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Definition of $Nth$ root:

$3rd$ order inverse group $1$ hyperoperation.

Division is how many times you can subtract a certain divisor from the dividend before it becomes negative.

Likewise Nth root is the result of repeated division by a certain divisor before it becomes $1$ or a decimal. The number of times you divide it before it becomes a decimal is the index.

Ex: $\sqrt [3]{8} = (8/2)/2$

Is the zeroth root even defined and if so what is $\sqrt [0]{x}$

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  • $\begingroup$ $\sqrt[0]{x}$ would be the answer to "Which number, when raised to the zeroth power, becomes $x$?". This is not an answerable question, in any practical interpretation of the word "answerable". $\endgroup$ – Arthur Oct 7 '14 at 21:25
  • $\begingroup$ Actually 1 would satisfy this because 1 to any power is 1. So there is a 0th root of 1. $\endgroup$ – Caters Oct 7 '14 at 22:04
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    $\begingroup$ @caters So would every other number... $\endgroup$ – Ali Caglayan Oct 7 '14 at 22:54
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The $n^\text{th}$ root of a real number $x$ is $$x^{1/n}$$

If $n=0$ then $1/0$ is undefined, so there is no such thing as the $0^{\text{th}}$ root.

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The $n$-th root is the inverse of the operation $x \mapsto x^n$. (Let's take $x>0$ to avoid complications.)

For $n=0$, the operation $x \mapsto x^0$ is not invertible, since it is a constant function.

So, there is no $0$-th root operation.

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No this is usually not defined. One definition of the $n$th root is $$ \sqrt[n]{x} = x^{\frac{1}{n}}. $$ So for a fixed $x>1$ you see that $$ \lim_{n\to 0^+} x^{\frac{1}{n}} = \infty. $$

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  • $\begingroup$ and there is also the definition of an nth root being repeated division because that is really what you are doing when taking the nth root is you are dividing repeatedly until your next division will lead to 1 or a decimal. Now of course this definition has limited divisors for any x. The divisors that will satisfy this are the factors(prime and composite both) of x. $\endgroup$ – Caters Oct 7 '14 at 21:27
  • $\begingroup$ @caters : Taking $n^{\text{th}}$ roots has nothing to do with repeated divisions... I don't know where you get this. $\endgroup$ – Patrick Da Silva Oct 7 '14 at 21:30
  • $\begingroup$ Wikipedia page about hyperoperations says that group 1 inverse hyperoperations are like this: Subtraction, Division, Nth root, Super root $\endgroup$ – Caters Oct 7 '14 at 21:34
  • $\begingroup$ Now according to this division is repeated subtraction, Nth root is repeated division, and Super root is repeated Nth root since a hyperoperation is an operation that can be or is defined as repeating the previous operation x times. $\endgroup$ – Caters Oct 7 '14 at 21:35
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If we reform this, it could be represented as $X^{\frac{1}{0}}$, so when used in standard functional mathematics, it would be undefined, because $\frac{1}{0}$ is undefined.

However, if used in the calculation of a limit and the function reduces to this form after substitution, I'd think this would be considered an indeterminate form, because not enough information is given to determine the original limit. It could be any number. It would end up then looking like $1^\infty$, which would itself be an indeterminate form.

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Hint: $$\sqrt[n]{x}=x^{1/n},$$ if $n=0$ then ...

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  • $\begingroup$ x^(1/0) is undefined because dividing by 0 itself is undefined $\endgroup$ – Caters Oct 7 '14 at 22:02
  • $\begingroup$ @caters Therefore $\sqrt[0]{x}$ isn't defined. $\endgroup$ – Hakim Oct 7 '14 at 22:18
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We can look for the value of zeroth root of 2 using limits.

Using trusty WolframAlpha to do this for us we arrive at the conclusion that the limit of 2^(1\n) as n approaches 0 is either 0 or infinity depending on the direction from which we arrive to 0 which means that the zeroth root is undefined.

Wolfram

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  • $\begingroup$ If someone can help me with the math code here, it would be great. $\endgroup$ – Cathier Dec 7 '16 at 8:57

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