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Questions like this, which asks to solve $$x^{\frac43} = \frac{16}{81}$$

confuse me. The solution for real $x$ is $\pm \frac8{27}$.

The question would presumably be a bit different to solve

$$x^{\frac42} = \frac{16}{81}$$

Because on the one hand $\frac42 = 2$ so the answer is $\pm\sqrt{\frac{16}{81}} = \pm\frac49$.

But on the other hand $x^{\frac{4}{2}} = x^{{\frac12}\cdot{4}}$ which, as a function of $x$, is not equal to $(x^{\frac12})^4$.

This is fine, but it raises the question, what is the proper way to define $x^{\frac{4}{3}}$? We have seen we cannot just arbitrarily factor the exponent and apply the multiplication rule. But that's the only way I can think to define $x^{4/3}$ without first defining $x^a$ for all real numbers probably by way of $\ln$.

This tends to come up for and similar questions on the site. It might be harder than first appears.

More precisely I am asking how to define $x^a$ for as general real $x$ and $a$ as possible, in the context of precalculus, or how to define it at any level of mathematics and adapt it to a precalculus or at least $\mathbb R$ concept.

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    $\begingroup$ $x^{\frac42}$ actually is equal to $x^2$ no matter how you define it provided you allow complex numbers in the intermediate calculations. $\endgroup$ – Matt Samuel Feb 16 '16 at 2:27
  • $\begingroup$ @user21820 because I didn't ask that question. $\endgroup$ – djechlin Feb 16 '16 at 17:25
  • $\begingroup$ Lol I must have seen this question before the other one and mistaken you for the other asker. Sorry. $\endgroup$ – user21820 Feb 16 '16 at 17:38
  • $\begingroup$ See if my answer satisfies your curiosity. There is no real escape from the nitty gritty details if you want to stick to the real exponentiation. I frankly believe complex exponentiation is much nicer, as a rigorous definition and proof of the basic properties of complex exponentiation is much shorter than for real exponentiation. Basically it is because complex exponentiation can be defined via Taylor series that only uses natural powers of complex numbers. Tell me if you are interested to learn about complex exponentiation. $\endgroup$ – user21820 Feb 16 '16 at 17:43
  • $\begingroup$ math.stackexchange.com/questions/1570221/… $\endgroup$ – Simply Beautiful Art Feb 16 '16 at 18:56
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Let's start with rational exponents, and take as our starting point the fact that $$ x^a\cdot x^b = x^{a+b} $$ OK, now look what that says about $x^{\frac{1}{q}}$ with integer $q$: $$ x^{\frac{1}{q}} \cdot x^{\frac{1}{q}} \cdot \cdots \cdot x^{\frac{1}{q}} \text{ (q factors) } = x^{q\cdot \frac1q}=x $$ So that forces $$x^{\frac{1}{q}} = \sqrt[q]{x}$$ And then we could multiply an arbitrary number $p$ factors of $x^{\frac{1}{q}}$ to find $$x^{\frac{p}{q}} = \sqrt[q]{x^p}$$ Now if $q$ is even and $p$ is odd, this does not work out for negative $x$, and we will say that (working within the reals) $(-|x|)^{\frac{2k+1}{2m}}$ is undefined.

Now, taking $x>0$, we can define $x^r$ for $r\in\Bbb{R}$ using just the concepts of $\sup$ (minimum number that is not lesser than any member of a set) or $\inf$ ((maximum number that is not greater than any member of a set):

$$x^r = \sup\left( \{ x^{\frac{p}{q}} : p,q \in \Bbb{Z}^+ \wedge \frac{p}{q} \leq r \} \right) $$

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  • $\begingroup$ Doesn't $\sup$ count as real analysis? However… defining exponential functions for real numbers needs to use real analysis, right? $\endgroup$ – Cehhiro Feb 16 '16 at 2:23
  • $\begingroup$ @OFRBG certainly. I do appreciate that this is scaffolded off $\mathbb Q$ without building on say $\exp$ or $\ln$ or complex numbers, so this is about what I was looking for. $\endgroup$ – djechlin Feb 16 '16 at 2:31
  • $\begingroup$ @djechlin: While the "starting fact" is of course the desired behaviour, it is certainly not a definition, until we prove that it is consistent to have such a behaviour. If you look at my answer, we define the real $n$-th root so that it has the desired behaviour, namely that the $n$-fold product of the radical is the original. But the definition itself requires non-trivial justification. The reason I mention this is that otherwise we can do all sorts of weird things by assuming some behaviour to be consistent, like $1 = \sqrt{1} = \sqrt{(-1)(-1)} = \sqrt{-1} \sqrt{-1} = -1$. $\endgroup$ – user21820 Feb 16 '16 at 18:07
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There is an interesting way for defining powers with real exponents that just uses the logarithm and the exponential. The logarithm can be defined, for $x>0$, by $$ \log x=\int_{1}^x \frac{1}{t}\,dt $$ and it's an easy exercise showing that, for $x,y>0$, $$ \log(xy)=\log x+\log y $$ By an easy induction, we can see that, for a nonnegative integer $n$ and $x>0$, $$ \log(x^n)=n\log x, $$ and this can be easily extended to any integer $n$: if $n<0$, $$ 0=\log 1=\log(x^nx^{-n})=\log(x^n)+\log(x^{-n})= \log(x^n)+(-n)\log x $$ so we get $\log(x^n)=n\log x$.

Moreover $\log$ is an increasing differentiable function with $$ \lim_{x\to\infty}\log x=\infty,\qquad \lim_{x\to0}\log x=-\infty $$ so its inverse function $\exp$ is defined on the whole real line, taking on each positive value.

The main property of $\exp$ is, of course, $\exp(x+y)=\exp x\cdot \exp y$ (it's the same as the property of the logarithm).

Now we want to define, for $a>0$, a function $f_a$ such that, for integer $n$, $$ f_a(n)=a^n $$ It's very easy: $$ f_a(x)=\exp(x\log a) $$ Indeed, if $n$ is integer, $$ f_a(n)=\exp(n\log a)=\exp(\log(a^n))=a^n $$

Note that \begin{align} f_a(x+y) &=\exp\bigl((x+y)\log a\bigr)\\ &=\exp(x\log a+y\log a)\\ &=\exp(x\log a)\exp(y\log a)\\ &=f_a(x)f_a(y) \end{align}

What about $f_a(m/n)$, where $m$ and $n$ are integers, with $n>0$? We have $$ a^m=f_a(m)=f_a\left(n\frac{m}{n}\right)=(f_a(m/n))^n $$ so it makes sense to define $$ a^{m/n}=f_a(m/n) $$ In particular $a^{1/2}=\sqrt{a}$ and similarly for the other roots. Note that we have deduced the roots from the logarithm: there was no need to define them in advance. (Actually, in the foundations of real numbers, the square root is necessary, in order to prove the uniqueness of the real number system.)

Now we do the last step and set $a^x=f_a(x)$, which is not ambiguous when $x$ is an integer.

What's $e$? Simple: $e=\exp1$. Since $\log'1=1$ by the fundamental theorem of calculus, we have $$ \lim_{t\to0^+}\frac{\log(1+t)}{t}=1 $$ and so also $$ 1=\lim_{x\to\infty}x\log\left(1+\frac{1}{x}\right)= \lim_{x\to\infty}\log\left(1+\frac{1}{x}\right)^{\!x} $$ from which we deduce, by applying $\exp$, $$ e=\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{\!x} $$


One could extend this to negative bases when the exponent is rational and, when reduced at minimal terms, has odd denominator; if $p$ and $q$ are integers, with $q>0$ and odd, set $$ a^{p/q}=(a^p)^{1/q} $$ Note, however, that one has to be very careful in doing algebraic manipulations with rational exponents, if the base is allowed to be negative.

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  • $\begingroup$ The disadvantage of this approach is that you need integration to even start, and it is not at all intuitive in the sense that there is no motivation to define the logarithm as that particular integral nor why one ought to define $a^x = \exp(x \ln(a))$. I know that this is commonly done in some textbooks but they are devoid of motivation, similar to what my physics professor said: "Just follow instructions and don't ask why. Mathematics is just your slave." In contrast, it is more meaningful to define $\exp$ as motivated by the defining differential equation, and define $\ln$ as an 'inverse'. $\endgroup$ – user21820 Feb 17 '16 at 7:23
  • $\begingroup$ And the definition of complex exponentiation just falls into place naturally because we would like $a^b = \exp(\ln(a))^b = \exp(\ln(a)b)$ where the second equality is what we wish to have so that it is as if $\exp(x) = e^x$ for some real $e$ that follows similar rules to integer exponentiation. $\endgroup$ – user21820 Feb 17 '16 at 7:27
  • $\begingroup$ Also, I'm not sure what you mean by saying that the square-root is necessary for proving the uniqueness of the real number system. The standard second-order theory of the reals is categorical, and there is nothing in that axiomatization concerning square roots. Rather it is the second-order nature of the completeness axiom that forces the theory to be categorical, since any first-order theory with an infinite model has different models. $\endgroup$ – user21820 Feb 17 '16 at 7:38
  • $\begingroup$ @user21820 Of course I'm not thinking that this approach is good for ninth-graders, so your objections are void. How complex exponentiation should work is a completely different matter, so that's not a problem. Finally, the square root is necessary for proving that the reals have no non identity field automorphisms. $\endgroup$ – egreg Feb 17 '16 at 8:39
  • $\begingroup$ Hmm the asker did explicitly ask "in the context of precalculus", so since you say that you don't thinking this integral-based approach is suitable for high-school, then it's not really answering the question. That said, I'm not saying your answer doesn't belong, because I think it's good to see the various approaches and compare them. As for my comment, I was not referring to the level of mathematics education as the hurdle, but rather the absence of motivation. $\endgroup$ – user21820 Feb 17 '16 at 11:06
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Natural powers

Usually we define natural number powers first. This works in any structure with multiplication.

$x^0 = 1$ for any real $x$.

$x^{n+1} = x^n x$ for any real $x$ and natural number $n$.

Integer powers

Then we define integer powers. This works for all elements with multiplicative inverse. For real numbers every element has a multiplicative inverse except $0$.

$x^{-n} = \frac{1}{x^n}$ for any real $x \ne 0$ and natural number $n$.

Rational powers

Then we define $n$-th roots. This is where it starts to get hairy, because there are two natural (and commonly used) definitions that are completely incompatible. You have to know exactly which one is used in any given situation. $\def\rr{\mathbb{R}}$ $\def\qq{\mathbb{Q}}$ $\def\zz{\mathbb{Z}}$

In this answer I only describe real exponentiation.

$\sqrt[n]{x}$ is the real number $y > 0$ such that $y^n = x$, for any real $x > 0$ and even integer $n > 0$.

$\sqrt[n]{x}$ is the unique real number $y$ such that $y^n = x$, for any real $x > 0$ and odd integer $n > 0$.

The first is valid because you can prove that $( t \mapsto t^n )$ is strictly increasing on $\rr_{\ge 0}$ and its range is $\rr_{\ge 0}$.

The second is valid because you can prove that $( t \mapsto t^n )$ is strictly increasing on $\rr$ and its range is $\rr$.

After that we define rational powers.

$x^q = \sqrt[n]{x}^m$ for any real $x$ and rational $q$ where $m,n$ are coprime integers such that $n > 0$ and $q = \frac{m}{n}$, and we require $x > 0$ if $n$ is even.

Note that this is valid because firstly we have defined the components above, and secondly every rational number can be uniquely written as the ratio of coprime integers with the denominator being positive. If it is not unique our definition is meaningless until proven otherwise.

Now we need some basic properties, the first two of which are proven by induction. (There is no escape from induction!)

$x^m \le x^n$ for any real $x \ge 1$ and integers $m,n$ such that $m \le n$.

$x^m \ge x^n$ for any real $x$ such that $0 < x \le 1$ and integers $m,n$ such that $m \le n$.

$x^q \le x^r$ for any real $x \ge 1$ and rational $q,r$ such that $q \le r$.

$x^q \ge x^r$ for any real $x$ such that $0 < x \le 1$ and rational $q,r$ such that $q \le r$.

Real powers

Finally we define real exponentiation.

$x^y = \sup( \{ x^q : q \in \qq \land q \le y \} )$ for any real $x \ge 1$ and real $y \ge 0$.

$x^y = \sup( \{ x^q : q \in \qq \land q \ge y \} )$ for any real $x$ such that $0 < x \le 1$ and real $y \ge 0$.

$x^{-y} = \frac{1}{x^y}$ for any real $x > 0$ and real $y \ge 0$.

This is not so trivial, and to prove it valid we need to prove that the set involved has an upper bound in $\rr$. We need:

$x^q \le x^{ceil(y)}$ for any real $x \ge 1$ and real $y$ and rational $q < y$ where $ceil(y)$ is the smallest integer $z \ge y$.

$x^q \le x^{floor(y)}$ for any real $x$ such that $0 < x \le 1$ and real $y$ and rational $q < y$ where $floor(y)$ is the largest integer $z \le y$.

These can be proven by using the above-mentioned properties and carefully considering all the cases. Note that rational powers of $x$ have already been defined, so this is not circular. The only ingredient that is missing (did you notice it?) is that it was not proven that $ceil$ and $floor$ are validly defined. Indeed to verify them, we need to use the completeness axiom of the real numbers. Here is a rough sketch:

If we have a real $y \ge 0$ such that there is no integer $z \ge y$, then by the completeness axiom $\zz$ has a supremum $s$ since it contains $0$ and has upper bound $y$ in $\rr$. By property of supremum, since $s-1 < s$ we can find an integer $k > s-1$ such that $k \in \zz$. But then $k+1 \in \zz$ and $k+1 > s$, contradicting the definition of $s$.

Therefore for any real $y \ge 0$ we have some integer $z \ge y$. And now we use induction to prove that there is a smallest such integer. This gives us $ceil(y)$. Likewise for $y \le 0$, and similarly for $floor$.

Thus we are done defining real powers. It remains a lot of tedious work to prove even basic properties of real exponentiation.

Comments

Notice that we did not define general real powers of non-positive real numbers, because there is no real meaningful way to do so.

I used $ceil$ and $floor$ for ease of reading the first time. Standard notation is $\lceil x \rceil = ceil(x)$ and $\lfloor x \rfloor = floor(x)$.

Details of all the above definitions and proofs should be carefully worked out, and ought to be present in any decent real analysis textbook.

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Here is a possible definition at the precalculus level, which relies on a theorem I won't prove.

Let $a > 0$. Then there is a unique function $f_a \colon \mathbb{R} \to (0,+\infty)$ which satisfies the following three conditions:

  • $f_a(1) = a$;
  • $f_a(x+y) = f_a(x) f_a(y)$ for all $x, y \in \mathbb{R}$;
  • $f_a$ is monotonic (i.e., monotonically increasing or decreasing).

We define $a^x = f_a(x)$.

One possible proof (not at the precalculus level) is to observe that $(\mathbb{R},+)$ and $((0,+\infty),\times)$ are both complete Archimedean groups. Therefore, if $a > 1$, there is a unique order isomorphism $f_a$ between them under which $1$ corresponds to $a$. For $0 < a < 1$, write $f_a(x) = f_{1/a}(-x)$.

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