0

I saw the following argument on Quora that provides which should actually be a false proof since $0^0$ is not defined to be $1$.

$(a+b)^{n}=\sum_{k=0}^{n}\binom n{ k}a^kb^{n-k}$. Put $a=0, b=1, n=0$. This gives $1^0=\binom0{0}1^00^0\implies 0^0=1$.

Also the person claiming this to be true says that:

"Also, $0^0$ is the number of mappings from the empty set to the empty set. Surely there’s no other answer to this number of mappings than $1$. Any arguments against setting $0^0=1$ are based on limiting forms $f(x)^{g(x)}$ where, as $x$ approaches some number, $f(x)$ and $g(x)$ approach zero but $f(x)^{g(x)}$ does not approach $1$. That doesn’t matter one bit: the limit of $⌊x⌋$ as $x$ tends to $0$ does not exist, but everyone is happy with defining $⌊0⌋=0$ all the same."

Why is this claim true if it is and why otherwise? What is the possible discrepancy in this proof and defining $0^0$ as $1$? Why is it beneficial to define it so?

Paras Khosla
  • 6,481
  • It's a common convention to define $a^0=1$ (where the domain for the exponent is restricted to only the integers) exactly so things like that binomial series expansion are valid. Plus, it's nice to have $a^{n+m}=a^n a^m$ for all integers $n$ and $m$, even if $a$ might be zero. The "proof" is merely a demonstration that this convention is assumed in the binomial expansion. – Kyle Miller Apr 11 '19 at 16:28

1 Answers1

7

One way to take this "proof" is as a reason to define $0^0=1.$ There are a lot of reasons to define $0^0=1,$ and only continuity reasons to leave $0^0$ undefined.

  1. In two basic ways to define exponentiation, set theory and lambda calculus, it is "easy" to define exponentiation if we have $0^0=1,$ but if we want to leave it undefined, particularly in lambda calculus, it becomes difficult.

  2. In a lot of notation about polynomials and power series (like in your "proof") we often implicitly take the value of $a_0x^0$ to be $a_0$ when $x=0.$ We'd have to either write a separate function or just implicitly accept we use $0^0$ in this case.

The disadvantage of taking $0^0=1$ is that the function $x^y$ would not be continuous at $(x,y)=(0,0).$ That's okay, as long as we remember the discontinuity. $0^0$ is an "indeterminate form" even if we define $0^0=1.$

One way to split the difference is to say that $0^0=1$ when the exponent $0$ is a natural number, and undefined when it is real. Our basic example in (1) both are about "types" of numbers that include the naturals, but not the reals.

This splitting the difference might make sense in programming languages, for example, when a "float" value of zero actually, after many computations, might represent a value approximated by $0.$ So we'd leave $0.0000^{0.0000}$ undefined because "estimating" the value with $1$ is not a useful estimation. (Whereas, if $x=0.0000, y=0.0000,$ then $x+y=0.00000,x\times y=0.00000$ are good estimations.)

Note that we define $0!=1,$ too. This is because an "empty product" is taken to be $1$. This is also a useful definition.

The value of $0^0$ is all about definitions, and what you want from that definition. If $0^0$ is undefined, then the standard binomial formula for $(a+b)^n$ is not exactly accurate when $a=0$ or $b=0.$

Thomas Andrews
  • 177,126