for any non-zero number n^(1/m) is the m-th root of n, which approaches 1 as m approaches infinity, so for n not equal to zero, continuity implies that you want n^0=1. Now, what about 0. We'd like continuity if we can have it, but we have two conflicting urges:
n^0=1 (except maybe at 0) and 0^m=0 for any m (except maybe 0), so, which is it? is 0^0=1 or is 0^0=0? There's no good way to decide. It's like 0/0--no good way to decide what it should be either. So, since we can't choose for it to have two answers(not helpful), we have to say that it is undefined.

In fact, in calculus you find that if you have functions f and g, both of which approach 0 as x approaches, say 0, then f^g as x approaches 0 can be...I think you can get any non-negative number out of that in the limit, depending on what the functions f and g happen to be (I'd need to do a couple of examples to make sure about the all non-negative numbers part, but that's what my intuition is telling me--I know for sure that there are infinitely many things you can get). Cool and weird (and occasionally useful).

Michael Weiss said...

These answers are more complicated than they really need to be.

The basic problem is that 0^x = 0 for all non-zero values of x, but x^0 = 1 for all non-zero values of x, and there is no way to define 0^0 in a way that is consistent with both of those rules at the same time.

One way to solve the problem is to figure out what x^x is as x approaches 0. It turns out that, indeed, as x approaches 0 from the right the limit is 1. On the other hand, if x approaches 0 from the left you have a more fundamental problem, because negative values raised to fractional powers are non-real (consider (-1/2)^(-1/2)).

Another approach is to consider the function x^y, as both x and y approach 0. (Imagine a continuous path of points (x,y) leading towards (0,) and calculate x^y at each point along the path.) But here the problem is that the limit depends on the path you take!

So, yeah. Sometimes 0^0 is defined to be 0 or 1 for conventional purposes, but there is no unambiguous way to say what it "should" be.

Webmaster said...

Yeah, that's right. This is wonderful blog. Success.
elementary school teacher

CyberChalky said...

0^0 is an indeterminate form, but L'hopitals rule allows us to devolve the two possible states (1 and 0) to a single result (1) by investigating the ratios of the derivates of the function as they approach zero from the negative and positive directions.
Dr. Math gives an intro to this concept here: