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Now we know that:

  • A number raised to n : xⁿ = x * x * x * x ...n
  • A number raised to -n : x⁻ⁿ = 1/x * x * x * x ...n

But what would a root of a number (i.e. x¹/ⁿ) be?

What would "a number raised to 1/n" be?

Or a better way to interpret the question : How would you represent a number raised to 1/n?

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    Related: https://math.stackexchange.com/questions/132703/what-does-2x-really-mean-when-x-is-not-an-integer/133238 – Dan Sep 01 '22 at 15:50

2 Answers2

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Unlike for integers, where $x^n$ is the result of some manipulation, $x^{1/n}$ is defined as the solution of an equation. Namely, for positive $x$, we define $y := x^{1/n}$ as the unique positive solution of the equation $y^n = x$. $x^{1/n}$ is then a convenient abuse of notation because it is compatible with the power law that we are used to: $$(x^a)^b = x^{ab}.$$ Consequently, we can then define $x^{m/n}$ as $(x^m)^{1/n}$ (or $(x^{1/n})^m$, which conveniently results in the same) for $m,n \in \mathbb{Z}$. It is important here that you don't think of it as some sort of fractional multiplication of $x$ with itself, but rather as a convenient definition that happens to coincide with the usual powers for integers.

A similar concept would be something like $\log_{10}(2)$. It is defined as the unique solution of $10^x = 2$ and, a priori, not obtained by some sort of manipulation.

Klaus
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Ok, so what I've understood,

If you have a number x and it is raised to 1/n, then that number can be represented as yⁿ = x, where y ∈ R.

Which can be also written as logy(x) = n, where y is base and x is the number and n is just the number of times y must be multiplied by itself.

So if you want to find any number raised to 1/n, you have to find the base of the logarithm.

Now after scouring the web I have found a formula to find the base of a logarithm. Even so, the formula requires the use of a calculator which you might as well use that calculator to find the root of that number.

Now I know that the above definition is really generalized, but it's a good start to understand roots. I guess?

Now Klaus has posted an in-depth answer to this and Dan has posted a frankly encyclopedic answer. It goes really deep into this topic.

Huge thanks to Klaus and Dan.

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    Correct, but your answer does not add anything new to the one already given or to Dan's comment, so it is not clear what its purpose is. – Alex M. Sep 01 '22 at 17:01
  • @AlexM. Its not supposed to add anything, its just simplifying. Its a simple answer. For that reason I have not marked as the best one. – Amrit Sanjeev Sep 01 '22 at 17:05