3

What is a proof for the rule $$\sqrt[n]{a}=a^{\frac{1}{n}}$$

Is this just a definition or can it be proven? If just a definition, how come it just happens to match perfectly in the series of other types of exponents:

$$ a^4=aaaa\\ a^3=aaa\\ a^2=aa\\ a^1=a\\ \vdots\\ a^{1/2}=\sqrt[2]{a}\\ a^{1/3}=\sqrt[3]{a}\\ a^{1/4}=\sqrt[4]{a}\\ \vdots \\ a^0=1\\ a^{-1}=\frac 1a\\ a^{-2}=\frac 1{aa}\\ a^{-3}=\frac 1{aaa}\\ a^{-4}=\frac 1{aaaa}$$

Has the definition just been chosen to match this sequence, and to match all other rules regarding fractions and all other work in mathematics? Can it be proven that this definition matches those other definitions, or how do we know that it all fits together?

Steeven
  • 749
  • 4
    That is the definition of LHS. – Kavi Rama Murthy Aug 12 '19 at 09:02
  • 1
    @KaviRamaMurthy I would say it's the definition of the RHS. Depending on the education level we are aiming for, of course. – Arthur Aug 12 '19 at 09:05
  • There is no clear pattern to your list: on the right hand side the exponents are $3,2,1,2,3,4,0,1,2,3$. – Toby Mak Aug 12 '19 at 09:12
  • 1
    @TobyMak I have updated the list. What breaks the pattern is of course that the fraction-exponents are non-whole exponents, where all the others are integers that fit in a pattern. The non-whole exponents are thus "inserted" between the $a^1$ and $a^0$ (and more could be inserted between any other pair, as we can always make up more examples of fraction-exponents.) – Steeven Aug 12 '19 at 09:31

2 Answers2

9

There are several views to take on this. My preferred one, at least at an introductory level, is this:

Given that $a^n$ for a natural number $n$ means $a\cdot a\cdots a$, what is the most sensible meaning we could give to $a^{1/n}$?

Clearly it depends on what you mean by "sensible". What do we want from the expression $a^{1/n}$?

Natural number exponents have two important properties / rules: $$ a^{m+n} = a^m\cdot a^n\\ (a^m)^n = a^{mn}$$ The conventional "sensible" choice has been to define $a^{1/n}$ in a way that makes these rules still apply (coincidentally, that's also the reason for why $a^{-n}$ is defined the way it is).

So, we want $a^{1/n}$ to be a number such that the above rules still apply. For instance, we want $(a^{1/n})^n = a$. Well, that's the definition of $\sqrt[n]{a}$. So setting those to be the same is sensible.

Of course, we have to prove that things don't break in the process. Just because this was the only (positive) choice that allowed $(a^{1/n})^n = a$, that doesn't automatically mean that everything works out in the end. We still have to check that the exponent rules mentioned above still hold for any other possible choice of rational numbers $m$ and $n$ while using this definition, and that a single expression can't be given two values depending on how you calculate it, and so on (this is why using fractional exponents requires $a>0$: using negative bases and fractional exponents together does break things).

Arthur
  • 199,419
1

It depends on how you define the 'nth root'.

According to Wikipedia's definition, $r = \sqrt[n]{a}$ is a real number that satisfies $r^n = a$. However, $(a^{1/n})^n = a$ as well due to the fact $(a^b)^c = a^{bc}$. Therefore $\sqrt[n]a = a^{1/n}$.

As an aside, the term 'root' comes from 'square root' (definition 3 in Lexico), which more formally speaking, is the $r$ needed for $r \cdot r = a$. The inverse of this function is exponentiation: the $a$ needed for $a \cdot a = r$.

So now we need to find a function where $f(1/x) = f^{-1} (x)$. The question of proving exponential functions are the only solutions to this equation is a question for another time. The most obvious way to start would be to define $e^x$ as the only function which satisfies $y'(x) = y(x)$ and $y(0) = 0$.

Toby Mak
  • 16,827