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How does one prove that $$ a^{n+m} = a^{n} a^{m} $$ for $a,n,m \in \mathbb R$.

I can prove this for integers but how can one prove this for real numbers?

gebruiker
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Ole Petersen
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    What is your definition of $a^n$ for non-integer $n$? – Chappers Jan 08 '16 at 13:19
  • Could it be $e^{n\log a}$... but then what of $a\leq 0$... – pshmath0 Jan 08 '16 at 13:25
  • Then my question would be: How can we prove that $e^{n+m}=e^{n}e^{m}$ – Ole Petersen Jan 08 '16 at 13:29
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    It will depend on what you are taking as given. You can easily prove this for rationals, and if you assume continuity, you can use the denseness of rationals in the reals to extend it to the reals. – MPW Jan 08 '16 at 13:30
  • Your statement is not true in general, because $a^{n}$ is not in general well-defined. You need either $a > 0$ or $n,m \in \mathbb Z$. – TonyK Jan 08 '16 at 13:33

3 Answers3

3

You need $a > 0$ for this to make sense.

  1. For $n, m$ integers, by induction
  2. For rational $n, m$, use the definition that $a^{n/m} = \sqrt[m]{a^n}$, and reduce to case (1)
  3. Extend to all real $n, m$ by continuity
vonbrand
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2

Hint:

If you meant for $a>0$, then

$$a^{n+m}=\exp\left[(n+m)\ln{a}\right]$$

Now use some knowledge of $\log$ to finish proving.

1

Define

$$F(z) := x^{z+a}x^{-z}x^{-a} \qquad a,x,z \in \Bbb R$$

and assume $x > 0$. Then

$$F'(z) = 0$$

so $F(z)$ is a constant function, which means $F(z) = F(0)$ for all $z$. Now set $a = 0$ and we get

$$F(z) = x^z x^{-z} = F(0) = x^0 x^0 = 1 \iff x^z x^{-z} = 1 \iff x^{-z} = \frac{1}{x^z} \qquad (1)$$

Furthermore, we encounter

$$F(z) = x^{z+a}x^{-z}x^{-a} = F(0) = x^a x^{-a} \underset{(1)}{=} 1$$

Hence

$$F(z) = x^{z+a}x^{-z}x^{-a} = 1$$

which is

$$x^{z+a} = x^z x^a$$

root
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