Zero to the power of zero: Undefined or 1?

[Dan Christensen]

In high school, they probably taught you that zero to the power of zero (0^0) is undefined. Your teachers would have argued something like:

We have: 0^4 = 0, 0^3 = 0, 0^2 = 0 and 0^1 = 0 (all equal to 0). Continuing this pattern, we would have 0^0 = 0.

But we also have: 4^0 = 1, 3^0 = 1, 2^0 = 1 and 1^0 = 1 (all equal to 1). Continuing this pattern, we would have 0^0 = 1.

Since 0 cannot equal 1, they told us to leave 0^0 undefined. That was good advice as it turns out, but readers here may be ready for a somewhat more sophisticated justification -- one based not on continuous variables as in many university textbooks (very complicated), but one based strictly on the use of the natural numbers.

We will construct or define a particular function ^ of 2 variables on the natural numbers that model the notion of ^ (exponentiation) being nothing more than the repeated multiplication of natural numbers.

Examples

3^2 = 3 x 3 = 9

3^3 = 3^2 x 3 = 3 x 3 x 3 = 27

3^4 = 3^3 x 3 = 3 x 3 x 3 x 3 = 81

3^5 = 3^4 x 3 = 3 x 3 x 3 x 3 x 3 = 243

and so on.

Working the pattern backwards and moving up this list of equations, we notice that with each step upward, we are dividing the value on the line below by 3. If we continue this pattern, we would have:

3 ^ 1 = 9 / 3 = 3 

3 ^ 0 = 3 / 3 = 1

Now, this would work for any base value, except for zero. Why doesn't it work for a base of zero? Only because we would not be able to "work the pattern backwards" by dividing by zero. We cannot divide 0^2 by 0 to get a value for 0^1. (Recall that we cannot divide by zero.)

What to do about 0^0 and 0^1, the only 2 missing pieces of the puzzle?  It turns out that any value whatsoever will work for 0^0. Pick any natural number p for 0^0. Since we are talking about repeated multiplication, 0^1 = 0^0 x 0 = p x 0 = 0. 

So, it seems that infinitely many different functions of 2 variables on the natural numbers will model repeated multiplications. Fortunately, they agree on every combination of values for the base and exponent except for the base and exponent both being zero!  Since 0^0 can be any number whatsoever, some mathematicians pick 1, mainly because it is so easy to work (a very controversial point, but nevertheless). Others think that, since we don't have a really good reason for picking any particular value, it should be left undefined -- just like your high school teachers told you!

If we want define 0^0 = 1, you might define ^ on the set N of natural numbers as follows:

1. For all n in N, we have n^0 = 1
   
2. For all n, m in N, we have n^(m+1)  = n^m x n
   

If you want to leave 0^0 undefined:

1. 0^1 = 0
   
2. For all n, m in N, if we do not have n=m=0, then n^(m+1) = n^m x n
   

Your comments or questions? Where do you stand on this controversial point? (It is  much written about on the internet.)

Dan

[Dan Christensen]

(Corrections)

On Sunday, 27 January 2019 00:31:41 UTC-5, Dan Christensen wrote:

In high school, they probably taught you that zero to the power of zero (0^0) is undefined. Your teachers would have argued something like:

We have: 0^4 = 0, 0^3 = 0, 0^2 = 0 and 0^1 = 0 (all equal to 0). Continuing this pattern, we would have 0^0 = 0.

But we also have: 4^0 = 1, 3^0 = 1, 2^0 = 1 and 1^0 = 1 (all equal to 1). Continuing this pattern, we would have 0^0 = 1.

Since 0 cannot equal 1, they told us to leave 0^0 undefined. That was good advice as it turns out, but readers here may be ready for a somewhat more sophisticated justification -- one based not on continuous variables as in many university textbooks (very complicated), but one based strictly on the use of the natural numbers.

We will construct or define a particular function ^ of 2 variables on the natural numbers that model the notion of ^ (exponentiation) being nothing more than the repeated multiplication of the natural numbers.

Examples

3^2 = 3 x 3 = 9

3^3 = 3^2 x 3 = 3 x 3 x 3 = 27

3^4 = 3^3 x 3 = 3 x 3 x 3 x 3 = 81

3^5 = 3^4 x 3 = 3 x 3 x 3 x 3 x 3 = 243

and so on.

Working the pattern backwards and moving up this list of equations, we notice that with each step upward, we are dividing the value on the line below by 3. If we continue this pattern, we would have:

3 ^ 1 = 9 / 3 = 3 

3 ^ 0 = 3 / 3 = 1

Now, this would work for any base value, except for zero. Why doesn't it work for a base of zero? Only because we would not be able to "work the pattern backwards" by dividing by zero. We cannot divide 0^2 by 0 to get a value for 0^1. (Recall that we cannot divide by zero.)

What to do about 0^0 and 0^1, the only 2 missing pieces of the puzzle?  It turns out that any value whatsoever will work for 0^0. Pick any natural number p for 0^0. Since we are talking about repeated multiplication, 0^1 = 0^0 x 0 = p x 0 = 0. 

So, it seems that infinitely many different functions of 2 variables on the natural numbers will model repeated multiplications. Fortunately, they agree on every combination of values for the base and exponent except for the base and exponent both being zero!  Since 0^0 can be any number whatsoever, some mathematicians pick 1, mainly because it is easier to work with (a very controversial point, but nevertheless). Others think that, since we don't have a really good reason for picking any particular value, it should be left undefined -- just like your high school teachers told you!

If we want define 0^0 = 1, you might define ^ on the set N of natural numbers as follows:

1. For all n in N, we have n^0 = 1
   
2. For all n, m in N, we have n^(m+1)  = n^m x n
   

If you want to leave 0^0 undefined:

1. 0^1 = 0
   

2.  For all m in N, if m >0, then 0^m = 0 

3.  For all n, m in N, if we do not have n=m=0, then n^(m+1) = n^m x n

[delaneyrm]

Dan,

Here’s another approach. Consider the function x^x  as x->0 through positive values. Let

y = x^x

then

ln(y) = ln(x^x) = x * ln(x) = ln(x) / (1/x)

The final expression has its numerator and denominator increase without limit as x->0+, so L'Hôspital's rule can be used.

lim(x->0+) ln(y) = lim(x->0+) ln(x) / (1/x) = lim(x->0+) (1/x) / (-1/x^2)

= lim(x->0+) (-x) = 0

So

lim(x->0+) ln(y) = 0

and thus

lim(x->0+) y = lim(x->0+) x*x = 1

which indicates the definition: 0^0 = 1

Numerical calculations support this.

.1^.1 = 0.7943282347242815

.01^.01 = 0.9549925860214359

.001^.001 = 0.9931160484209338

.0001^.0001 = 0.9990793899844618

.00001^.00001 = 0.9998848773724686

.

.

.

.0000000001^.0000000001 = 0.9999999976974149

No proof, but satisfying.

 

More generally, consider two continuous functions f(x) and g(x) where

lim(x->0+) f(x) = lim(x->0+) g(x) = 0

Then f(x)^g(x) takes the form 0^0 as x->0+. Let

y = f(x)^g(x)

so

ln(y) = ln(f(x)^g(x)) = g(x) * ln(f(x)) = ln(f(x)) / (1/g(x))

Apply L'Hôspital's rule.

lim(x->0+) ln(y) = lim(x->0+) ln(f(x)) / (1/g(x))

= lim(x->0+) (f’(x)/f(x)) / (-g’(x) / g(x)^2)

= lim(x->0+) [- (f’(x) / g’(x)) (g(x)^2 / f(x))]

If f(x) = x^m and g(x) = x^n where m and n are positive integers, then

lim(x->0+) ln(y) = lim(x->0+) -(m/n) x^(m-n) x^(2n-m) = lim(x->0+) -(m/n) x^n = 0

So one again finds

lim(x->0+) y = lim(x->0+) (x^m)^(x^n) = 1

no matter what values the positive integers m and n have.

Can one find such functions f(x) and g(x) which do not give lim(x->0+) f(x)^g(x) = 1 ?

Bob

[Dan Christensen]

On Wednesday, 30 January 2019 00:27:13 UTC-5, delaneyrm wrote:

Dan,

Here’s another approach. Consider the function x^x  as x->0 through positive values. Let

y = x^x

Have look at  Continuous Exponents at Wikipeda. There they talk about limits of z=x^y along different curves toward the origin in the x-y plane. You essentially chose a path along the y=x (with base = exponent). That does indeed tend to a limit of 1. Apparently, by choosing the right path, however, the limit can be any other non-negative value. (See the diagram at this link.) As I understand it, that is why, for continuous variables 0^0 is left undefined. Similarly in the discrete case that I have written about, it can be argued that any value will work for 0^0, so it could be left undefined. AFAIK, there is no purely arithmetical justification for choosing any one particular value. 

Dan

[Dan Christensen]

On Sunday, 27 January 2019 22:12:19 UTC-5, Dan Christensen wrote:

(Corrections)

On Sunday, 27 January 2019 00:31:41 UTC-5, Dan Christensen wrote:

In high school, they probably taught you that zero to the power of zero (0^0) is undefined. Your teachers would have argued something like:

We have: 0^4 = 0, 0^3 = 0, 0^2 = 0 and 0^1 = 0 (all equal to 0). Continuing this pattern, we would have 0^0 = 0.

But we also have: 4^0 = 1, 3^0 = 1, 2^0 = 1 and 1^0 = 1 (all equal to 1). Continuing this pattern, we would have 0^0 = 1.

Since 0 cannot equal 1, they told us to leave 0^0 undefined. That was good advice as it turns out, but readers here may be ready for a somewhat more sophisticated justification -- one based not on continuous variables as in many university textbooks (very complicated), but one based strictly on the use of the natural numbers.

We will construct or define a particular function ^ of 2 variables on the natural numbers that model the notion of ^ (exponentiation) being nothing more than the repeated multiplication of the natural numbers.

Examples

3^2 = 3 x 3 = 9

3^3 = 3^2 x 3 = 3 x 3 x 3 = 27

3^4 = 3^3 x 3 = 3 x 3 x 3 x 3 = 81

3^5 = 3^4 x 3 = 3 x 3 x 3 x 3 x 3 = 243

and so on.

Working the pattern backwards and moving up this list of equations, we notice that with each step upward, we are dividing the value on the line below by 3. If we continue this pattern, we would have:

3 ^ 1 = 9 / 3 = 3 

3 ^ 0 = 3 / 3 = 1

Now, this would work for any base value, except for zero. Why doesn't it work for a base of zero? Only because we would not be able to "work the pattern backwards" by dividing by zero. We cannot divide 0^2 by 0 to get a value for 0^1. (Recall that we cannot divide by zero.)

What to do about 0^0 and 0^1, the only 2 missing pieces of the puzzle?  It turns out that any value whatsoever will work for 0^0. Pick any natural number p for 0^0. Since we are talking about repeated multiplication, 0^1 = 0^0 x 0 = p x 0 = 0. 

More formally, for all x0 in N, there exists a binary function exp on N such that:

     1.  exp(0, 0) = x0

     2.  For all x in N: exp(x, 2) = x*x

     3. For all x, y in N: exp(x, y+1) = exp(x, y) * x

Also, for any binary functions exp and exp' on N and x1, x2 in N such that:

     1.  exp(0, 0) = x1

     2.  For all x in N: exp(x, 2) = x*x

     3. For all x, y in N: exp(x, y+1) = exp(x, y) * x

and

     1.  exp'(0, 0) = x2

     2.  For all x in N: exp'(x, 2) = x*x

     3. For x, y in N: exp'(x, y+1) = exp'(x, y) * x

we must then have:

     For all x, y in N: [~[x=0 & y=0] => exp(x, y) = exp'(x,y)]

In other words, the functions exp and exp' agree on every combination of base and exponent values with only the possible exception of base = exponent = 0.

Dan

[Dan Christensen]

On Monday, February 11, 2019 at 1:43:27 AM UTC-5, Dan Christensen wrote:

 

More formally, for all x0 in N, there exists a binary function exp on N such that:

     1.  exp(0, 0) = x0

     2.  For all x in N: exp(x, 2) = x*x

     3. For all x, y in N: exp(x, y+1) = exp(x, y) * x

Also, for any binary functions exp and exp' on N and x1, x2 in N such that:

     1.  exp(0, 0) = x1

     2.  For all x in N: exp(x, 2) = x*x

     3. For all x, y in N: exp(x, y+1) = exp(x, y) * x

and

     1.  exp'(0, 0) = x2

     2.  For all x in N: exp'(x, 2) = x*x

     3. For x, y in N: exp'(x, y+1) = exp'(x, y) * x

we must then have:

     For all x, y in N: [~[x=0 & y=0] => exp(x, y) = exp'(x,y)]

In other words, the functions exp and exp' agree on every combination of base and exponent values with only the possible exception of base = exponent = 0.

  

Here is a formal proof of this. 

Compared to the informal proofs that you would see in most math textbooks, formal proofs go into much greater detail and are usually much longer.

Dan