# Definitive answer to existence of real exponents

Well, I've been searching through this fórum and I know this question has been answered many times. But the answers I see, are kinda circular (I think). Let's start by the natural case.

• Natural case

As the name says, this is natural to think. We can define $a^x \cdot a^y$ as $a^{x+y}$ because it is something we can observe and that is definitively true. When we think about the $0$ case for na expoent, and define it by being $1$ for every number except $0$ itself (not gonna enter in the $0^0$ discussion in this post) we can think in many ways. The best way to define this strongly as being equals to one, should be by taking the limit from both sides, but we don't have a definition yet, for a continuous function $f(x) = a^x$ so we're closed in the natural case. If we observe this behavior:

$$2^3 = 8$$ $$2^2 = 4$$ $$2^1 = 2$$ Then maybe $2^0$ should be $1$ because in the left side we're subtracting $1$ for the exponent, and in the right side we are dividing by the base of the exponent. So let's proceed to other definitions and see if we can still stay with the definition of the zero power.

• Integer case

If we think the same way, we can create a definition that works for both the natural case, and our new case: the integers. If we keep subtracting that table in the left side, we're gonna get to the negative case. So if we accept: $$2^0 = 1$$ And subtract 1 from the exponent on the left side, and divide by the base in the right side, we reach: $$2^{-1} = \frac{1}{2}$$ And if we continue, we'll see that we can define, for every integer exponent:

$$a^y = a\cdot a \cdots a \qquad\text{a times itself y times, for every y>0, y \in \mathbb Z }$$ $$a^y = 1 \qquad\text{for y = 0}$$ $$a^y = \frac{1}{a\cdot a \cdots a} \qquad\text{1 over a times itself y times, for every y<0, y \in \mathbb Z}$$

• Rational case Well, here's what I've been reading:

Define $a^{\frac{1}{n}} = c$ as the n-th root of $c$. So, for example, $a^{\frac{1}{2}} = \sqrt{a}$ This definition only makes sense to me if we ACCEPT that $$a^{b} \cdot a^{c} = a^{b+c}\tag {for b and c \in R}$$ then $$a^{\frac{1}{n}} = b$$ so $$(a^{\frac{1}{n}})^n = c^n$$ then $$a = c^n$$ As we can see, $a$ was the n-th root of $c$, but this only makes sense if we accept that the law for natural numbers hold for the rational case. We can, of course, 'take the limit' (we cannot take the limit yet, because it's not in the reals) of both sides and verify that by our acception, when the exponent $\frac{1}{n}$ gets closer to $1$ (in other words, $n$ gets closer to $1$) we get the number that multiplied by itself $1$ times is equal to itself, wich is just the number itself. In other words: $a^{\frac{1}{n}} = a$ when $n$ gets closer and closer to $1$. So it seems that we have good reasons to believe that the sum property holds for rational exponentes. But since math is highly rigorous we cannot ever accept this only because we have good reasons. Just because something don't makes sense under some sets, we cannot define it. So is there something that really proves that the sum rule rolds for rationals, and even real exponents? And what about imaginary and complex exponents? What can we say about it?

• Well, if you define $$e^x = \sum\limits_{n = 0}^{\infty} \frac{x^n}{n!}$$ and prove the rules of exponents from there, the concerns go away. Alternatively, you can define exponents with real numbers via a limit, based on the rational case. – user61527 Jan 2 '14 at 2:06
• @T.Bongers but I need to prove the rational case first – user108425 Jan 2 '14 at 2:10
• Why does the definition $a^{1/n} = \sqrt[n]{a}$ only make sense if we accept that $a^{b+c} = a^c \cdot b^c$ for $a$, $b$ rational? I didn't quite get that. – Ulrik Jan 2 '14 at 2:13
• @Svinepels I though i wanted to say that $a^{\frac{1}{n}}$ is the number that can be raised to the $n$ exponente. Don't know why I wrote this D; – user108425 Jan 2 '14 at 2:15

I'm not completely sure if I got your point, but here's how I would continue defining rational exponents.

Definition: For $r = m/n \in \mathbb{Q}$, define $a^r = (\sqrt[n]{a})^{m}$. This definition is easily seen to be compatible with the special case where $r$ is an integer.

We can now prove the following: $a^{r+s} = a^r a^s$ for $r,s \in \mathbb{Q}$ (1):

Let $r = m/n$ and $s = p/q$. Then $$a^{r+s} = a^{(mq+np)/nq} = \left( \sqrt[nq]{a} \right)^{mq+np} = \left( \sqrt[nq]{a} \right)^{mq} \left( \sqrt[nq]{a} \right)^{np} = a^{mq/nq}a^{nq/np} = a^{m/n}a^{q/p} = a^ra^s$$

As you can see, we didn't need to assume beforehand that (1) was true, but we could prove it right from the definition.

Extending the definition to real exponents is worse though...

• In order for $a^{m/n}$ to be well defined we need to prove that if $m/n=p/q$ then $a^{m/n}=a^{p/q}$. (This is not hard, of course). I would roll it into the definition of $a^{1/r}$, though, and say that when $a\ge 0$, $a^{m/n}$ is the unique nonnegative solution for $x$ to $x^n=a^m$. So roots would simply be an abbreviation for $a^{1/r}$. – Henning Makholm Jan 2 '14 at 3:30
• But isn't this the same as accepting that: ${(a^{\frac{1}{n}})}^n = a$? How can you prove it? – Lucas Zanella Jan 2 '14 at 3:39
• It can be proved that there exists a unique, positive solution to the equation $x^n = a$ for every $a > 0$. Since it is unique, we give it a name: $\sqrt[n]{a}$. And now, since we have defined $a^{1/n} = \sqrt[n]{a}$, the number $a^{1/n}$ is by definition solution to the equation $x^n = a$, i.e. $\left( a^{1/n} \right)^n = a$. – Ulrik Jan 2 '14 at 3:53
• @Svinepels what's the reason to say that $a^{1/n} = \sqrt[n]{a}$? – Lucas Zanella Jan 2 '14 at 16:56
• @LucasZanella What do you mean by "reason"? It certainly is a special case of our definition of $a^{m/n}$. – Ulrik Jan 2 '14 at 23:00

It's not a question of “belief” or “acceptance”, it's a matter of being self-consistent and self-coherent, inasmuch as the latter are characteristics of truth, of which mathematics is intended to be part of, as opposed to it being a useless and incoherent system of falsehoods and inconsistencies.