# [Debate] The Infamous Magic of Zero Debate in the comments if $0^0$ should be $1$, $0$ or undef

$0^0$ is defined to be 1 by the IEEE.

That is so because that would be more useful.

What it should be essentially boils down to the definition of exponentiation you're using.

This is an indeterminate form:

$\lim_{x \to a}f(x)^{g(x)} \quad \text{where } (\lim_{x \to a} f(x), \lim_{x \to a} g(x)) = (0,0)$

But please understand that this is not the same expression at what we are looking 5 years, 5 months ago

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"Indeterminate" means that there exists more than one alternative evaluation of the expression, which is clearly the case here. Hence, it's indeterminate. Given any value $x$, some argument can be made that ${ 0 }^{ 0 }=x$. So, people can choose one value of $x$ as a "convention", but that's all it is, a convention, and not a derived mathematical fact.

Mathematics does not say that every expression necessarily evaluates to an unique value.

- 5 years, 5 months ago

I have to agree with this. We can work backwards. If

$2^0=\dfrac{2^1}{2}=1$

$0^0=\dfrac{0^1}{0}=$ undefined

- 5 years, 5 months ago

By that argument, zero has no powers, because, for example, $\frac{0^3}{0} = 0^2$, but division by zero is undefined, so $0^2$ is also undefined.

Or, $\frac{0^2}{0} = 0$, so zero doesn't even exist!

- 5 years, 5 months ago

I completely agree that it is indeterminate. A few years ago, I came up with an example involving integer addition which shows that $0^0$ can be either 0 or 1, and I put it in this post.

- 5 years, 4 months ago

Since $0$ is such a tricky number, as it cannot be algebraically manipulated, I like to try to take a look at things in a basic (caveman) perspective for once. $n^{m}=\begin{matrix} m \text{ times} \\ \overleftrightarrow { n \times n \times ... n \times n} \end{matrix}$ So $0^{0}=\begin{matrix} 0 \text{ times} \\ \overleftrightarrow { 0 } \end{matrix}$ It's like dimensions. $2^{2}$ gives a square of length $2$, $3^{3}$ gives a cube of side $3$.

So, $0^{0}$ would give a point. That point, in our $3D$ world, it's volume is $0$. However, measuring the "value" of that point in $0$ dimension would give...1? One could argue that since it's base is $0$, the "value" of that point would be $0$ even in $0$ dimensions. This would explain why something like $3^{0}=1$, because of the $3$ as the base, in the $0$ dimension it is still worth some value which would be $1$.

I would say $0$ for this case.

- 5 years, 5 months ago

I really like your "caveman" logic. So, can we simply say the following:

${ m }^{ n }=\begin{matrix} \quad \quad n\quad times \\ 1\times \overleftrightarrow { m\times m\times ...\times m } \end{matrix}\\ { 0 }^{ 0 }=\begin{matrix} \quad \quad 0\quad times \\ 1\times \overleftrightarrow { \quad \quad \quad 0\quad \quad \quad } \end{matrix}\\ \Rightarrow { 0 }^{ 0 }=1\quad ?$

Since multiplying by zero, zero times is equivalent to not multiplying anything at all.

- 5 years, 5 months ago

waaaat.... But by doing this you are assuming that $0^{0}=1$ at the second line to the third line.

- 5 years, 5 months ago

No, I think i just made the assumption that $0^0=1\times 0^0$. Which seems right to me.

- 5 years, 5 months ago

Im starting to hate $0$ - 5 years, 5 months ago

No, no, it's not an assumption. You can't get $0^0 = 0$ by multiplying no zeroes, because there are no zeroes to multiply, like you can't get $3^0 = 3$ since there are no threes to multiply.

- 5 years, 5 months ago

Well, he is trying to get a geometric interpretation

- 5 years, 5 months ago

But in the 0 dimensional universe, wouldn't 0=1=2=... because that 0 is all that exists

- 5 years, 5 months ago

Im not sure about that. It's like asking for the thickness of a $2D$ object. That's why I tried to assign a value for a $0D$ object, which I am not sure is the correct thing to do.

- 5 years, 5 months ago

That is a nice way of looking at the problem, although it is only limited to non-negative integers

- 5 years, 5 months ago

It's like $0!$. For example, in binomial theorem, we have the form $(x+y)^n$. If it is to be extended to the form $(x+0)^n$ or simply $x^n$, we end up with the conclusion that $0^0=1$

- 5 years, 5 months ago

There is a big difference between saying $f(a)=b$ and $\lim_{x \to a}f(x)=b$. In the former, the function is defined at a, whereas this requirement is not necessary for the later relation.

$\sin(0)/0$ is indeterminate. However, $\lim_{x \to 0} \sin(x)/x =1$

- 5 years, 5 months ago

In case of a field with $0$ as its sole element. With the operations $0+0=0$ and $0*0=0$, we will have $1=0$, $0^{-1}=0$ and other weird things.

- 5 years, 5 months ago

Why do you want to work in this field?

- 5 years, 5 months ago image

- 5 years, 5 months ago

Note that it states that the result is indeterminate, even though it appears to give you the value of 1.

Staff - 5 years, 5 months ago

That's why I posted this

- 5 years, 5 months ago

I like 0^0 = 1 but the mighty alpha says indeterminate. link Them fellows over there are a might bit clever and they probably have a dang good reason for disagreeing with me.

- 5 years, 5 months ago

What does pappym think about disagreeing with one's own self?

- 5 years, 5 months ago

Great, the argument is heating up! I got the fastest 8 reshares ever

- 5 years, 5 months ago

yay! Im the $8^{th}$ reshare!

- 5 years, 5 months ago

@Agnishom Chattopadhyay You didn't... the TKC, 'twas... :-P

- 5 years, 4 months ago

$0^{0}$ would be undefined.

First of all, any number raised to the power of $0$ is equal to $1$. If you would logically think of it, $n^{0}=\frac{n}{n}$ . However, $\frac{0}{0}$ would be undefined, because it would give many contradictions.

However, on what I am saying that $0^{0}$ is undefined, it is very unstable too, giving more contradictions. For example, $0^{5} = 0^{0}$. Surprised? Well, we can say that $0^{5} = 0^{(6-1)}$. And $0^{(6-1)} = \frac{0^6}{0^1}$, which would still be equal to $\frac{0}{0}$.

One thing's for sure: zero is very, very, very full of black magic.

- 5 years, 5 months ago

You say $0^0$ is undefined, AND anything to the power 0 is equal to 1? That itself sounds like a contradiction.

- 5 years, 4 months ago

One should ask to oneself, does it correspond to a "real" life situation? We may look at this like this : I define factorials for all +ve integers (>0). In physical context, it corresponds to 'no. of arrangements of n things'. So, logically, 0!=1.(only one way!) . Then I extend this whole definition to +ve non integers too(Gamma Function) and that too tells that 0!=1. So, I can proudly say that 0!=1. Now, what's the need to define 0^0? It doesn't correspond to a "real" life situation. And speaking mathematically, it is an undefined form(strictly, it's undefined and not indeterminate. For(->0)^(->0) is Indeterminate),so it may be given any value. But what's the point in doing this? And if you do, then also define 0/0......

- 5 years, 3 months ago

If we assum 0^0 as one .then how would the graph be seen ,would it be a non continuous and what will its range.

- 4 years, 6 months ago

It would be discontinuous at zero, I think.

- 4 years, 6 months ago

I do agree ,but would it be a point size gap will there be some range of ,ie.lim to 0 is still 1 but not for 0 so it would be unimaginable small that it would virtue to be a lin

- 4 years, 6 months ago

I think that 0^0 should be 1

taking$h\left( x \right) ={ x }^{ x }$

$\lim _{ x\rightarrow { 0 }^{ + } }{ { x }^{ x } } =\lim _{ x\rightarrow { 0 }^{ + } }{ { e }^{ xlogx } }$

since $\lim _{ x\rightarrow { 0 }^{ + } }{ xlogx } =0$ The above limit should be 1 i'm not sure if this is right because i'm just taking two functions as f(x),g(x) as x when they could be any other functions tending to zero and h(x)=f(x)^g(x)

- 5 years, 5 months ago

I agree that $0^0$ should be 1, but not for those reasons. Multiplying by five squared is like multiplying by five twice, multiplying by 12 cubed is like multiplying by 12 thrice, multiplying by anything to the zeroth is like not multiplying by that base at all. It shouldn't matter what the limits are.

- 5 years, 4 months ago

${ 2 }^{ -\infty }=0\\ 2={ 0 }^{ \frac { 1 }{ -\infty } }={ 0 }^{ 0 }$

So this way ${ 0 }^{ 0 }$ can be anything

- 5 years, 5 months ago

Division by infinity is not a valid operation. Infinity is not a real number

- 5 years, 5 months ago

What if we put "n" instead of infinity there and write lim n->infinity

- 5 years, 4 months ago

That doesn't work either. $\displaystyle \lim_{x -> 0}\frac xx =1$, and $\displaystyle \lim_{x -> 0}\frac 0x =0$, but $\frac 00$ is undefined.

- 5 years, 4 months ago

If you're saying $0^{0}$ can be anything, then it's undefined/indeterminate. That is that. Nothing can define it

- 5 years, 5 months ago

Yes, I'm saying that only

- 5 years, 5 months ago

How can you say that ${ 2 }^{ -\infty }=0$? Although I get your point, the expression is indeterminate. You can rather put it as

$\huge{\lim _{ x\rightarrow -\infty }{ { 2 }^{ x } } =0}$

- 5 years, 4 months ago

Exponentiation is supposed to be a function, and functions can only have one value. Thus, 0x = 3 has no solution, 0x = 0 has everything for a solution, but both 0/0 and 3/0 are undefined.

- 5 years, 4 months ago

By the reflexive property of equality, it can only be at most one thing.

- 5 years, 3 months ago

It is not correct to assume ${0}^{0}=1$ because $0/0$ is not defined. If we think in terms of calculus, For example, $\lim _{ a\rightarrow 0 }{ (a/a) } =1$ because the value of $a$ tends to $0$ ,not equal to zero. This means its infinitely small but not equal to zero.

$Conclusion$: ${0}^{0}=1$is wrong.

- 5 years ago

There's a problem with that kinda thinking: 0/0 is not defined, but $\lim_{x \to 0} \frac{x}{x} = 1$. The limit of the fraction is defined, even though the fraction is not.

- 5 years ago

undefined

- 5 years, 5 months ago