What is 0 (zero) to the power of 0 (zero)?

[AsleepInYorkshire]

Spoiler

The spoiler is a YouTube fourteen minute explanation of how to get to the answer.

AiY

[GoSeigen]

Spoiler

The spoiler is a YouTube fourteen minute explanation of how to get to the answer.

AiY

Did Knuth lead you here?

GS

[absolutezero]

There is no agreed value but I go with 1.
Anything to the power of 0 is 1.

[scrumpyjack]

What is 0 to the power 0?
I assume you are familiar with powers.

The problem is similar to that with division by zero. No value can be assigned to 0 to the power 0 without running into contradictions. Thus 0 to the power 0 is undefined!

How could we define it? 0 to any positive power is 0, so 0 to the power 0 should be 0. But any positive number to the power 0 is 1, so 0 to the power 0 should be 1. We can't have it both ways.

Underlying this argument is the same idea as was used in the attempt to define 0 divided by 0. Consider a to the power b and ask what happens as a and b both approach 0. Depending on the precise way this happens the power may assume any value in the limit.

(per the University of Utah maths department) (Excel agrees and returns an error, not 1)

[Mike4]

There is no agreed value but I go with 1.
Anything to the power of 0 is 1.

On the other hand, anything multiplied by 0 is 0, including itself I'd have thought.

[AsleepInYorkshire]

What is 0 to the power 0?
I assume you are familiar with powers.

The problem is similar to that with division by zero. No value can be assigned to 0 to the power 0 without running into contradictions. Thus 0 to the power 0 is undefined!

How could we define it? 0 to any positive power is 0, so 0 to the power 0 should be 0. But any positive number to the power 0 is 1, so 0 to the power 0 should be 1. We can't have it both ways.

Underlying this argument is the same idea as was used in the attempt to define 0 divided by 0. Consider a to the power b and ask what happens as a and b both approach 0. Depending on the precise way this happens the power may assume any value in the limit.

(per the University of Utah maths department) (Excel agrees and returns an error, not 1)

Is the argument more about the relative value of 0? I'm not entirely sure I even understand my own question

But if we were to divide zero by zero then surely the answer is 1? Failing which I'd suggest we have to prove that zero exists at all. Because if it doesn't exist then I can't prove it does. If it does then it must have value?

No I've not had any sherry

AiY

[scrumpyjack]

I think you should have a very large glass of sherry. Zero in on the wine cupboard and be noughty!

[AsleepInYorkshire]

I think you should have a very large glass of sherry. Zero in on the wine cupboard and be noughty!

[Dod101]

Excellent youtube video. I wish I had had a guy like him to teach me! Absolute 0 to the power absolute 0 is surely 0?

Dod

[jfgw]

There is no agreed value but I go with 1.
Anything to the power of 0 is 1.

On the other hand, anything multiplied by 0 is 0, including itself I'd have thought.

Except that it is multiplied by itself exactly zero times. Zero to a positive power is 0; zero to a negative power is 1/0. I haven't watched the video yet but I understood 0^0 to be 1.


Julian F. G. W.

[johnhemming]

On the other hand, anything multiplied by 0 is 0, including itself I'd have thought.

The way I looked at it before watching the video is that when it comes to powers if a is to the power of b one starts with 1 and you multiply that by a b times.

Hence for the number a^0 you start with 1 and multiply that by a zero times which leaves 1.

The video sort of proves that is the correct answer as one goes from x ^ x as x tends towards zero from positive numbers.

From negative numbers it is complex or even imaginary.

[AsleepInYorkshire]

Excellent youtube video. I wish I had had a guy like him to teach me! Absolute 0 to the power absolute 0 is surely 0?

Dod

Yup. His passion and enthusiasm is absolutely something to witness. Many of my daughters teachers have a similar presence and it works. They really do inspire her to go that extra mile.

Now as for the answer being 1 or nearly 1 or a total figment of my imagination, that's just mind blowing stuff.

AiY

[Gengulphus]

What is 0 to the power 0?
I assume you are familiar with powers.

The problem is similar to that with division by zero. No value can be assigned to 0 to the power 0 without running into contradictions. Thus 0 to the power 0 is undefined!

How could we define it? 0 to any positive power is 0, so 0 to the power 0 should be 0. But any positive number to the power 0 is 1, so 0 to the power 0 should be 1. We can't have it both ways.

Underlying this argument is the same idea as was used in the attempt to define 0 divided by 0. Consider a to the power b and ask what happens as a and b both approach 0. Depending on the precise way this happens the power may assume any value in the limit.

(per the University of Utah maths department) (Excel agrees and returns an error, not 1)

That isn't quite right, because what it's run into is a discontinuity, not a contradiction: the function x^y cannot be made continuous at (x,y) = (0,0).

E.g. if I define a function f(x) by f(x) = 0 if x <= 0, 1 if x > 0, then if I approach x = 0 from below without actually getting to it, f(x) is always 0, and if I approach x = 0 from above without actually getting to it, f(x) is always 1. If the quoted argument were entirely correct, I would have to conclude that f(0) is undefined because giving it any value runs into a contradiction. But that conclusion that f(0) is undefined itself contradicts my definition of f(x), which says that f(0) is defined to be 0. I.e. the attempt to avoid a 'contradiction' itself runs into a contradiction!

The resolution of that issue is that functions do not need to have the property that as x approaches y, f(x) approaches f(y). That property (which needs to be defined a bit more formally - "approaches" is rather vague - but I won't bore people with the formal definitions!) is known as continuity, and the resolution of what is said in the last paragraph is that f(0) is defined and equal to 0, but the function f(x) is discontinuous at x = 0. Furthermore, it cannot be made continuous at x = 0 by defining f(0) differently, because f(x) approaches two different values as x approaches 0, depending on how it approaches 0. (As a contrast, consider the function g(x) defined by g(x) = 0 if x = 0, 1 if x is not equal to 0. Since g(x) is always 1 as x approaches 0, and g(0) is not 1, the function g(x) is discontinuous at x = 0. But since g(x) only ever approaches 1 as x approaches 0, it can be made continuous by choosing g(0) = 1 instead.)

Anyway, the quoted argument says that the function x^y approaches different values as (x,y) approaches (0,0), depending on how the approach is done. So x^y is discontinuous at (x,y) = (0,0), and cannot be made continuous there by choosing any particular value for 0^0. That doesn't say that 0^0 cannot be defined - it just says that whether one defines it or not, one cannot use continuity arguments about it.

So how, if at all, should one define 0^0? At this point, an issue comes up that is quite common in mathematics, but not particularly well-known outside mathematics: often, the question of how one best defines a mathematical concept depends on the mathematical area in which one is operating. In particular, in the case of 0^0, it makes a big difference whether one is working with real numbers or with integers (i.e. whole numbers).

When working with real numbers, the issue of x^y being discontinuous at (x,y) = (0,0) is a fairly important one, so that there is good reason for 0^0 to be left undefined, because one then has the fairly simple theorem "x^y is continuous everywhere where it is defined" rather than the somewhat more complex "x^y is continuous everywhere where it is defined, except at 0^0". In particular, anyone trying to apply the former theorem needs to apply it for the two cases "x^y is defined" and "x^y is not defined", rather than the three cases "x^y is 0^0", "x^y is defined but not 0^0" and "x^y is not defined", so arguments using the theorem will tend to be simpler.

When working with integers, and especially for questions in the mathematical area of combinatorics where the numbers are generally non-negative integers, continuity is essentially irrelevant, and the definition used for X^Y is basically "the number of ways one can choose an ordered sequence of Y objects from X different types of object, with duplicate choices allowed". For example, 2^3 = 8 because I can choose an ordered sequence of 3 objects from objects of types A and B in eight different ways: (A,A,A), (A,A,B), (A,B,A), (A,B,B), (B,A,A), (B,A,B), (B,B,A) and (B,B,B). And 3^2 = 9 because I can choose an ordered sequence of 2 objects from objects of types X, Y and Z in nine different ways: (X,X), (X,Y), (X,Z), (Y,X), (Y,Y), (Y,Z), (Z,X), (Z,Y) and (Z,Z).

Using that definition, n^0 = 1 for all non-negative n, because there is always just one way of making no choices, and 0^n = 0 for all positive n but not for n = 0, because we are faced with the impossible task of choosing an object from no types of object as soon as we are asked to choose an object. So 0^0 is unambiguously 1 in the mathematical area of combinatorics.

I don't know of any mathematical areas that would favour any other values for 0^0, so I think the answer to the question in this thread's subject is basically "Either 1 or undefined, depending on the mathematical area one is working in".

Gengulphus

[servodude]

What is 0 to the power 0?
I assume you are familiar with powers.

The problem is similar to that with division by zero. No value can be assigned to 0 to the power 0 without running into contradictions. Thus 0 to the power 0 is undefined!

How could we define it? 0 to any positive power is 0, so 0 to the power 0 should be 0. But any positive number to the power 0 is 1, so 0 to the power 0 should be 1. We can't have it both ways.

Underlying this argument is the same idea as was used in the attempt to define 0 divided by 0. Consider a to the power b and ask what happens as a and b both approach 0. Depending on the precise way this happens the power may assume any value in the limit.

(per the University of Utah maths department) (Excel agrees and returns an error, not 1)

That isn't quite right, because what it's run into is a discontinuity, not a contradiction: the function x^y cannot be made continuous at (x,y) = (0,0).

E.g. if I define a function f(x) by f(x) = 0 if x <= 0, 1 if x > 0, then if I approach x = 0 from below without actually getting to it, f(x) is always 0, and if I approach x = 0 from above without actually getting to it, f(x) is always 1. If the quoted argument were entirely correct, I would have to conclude that f(0) is undefined because giving it any value runs into a contradiction. But that conclusion that f(0) is undefined itself contradicts my definition of f(x), which says that f(0) is defined to be 0. I.e. the attempt to avoid a 'contradiction' itself runs into a contradiction!

The resolution of that issue is that functions do not need to have the property that as x approaches y, f(x) approaches f(y). That property (which needs to be defined a bit more formally - "approaches" is rather vague - but I won't bore people with the formal definitions!) is known as continuity, and the resolution of what is said in the last paragraph is that f(0) is defined and equal to 0, but the function f(x) is discontinuous at x = 0. Furthermore, it cannot be made continuous at x = 0 by defining f(0) differently, because f(x) approaches two different values as x approaches 0, depending on how it approaches 0. (As a contrast, consider the function g(x) defined by g(x) = 0 if x = 0, 1 if x is not equal to 0. Since g(x) is always 1 as x approaches 0, and g(0) is not 1, the function g(x) is discontinuous at x = 0. But since g(x) only ever approaches 1 as x approaches 0, it can be made continuous by choosing g(0) = 1 instead.)

Anyway, the quoted argument says that the function x^y approaches different values as (x,y) approaches (0,0), depending on how the approach is done. So x^y is discontinuous at (x,y) = (0,0), and cannot be made continuous there by choosing any particular value for 0^0. That doesn't say that 0^0 cannot be defined - it just says that whether one defines it or not, one cannot use continuity arguments about it.

So how, if at all, should one define 0^0? At this point, an issue comes up that is quite common in mathematics, but not particularly well-known outside mathematics: often, the question of how one best defines a mathematical concept depends on the mathematical area in which one is operating. In particular, in the case of 0^0, it makes a big difference whether one is working with real numbers or with integers (i.e. whole numbers).

When working with real numbers, the issue of x^y being discontinuous at (x,y) = (0,0) is a fairly important one, so that there is good reason for 0^0 to be left undefined, because one then has the fairly simple theorem "x^y is continuous everywhere where it is defined" rather than the somewhat more complex "x^y is continuous everywhere where it is defined, except at 0^0". In particular, anyone trying to apply the former theorem needs to apply it for the two cases "x^y is defined" and "x^y is not defined", rather than the three cases "x^y is 0^0", "x^y is defined but not 0^0" and "x^y is not defined", so arguments using the theorem will tend to be simpler.

When working with integers, and especially for questions in the mathematical area of combinatorics where the numbers are generally non-negative integers, continuity is essentially irrelevant, and the definition used for X^Y is basically "the number of ways one can choose an ordered sequence of Y objects from X different types of object, with duplicate choices allowed". For example, 2^3 = 8 because I can choose an ordered sequence of 3 objects from objects of types A and B in eight different ways: (A,A,A), (A,A,B), (A,B,A), (A,B,B), (B,A,A), (B,A,B), (B,B,A) and (B,B,B). And 3^2 = 9 because I can choose an ordered sequence of 2 objects from objects of types X, Y and Z in nine different ways: (X,X), (X,Y), (X,Z), (Y,X), (Y,Y), (Y,Z), (Z,X), (Z,Y) and (Z,Z).

Using that definition, n^0 = 1 for all non-negative n, because there is always just one way of making no choices, and 0^n = 0 for all positive n but not for n = 0, because we are faced with the impossible task of choosing an object from no types of object as soon as we are asked to choose an object. So 0^0 is unambiguously 1 in the mathematical area of combinatorics.

I don't know of any mathematical areas that would favour any other values for 0^0, so I think the answer to the question in this thread's subject is basically "Either 1 or undefined, depending on the mathematical area one is working in".

Gengulphus

reading that made my heart sinc

[9873210]

Really awful video. I'm really glad my math teachers were nothing like that.

Enthusiasm is no substitute for rigour.

For a start he jumps from the definition of N^M for positive integers and does not even bother to extend it for reals. (Or rationals which would be an even deeper pit.)

To define x^y for reals you need limits (or something similar) and need to understand uniform convergence. At which point the argument in the video does not work. It does not matter if the answer is correct or not if the argument is wrong.

[9873210]

Less crustily:

If we want to define the value as a limit, and we want it to be well behaved we need

lim x,y->0 { x^y }

to exist. For this to exist we need the limit to be the same for all paths that end up at (0,0). But that is not the case here.

If we approach along x = 0 we get

lim y->0 { 0^y } = 0

If we approach along y = 0 we get

lim x->0 { x^0 } = 1

These can also be considered as
lim y->0 { lim x->0 { x^y } } = 0
lim x->0 { lim y->0 { x^y } } = 1

Which shows you cannot, in general, swap the order of the limits and get the same result.

Other paths give other answers, in the video he approaches along x = y
lim x->0 { x^x } = 1

More exotically we can approach (0,0) along the curve y = -1/ln(x) and get

lim x->0 { x ^ (-1/ln(x) ) } = lim x->0 { 1/e } = 1/e = 0.367879...

Using a different base for the logarithm get's you anything you want.

So considering it as a function of two variables f(x,y) = x^y
we can find points arbitrarily close to (0,0) where x^y takes on any value.
The easy way to deal with this singularity is to put up the "here be dragons" sign and call it "undefined".
One of the hard ways to deal with it is to define it as 1 and be eaten by the dragons.

[johnhemming]

I am confused myself because I thought it was y=x^x that we were considering.

[9873210]

I am confused myself because I thought it was y=x^x that we were considering.

We were considering 0^0. We can approach that in various ways. x^x, 0^x, x^0 etc. None of them is privileged.

lim x->0 ( x^x ) = 1 is uncontested. What is contested is its bearing on the value of 0^0.

[johnhemming]

I think we are looking at fn(x) hence only fn(x) viz x^x is privileged

[johnhemming]

I did a bit of driving around Birmingham and I thought actually the nature of the process of fn(x) for x^x and x^0 was privileged, but 0^x is not so privileged because of the way in which the function operates.

That's what taxi driving for the family does for you.