## Zero Exponent and Roots

The number with exponent zero is highly nonintuitive, and I have two ways of looking at it. Using the algebra of exponents we know that $a^ma^n$ is $a$ with exponent $m+n$. And, since $0=1-1=(1)+(-1)$, we can construct an alternative and equivalent form of $a^0$. Let $a$ and $n$ be fixed numbers, other than zero, then: $$a^0=a^{n+(-n)}=a^na^{-n}=\frac{a^n}{a^n}=1$$

The second way of deriving what the zero exponent gives you, is by using an exponent that one can make arbitrarily close to zero, along with theorems governing their characteristics. From inspecting the quantity of $f(x)=1/x$ for large $x$ we use first the algebraic tool to make $1/x$ arbitrarily small (like zero), and secondly we recognize this as the $n$'th root where the $1/x$ is in the exponent position of a (positive, for simplicity) number $a$, where $x\rightarrow n$. $$b=a^{1/n}$$ $b$ is the root of $a$, and the name of the construction is from the very first scholars of the exponents (16th century). Roots and their trends will be studied by argued statement and graphic presentation. $b$ approximates $a^0$ for large $n$, which is argued by the following:

1. The roots are magnitudinally ordered according to $n$. Which is to say $b=b(n)$ is the function $b$ of $n$, where $a$ is implicitly constant, and that for a given $n=n_1$, we can say $b(n_1)$ is either greater or lesser than another $b(n_2)$ without calculating either the $n_1$'th or the $n_2$'th root.
2. The roots $b(n)$ are monotonic on either side of $1$. Which is to say the curving value $b(n)$ goes in one direction, at some variable rate (without specificity constraining the rate of rate), depending upon the mathematical order of $a$ and $1$.
To be demonstrated is that in the large $n$ limit, $b$ gets arbitrarily (limited only by Infinity) close to $1$.

The formula for $b$ is known as the $n$'th root of $a$, where $b$ has the exponent of inverse $n$ ($1$ in the numerator and $n$ in the denominator of the exponent). To calculate a root, $b$, the only test of verification is that $b$ is some number which when multiplied by $n$ factors of itself, equals $a$: $$\left( b \right)^n = \left( b \right) \left( b \right)\ldots\left( b \right) = \left(a^{1/n}\right)^n = a^{n/n} = a^1 = a$$ Where the ellipsis substitutes $(b)^{n-3}$ terms in the product, since $3$ are displayed inline (assuming $n\ge 3$).

For example, if we choose $n=5$, and $a=10$, then the (fifth) root of $10$ is given by a calculator to be about (rounded) $b=1.5849$. And, $b^5$ = $10.0002$. If we don't round the number and truncate the fraction to $b=1.5848$ then $b^5$=$9.9970$. In practice, it never comes out perfect ($b^n\approx a$) because the root ($b$) is an irrational number (a.k.a. a "surd", ibid.) meaning no truncated number will be exact and complete to serve as $b$.

Before going on with roots, we need to be familiar with number order. When two numbers are multiplied you can compare the product, $p$, and a factor, $x$, and say something about its coefficient, $c$. If the coefficient ($c$) of a factor ($x$) is compared to the unit number 1, it can be said immediately what the order of the product and the factor will be, without doing multiplication. $$p=c x$$ Let $c$ be the coefficient to test.

coefficient order product order
$c \lt 1$ $p \lt x$
$c = 1$ $p = x$
$c \gt 1$ $p \gt x$

Table 1: product ordering; ordering with three conditions respecting unity.

This magnitude ordering is posited axiomatically and will not be proven.

There are three conditions pertaining to the order of $a$ and $1$ which determine three orderings of $b$, which will be proven.

$a$ $n$'th root
$a \lt 1$ $a \lt b \lt 1$
$a = 1$ $b = 1$
$a \gt 1$ $a \gt b \gt 1$

Table 2. $n$'th roots ordering; ordering with three conditions regarding $1$.

Using only the coefficient ordering table above, posited axiomatically, and the definition of $n$'th root, we shall prove this table.

First row, if $a$ is less than one (and greater than zero, for simplicity) then the $n$'th root is less than one: $b\lt 1$, because if $b\ge 1$, then $bb\gt b$, by product ordering, and by repeating the product ordering rule $n-1$ times, we have $b^n\gt 1$ which contradicts $a\lt 1$. Also, $b\gt a$ if $a\lt 1$, because $b\lt 1$ means $$b^n\lt b^{n-1}\lt\dots\lt b$$ where we are using the first table of product ordering, as $c\rightarrow b$ and $x\rightarrow b$, meaning the relation between the product and factor is $bb\lt b$, and then repeating for $x\rightarrow b^2$ to get $(b)(b^2)\lt b^2$, and so on, leaving $a=b^n \lt b$.

For the second row of table 2, if $a$ is one then the $n$'th root is one, because if $b\lt 1$ then $b^n\lt 1$, and if $b\gt 1$ then $b^n\gt 1$.

Third row, if $a$ is greater than one (like $11$), then any root of $a$ is greater than one, too, because if $b\le 1$ then $b^n\lt b$, but since $a=b^n$ $b^n\gt 1$, so $b\gt 1$ if $a\gt 1$. Next, if $a\gt 1$ then $b$ is less than $a$, because since $b\ge 1$ then $b^n\gt b^{n-1}\gt\dots\gt b$ (separated into $cx$ form, product ordering for $c\gt 1$), and so $a=b^n\gt b$.

Further, regardless of where $a$ lies (somewhere between zero and infinity), the square root ($a^{1/2}$) is closer to one than $a$, so the fourth root ($(a^{1/2})^{1/2}$) is closer to one than $a^{1/2}$, so that the larger $n$ the closer to one. $$1 \le \left(\left(a^{1/2}\right)^{1/2}\right)^{1/2} \lt \left(a^{1/2}\right)^{1/2} \lt a^{1/2} \lt a$$ Which is obtained by using the ordering rule for the square root ($n=2$), when $a\gt 1$, repeatedly.

To demonstrate this graphically, we'll let $a=x$ and look at the curves of $b(a)=a^{1/n}$ for $n=\{2, 4, 8\}$.

Plot of the square, fourth, and eighth roots ($b(2)$, $b(4)$, and $b(8)$), over [1, 10]. The trend is to move closer/down to the flat line of $b=1$ with increasing root number $n$.

Plot of the square, fourth, and eighth roots, over [0, 1]. For $a\lt 1$, the higher root is closer to the flat line above the curves at $b=1$.

In the case $n$ goes to infinity (which is the desired end, $0=1/n$) the infinite root is just $1$, for any base number $a$.

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