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Let x be any positive real number, and m, n be real numbers. Is it true that $x^{m+n} = x^m \times x^n$? If so, how do you argue this? What other definitions are required save for $x^0 = 1$ and $x^{m+1} = x \times x^m$, with m being a real number?

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Anirban Mandal
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    We define $\forall (x,n) \in \mathbb{R}^2, x>0,x^n=\exp(n\ln(x))$, which extends the definition for integers. So yes, it is true – Vincent Dec 02 '16 at 12:03

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Yes it's true. For defining the power $x^m$ with general $m \in \mathbf R$ and $x \in (0,\infty)$, there are two possibilities:

  1. The direct way (using some definition of $\exp$ and $\log$ that does not use non-integral powers, e.g. the power series): We define $$ x^m := \exp(m\log x) $$ then the proof uses the known properties of $\exp$ and $\log$, giving $$ x^m x^n = \exp(m\log x)\exp(n\log x) = \exp\bigl(m\log x + n \log x\bigr) = \exp\bigl((m+n)\log x\bigr) = x^{n+m} $$ Doing it this way we have to show that for integral $n\in \mathbf N$, we get the same $x^n$ as with the classical inductive definition (this is a simple induction).

  2. The "step by step"-way. We extend our definition of $x^m$ along with the equation from $m \in \mathbf N$ over $m \in \mathbf Z$ and $m \in \mathbf Q$ to $m \in \mathbf R$.

    • From $\mathbf N$ to $\mathbf Z$: For negative $m \in \mathbf Z \setminus \mathbf N$ define $$ x^{m} := \frac 1{x^{-m}} $$ Then (just consider all possible signs of $m$ and $n$): $x^{m+n} = x^mx^n$, for all $m,n \in \mathbf Z$.
    • From $\mathbf Z$ to $\mathbf Q$: For $m \in \mathbf Q$, write $m = \frac pq$ with $p \in \mathbf Z$ and $q \in \mathbf N^\times$. Define $$ x^{m} = x^{p/q} := \sqrt[q]{x^p} $$ Then, we have for $m,n \in \mathbf Q$, $m =\frac pq$, $n = \frac st$, that \begin{align*} x^{m+n} &= x^{\frac{pt + qs}{qt}}\\ &= \sqrt[qt]{x^{pt+qs}}\\ &= \sqrt[qt]{x^{pt}x^{qs}}\\ &= \sqrt[qt]{x^{pt}}\sqrt[qt]{x^{qs}}\\ &= \sqrt[q]{x^p}\sqrt[t]{x^s}\\ &= x^m x^n \end{align*}
    • From $\mathbf Q$ to $\mathbf R$: For $m \in \mathbf R$ and $x\ge 1$, define $$ x^m := \sup_{q\in \mathbf Q, q \le m} x^q $$ for $x \in (0,1)$, define $$ x^m := \inf_{q\in \mathbf Q, q \le m} x^q $$ Then $x^{m+n}= x^m x^n$ for all $m,n \in \mathbf R$ (that is straightforward using the definition and properties of the sup).
martini
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  • I agree with the second form but I have doubt about the first. At the first, are you not using what you have to prove? – Arnaldo Dec 02 '16 at 12:24
  • @ArnaldoNascimento It depends on the way you define $\exp$. As I wrote above, the cruical point is that in defining $\exp$, you must not use general powers. For example: If we define $\exp\colon \mathbf R \to \mathbf R$ as the one and only solution to $f'=f$, $f(0)=1$. Then one can show that $\exp$ is a group isomorphism $\exp \colon (\mathbf R, +) \to \bigl((0,\infty), \cdot\bigr)$. Define $\log := \exp^{-1}$. Then define the power by $x^m := \exp(m \log x)$ and prove the power law.
    If we now define $e := \exp(1)$, we have $$ e^x = \exp(x \log e) = \exp(x) $$ as a theorem.     – martini Dec 04 '16 at 07:58
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We fix $n$ and prove induction on $m$. When $m=0$, we have to show that $x^{n+0}=x^{n}\times x^{0}$. But the left hand side is $x^{n+0}=x^{n}$ while the right hand side is $x^{n}\times 1= x^{n}$, so we are done with $m=0$.

Now suppose inductively we have already proven that $x^{n+m}=x^{n}\times x^{m}$; we know wish to show that $x^{n+(m++)}=x^{n}\times x^{m++}$. The left hand side is $x^{(n+m)++}= x^{n+m}\times x$ by definition of exponentiation; by induction hypothesis this is equal to $(x^{n}x^{m})x$. Meanwhile the right hand side is $x^{n}\times x^{m++}= x^{n}(x^{m}x) $. The two sides are thus equal due to the associativity of multiplication. Hope it helps.

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Warning: very hand-wavy argument :D

1) Real number to the real power can be defined as the limit of a real number to the quotient power. If some law holds for quotient powers, it holds for real powers in the limit

$x^r = \lim_{\frac{a}{b} \rightarrow r} (x^{\frac{a}{b}}) $

2) Real number to quotient power is defined as taking the denominator order square root, followed by numerator order power (or vice versa).

$x^{\frac{a}{b}} = (\sqrt[b]{x})^a$

3) For integer powers, the additivity is simply justified by the count of arguments to be multiplied with itself. For the integer roots it is a little bit more tricky

$(\sqrt[b1]{x})^{a1} \cdot (\sqrt[b2]{x})^{a2} = \sqrt[b1b2]{((\sqrt[b1]{x})^{a1} \cdot (\sqrt[b2]{x})^{a2})^{b1b2}} = \sqrt[b1b2]{x^{a1b2} \cdot x^{a2b1}} = \sqrt[b1b2]{x^{a1b2+a2b1}} = x^{\frac{a1}{b1} + \frac{a2}{b2}}$

Aleksejs Fomins
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