[Debate] The Infamous Magic of Zero

"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.

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.

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 $0^0=1$

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, $\lim_{x \to 0} \sin(x)/x =1$

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.

The answer is here LOL

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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.

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

$0^{0}$ 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.

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......

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.

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)

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

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

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.

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