The validity of the index laws for real numbers

January 28, 2014

The validity of the index laws for real numbers

Filed under: mathematics — ckrao @ 12:51 pm

The fundamental index laws are given by

$\begin{aligned} x^a x^b &= x^{a + b}\quad\quad &(1)\\(x^a)^b &= x^{ab}\quad \quad &(2)\\(xy)^a &= x^a y^a\quad\quad &(3)\end{aligned}$

If these are true we can also deduce

$\begin{aligned} x^0 &= 1 \ \text{ for }x \neq 0\quad \text{(setting } b = 0\text{ in (1))}\quad\quad\quad\quad&(4)\\1^a &= 1 \ \text{ for real numbers }a \quad \text{(setting } y = 1\text{ in (3))}\quad\quad\quad\quad&(5)\\x^{-a} &= 1/x^{a}\quad\text{(setting } b=-a \text{ in (1) and using (4))}\quad\quad&(6)\\x^{a}/x^{b} &= x^{a-b}\quad\text{(replacing } b \text{ with }-b \text{ in (1) and using (6))}\quad\quad&(7)\\(x/y)^a &= x^a/y^a \ \text{ where }y \neq 0\quad\text{(replacing }y \text{ with }1/y\text{ in (3))}\quad\quad&(8)\end{aligned}$

In this post we discuss the following conditions under which these laws hold.

Case 1: The laws are true if $x$ and $y$ are positive real numbers and $a$ , $b$ are real numbers.

Case 2: If $x$ or $y$ is negative we require $a$ and $b$ to be rational with odd denominator for $x^a$ or $x^b$ to be defined (this includes $a$ or $b$ being integers as the denominator in this case is 1). All the laws except (2) hold in this case. For (2) to be true we require the denominators of $a$ and $ab$ in reduced form to be odd and that either (a) the numerators and denominators of $a$ and $b$ are all odd or (b) the numerator of $ab$ in reduced form is even.
Case 3: The laws are mostly true if $x = 0$ , except $a,b$ need to be positive and laws (6), (7) do not apply.

Case 1: Positive bases

Let us seek to define $x^a$ from first principles for arbitrary $x$ positive and $a$ real. We can be motivated by the continuity of the function $x^t$ as a function of $t$ (where $x$ is fixed) and think of $x^a$ as the limit of values $x^r$ as rational numbers $r$ approach $a$ . To me it is easier to proceed via the logarithm function which may be defined as

$\displaystyle \log(x) := \int_1^x \frac{1}{t}\ \text{d}t\quad x > 0.\quad\quad(9)$

Being the area under a continuous positive-valued function, this function is continuous and monotonically increasing. It is also apparent that $\log(1) = 0$ . From the definition the important identity $\log (xy) = \log(x) + \log(y)$ (where $x, y > 0$ ) can be derived via a change of variable $u=t/x$ as follows:

$\begin{aligned} \log (xy) &= \int_1^{xy} \frac{1}{t}\ \text{d}t\\&=\int_1^{x} \frac{1}{t}\ \text{d}t + \int_x^{xy} \frac{1}{t}\ \text{d}t\\&= \int_1^{x} \frac{1}{t}\ \text{d}t + \int_1^{y} \frac{1}{ux}x\ \text{d}u\\&= \int_1^{x} \frac{1}{t}\ \text{d}t + \int_1^{y} \frac{1}{u}\ \text{d}u\\&=\log(x) + \log(y). \quad \quad (10)\end{aligned}$

By repeated use of this rule, $\log(x^n) = n \log(x)$ for positive integers $n$ . As $\log(2) > 0$ , $\log(2^n) = n \log 2$ grows without bound as $n$ increases, which shows that the log function is unbounded above. It is also unbounded below due to the relationship $\log(1/x) = -\log(x)$ (which follows from $0 = \log (1) = \log(x.(1/x)) = \log x + \log (1/x)$ ). Hence the log function is monotonic, continuous and has range $(-\infty, \infty)$ . It thus has a continuous inverse which is how we may define the exponential function: $\exp(x)$ is the unique positive value $y$ satisfying

$\displaystyle x = \int_1^y \frac{1}{t}\ \text{d}t.$

From the relationship $\log(uv) = \log(u) + \log(v)$ follows

$\exp(x+y) = \exp(x) \exp(y) \quad\quad(11)$

where $x$ and $y$ are real. Repeated use of this gives

$\displaystyle \exp(nx) = \exp(x)^n \quad \quad (12)$

for $n$ a positive integer. Defining $e:= \exp(1)$ , we have the familiar form $\exp(n) = e^n$ .

We can extend (12) to rational values of $n$ . Let us recall what we mean by a rational power. For $a > 0$ and relatively prime positive integers $p, q$ we define $a^{p/q}$ to be the unique positive number $y$ such that $y^q = a^p$ . (That the $q$ ‘th root exists follows from the fact that the function $x^q$ has an inverse for $x > 0$ ).

From (12) we then have

$\displaystyle (\exp(x))^p = \exp(px) = \exp((px/q).q) = \exp(px/q)^q.$

Setting $\exp(x)$ to $a$ and $\exp(px/q)$ to $y$ we have thus shown $a^p = y^q$ , so $\exp(x)^{p/q} = \exp(px/q)$ . Therefore (12) is satisfied when $n$ is positive rational. Finally we use $\exp(-x) = 1/\exp(x)$ (which can be proved using (11)) to extend (12) to negative rationals:

$\displaystyle \exp(nx) = 1/\exp(-nx) = 1/\exp(x)^{-n} = \exp(x)^n.$

We are finally ready to define $x^a$ for arbitrary $x$ positive and $a$ real relying on continuity. We simply extend (12) replacing $\exp(x)$ with $x$ (also replacing $x$ with $\log(x)$ ) and $n$ with real values $a$ :

$\displaystyle x^a := \exp(a \log(x)), \quad x > 0, a \in \mathbb{R}.\quad\quad(13)$

From this definition it is quick to verify laws (1) and (2):

$\begin{aligned} x^a x^b &= \exp(a \log(x))\exp(b \log (x))\\ &= \exp(a \log x + b\log x)\\ &= \exp((a+b)\log x)\\ &= x^{a+b}\quad\quad(14)\\\text{and }\ (x^a)^b &= \exp(b \log(x^a))\\ &= \exp(b a \log(x))\\ &= x^{ab}\quad\quad(15)\end{aligned}$

Law (3) is immediate if either $x$ or $y$ is equal to 1. Otherwise, we can find $c$ so that $y = x^c$ (choose $c = \log(y)/\log(x)$ ) and then we have from (1) and (2) the following:

$\begin{aligned} (xy)^a &= (x.x^c)^a\\ &= (x^{1+c})^a\\ &= x^{a + ac}\\ &= x^a x^{ac}\\ &= x^a (x^c)^a\\ &= x^a y^a.\quad\quad (16) \end{aligned}$

Case 2: Negative bases

When the base $x$ is negative we can define $x^a$ by the real number $(-1)^a(-x)^a$ provided $(-1)^a$ exists. This will be the case if $a$ is rational and has odd-valued denominator (square roots of -1 are not real-valued nor are irrational powers of -1). Writing $a = p/q$ where $p$ and $q$ are relatively prime ( $q$ odd), we then have $(-1)^{p/q} = (-1)^p$ . Since odd denominators are preserved under addition, law (1) can be shown to hold.

To check law (3) we consider the cases of both $x,y$ being negative or only one (say $y$ ) being negative.

a) If $x$ and $y$ are both negative:

$\begin{aligned} (xy)^{p/q} &= ((-x)(-y))^{p/q}\\&= (-x)^{p/q} (-y)^{p/q}\\ &= (-1)^{2p/q}(-x)^{p/q} (-y)^{p/q}\\&= (-1)^{p/q}(-x)^{p/q}(-1)^{p/q}(-y)^{p/q}\\&= x^{p/q}y^{p/q}\end{aligned}$

b) If $x$ is positive and $y$ is negative:

$\begin{aligned} (xy)^{p/q} &= (-1)^{p/q}(x(-y))^{p/q}\\&= (-1)^{p/q}(x)^{p/q} (-y)^{p/q}\\&= x^{p/q}y^{p/q}\end{aligned}$

However rule (2) is a little more complicated. For example, $(x^2)^{1/2} \neq x$ if $x <0$ so we cannot say

$\displaystyle 1 = 1^{1/2} = ((-1)^2)^{1/2} = (-1)^{2/2} = (-1)^1 = -1$ .

Issues arise when even denominators in the exponents appear because positive square roots will be taken when negative numbers may be required. It is also possible that $(x^a)^b = x^{ab}$ but for $x^b$ not to be real-valued (e.g. $(x^4)^{1/2} = x^2$ but $x^{1/2}$ is not real-valued).

For $(x^a)^b = x^{ab}$ to make sense when $x < 0$ we require that the left and right sides are defined and are equal in the equality $((-1)^a)^b = (-1)^{ab}$ . Hence:

for the left side to exist, $a$ must have odd denominator and if $(-1)^a <0$ then $b$ must also have odd denominator
for the right side to exist, $ab$ in reduced form must have odd denominator

Also:

(a) the left side is negative iff $a,b$ both have odd numerators and denominators and the same is true for the right side.
(b) the left side is positive iff either of $a,b$ has even numerator and the right side is positive iff $ab$ has even numerator and odd denominator in reduced form.

We conclude that (2) holds when either (a) the numerators and denominators of $a$ and $b$ are all odd (in which case the result is negative) or (b) the product $ab$ has even numerator and odd denominator in reduced form (in which case the product is positive).

This allows the possibility of an even numerator cropping up such as $(x^{2/3})^3 = x^2$ . It is interesting to see that $(x^{4/3})^{1/2} = x^{2/3}$ is valid but $(x^{2/3})^{1/2} = x^{1/3}$ is not for $x < 0$ .

Also, reduced form (cancelling out even factors) is important to avoid erroneous calculations such as $(-1)^{2/4} = ((-1)^2)^{1/4} = 1$ .

Finally we remark that for negative bases we lose continuity: for example $(-1)^q$ switches between $+1$ and $-1$ depending on whether the numerator of $q$ is odd or even.

Case 3: Base 0

Note that $0^a$ is $0$ for positive $a$ (again defined firstly for rationals then extended to positive reals by continuity). However it is undefined for $a$ negative which is why the laws only hold for positive indices and neither (5) nor (6) apply. In the discrete world it is common to define $0^0$ to be 1 for convenience but the function $x^y$ fails to be continuous at $(x,y) = (0,0)$ for any choice of value of $0^0$ .

Postscript

I’ll end this post with a cool-looking exponential of logarithm identity that is not as well known as it perhaps ought to be. For $x,y >0$ we have

$\displaystyle x^{\log y} = y^{\log x}.$

(The logarithm of both sides is $\log y \log x = \log x \log y$ !)

For example, $4^{\log_2 9} = 9^{\log_2 4} = 9^2 = 81$ .

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[…] an earlier post we saw that some of the index laws fail when the base is negative or zero. Now we shall see what […]

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January 28, 2014