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

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TopNewest"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. – Michael Mendrin · 1 year, 9 months ago

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\(2^0=\dfrac{2^1}{2}=1\)

\(0^0=\dfrac{0^1}{0}=\) undefined – Trevor Arashiro · 1 year, 9 months ago

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Or, \(\frac{0^2}{0} = 0\), so zero doesn't even exist! – Whitney Clark · 1 year, 9 months ago

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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. – Christopher Mowla · 1 year, 8 months agoLog in to reply

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. – Julian Poon · 1 year, 9 months ago

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

Since multiplying by zero, zero times is equivalent to not multiplying anything at all. – Raghav Vaidyanathan · 1 year, 9 months ago

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– Julian Poon · 1 year, 9 months ago

waaaat.... But by doing this you are assuming that \(0^{0}=1\) at the second line to the third line.Log in to reply

– Raghav Vaidyanathan · 1 year, 9 months ago

No, I think i just made the assumption that \(0^0=1\times 0^0\). Which seems right to me.Log in to reply

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– Whitney Clark · 1 year, 9 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.Log in to reply

– Agnishom Chattopadhyay · 1 year, 9 months ago

Well, he is trying to get a geometric interpretationLog in to reply

– Curtis Clement · 1 year, 9 months ago

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

– Agnishom Chattopadhyay · 1 year, 9 months ago

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

– Julian Poon · 1 year, 9 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.Log in to reply

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 \(0^0=1\) – Raghav Vaidyanathan · 1 year, 9 months ago

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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, \(\lim_{x \to 0} \sin(x)/x =1\) – Janardhanan Sivaramakrishnan · 1 year, 9 months ago

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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. – Janardhanan Sivaramakrishnan · 1 year, 9 months ago

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– Agnishom Chattopadhyay · 1 year, 9 months ago

Why do you want to work in this field?Log in to reply

The answer is here LOL

image

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– Calvin Lin Staff · 1 year, 9 months ago

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

– Krishna Sharma · 1 year, 9 months ago

That's why I posted thisLog in to reply

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. – Bobbym None · 1 year, 9 months ago

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– Agnishom Chattopadhyay · 1 year, 9 months ago

What does pappym think about disagreeing with one's own self?Log in to reply

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. – Jash Shah · 10 months, 3 weeks ago

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– Whitney Clark · 10 months, 3 weeks ago

It would be discontinuous at zero, I think.Log in to reply

– Jash Shah · 10 months, 3 weeks 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 linLog in to reply

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...... – Shubham D Man · 1 year, 7 months ago

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\(0^{0}\) 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. – Jeremy Bansil · 1 year, 9 months ago

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– Whitney Clark · 1 year, 8 months ago

You say \(0^0\) is undefined, AND anything to the power 0 is equal to 1? That itself sounds like a contradiction.Log in to reply

Great, the argument is heating up! I got the fastest 8 reshares ever – Agnishom Chattopadhyay · 1 year, 9 months ago

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@Agnishom Chattopadhyay You didn't... the TKC, 'twas... :-P – Satvik Golechha · 1 year, 8 months ago

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– Julian Poon · 1 year, 9 months ago

yay! Im the \(8^{th}\) reshare!Log in to reply

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. – Saarthak Marathe · 1 year, 4 months ago

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– Whitney Clark · 1 year, 4 months 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.Log in to reply

\[{ 2 }^{ -\infty }=0\\ 2={ 0 }^{ \frac { 1 }{ -\infty } }={ 0 }^{ 0 }\]

So this way \({ 0 }^{ 0 }\) can be anything – Archit Boobna · 1 year, 9 months ago

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– Agnishom Chattopadhyay · 1 year, 9 months ago

Division by infinity is not a valid operation. Infinity is not a real numberLog in to reply

– Archit Boobna · 1 year, 8 months ago

What if we put "n" instead of infinity there and write lim n->infinityLog in to reply

– Whitney Clark · 1 year, 8 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.Log in to reply

– Jeremy Bansil · 1 year, 9 months ago

If you're saying \(0^{0}\) can be anything, then it's undefined/indeterminate. That is that. Nothing can define itLog in to reply

– Archit Boobna · 1 year, 9 months ago

Yes, I'm saying that onlyLog in to reply

– Whitney Clark · 1 year, 7 months ago

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

– Whitney Clark · 1 year, 8 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.Log in to reply

\(\huge{\lim _{ x\rightarrow -\infty }{ { 2 }^{ x } } =0}\) – Arulx Z · 1 year, 8 months ago

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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) – Atul Antony Zachariahs · 1 year, 9 months ago

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– Whitney Clark · 1 year, 8 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.Log in to reply

undefined – Ramanathan K · 1 year, 9 months ago

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