# Property of real exponent

I'm trying to prove that even for real exponent we have that $a^{x_1+x_2}=a^{x_1}a^{x_2}$, for every $x_1,x_2\in\mathbb{R}$ and $a>0$.

In other words, I have to show this:

$$\left(\lim_{\mathbb{Q}\ni r_1\to x_1}a^{r_1}\right) \cdot \left(\lim_{\mathbb{Q}\ni r_2\to x_2}a^{r_2}\right)=\left(\lim_{\mathbb{Q}\ni r\to x_1+x_2}a^{r}\right)$$

So, considering the left member, I can say that, given $\epsilon>0$, there exist $\delta_1$ and $\delta_2$ such that:

$$\left\{\begin{matrix} \left | r_1-x_1 \right |<\delta_1\Rightarrow \left | a^{x_1}-a^{r_1}\right |<\epsilon \\ \left | r_2-x_2 \right |<\delta_2\Rightarrow \left | a^{x_2}-a^{r_2}\right |<\epsilon \end{matrix}\right.$$

Multiplying the first inequality by $|a^{x_2}|$ and the second by $|a^{r_1}|$, I can assert that for any $\epsilon>0$ exists $\delta$ ($\delta=\min\{\delta_1,\delta_2\}$) such that:

$$\left | a^{x_1}a^{x_2}-a^{r_1+r_2} \right |<\epsilon\left ( |a^{r_1}|+|a^{x_2}| \right )$$

for any $r_1+r_2$ in a deleted neighborhood (in $\mathbb{Q}$) of $x_1+x_2$ with radius $\delta$.

I would terminate here the proof, but the quantity $\epsilon\left ( |a^{r_1}|+|a^{x_2}| \right )$ has a $r_1$ dependence. I can't remove it.

Does someone have an hint?

• What exactly do you mean by limit. Do you mean supremum? – Don Thousand Sep 10 '18 at 14:37
• With $\lim_{\mathbb{Q}\ni r \to x}f(r)=A$ I mean that for every $\epsilon>0$ there exists a $\delta>0$ such that $|r-x|<\delta$ ($r \in \mathbb{Q}$) $\Rightarrow |f(r)-A|<\epsilon$. The classical definition of limit of a function. – Nameless Sep 10 '18 at 14:40
• I mean sure, Real Numbers, at least in the Dedekind construction, give you this off the bat... – Don Thousand Sep 10 '18 at 14:41
• But you need to show more than simply the limit. You need to show that the supremums are equal – Don Thousand Sep 10 '18 at 14:41
• No, I already proved that $\lim_{\mathbb{Q}\ni r\to x}a^r=a^x:=\sup_{r<x}a^r=\inf_{r>x}a^r$. Now, I want to prove the exponent property for the product. – Nameless Sep 10 '18 at 14:46

Copy the demonstration of the limit of a product of functions : $$\left|a^x-a^{x_1}a^{x_2}\right| \le \left|a^x-a^{r_1+r_2}\right| + \left|a^{r_1+r_2}-a^{x_1}a^{x_2}\right|$$ As you claimed, you know how to get the first term lower than any $\epsilon>0$. Problem is the second term. \begin{align}\left|a^{r_1+r_2}-a^{x_1}a^{x_2}\right| &= \left|a^{r_1}a^{r_2}-a^{x_1}a^{x_2}\right| \\ &= \left|a^{r_1}a^{r_2}-a^{x_1}a^{r_2}+a^{x_1}a^{r_2}-a^{x_1}a^{x_2}\right| \\ &\le a^{r_2}\left|a^{r_1}-a^{x_1}\right| + a^{x_1}\left|a^{r_2}-a^{x_2}\right| \end{align} Now all you have to do is note $a^{x_1}$ is fixed, and as $a^{r_2}$ has limit $a^{x_2}$, it is bounded.
• So, as we chose $\delta=\min\{\delta_1,\delta_2\}$, then in my last inequality I can say that: $$\left | a^{x_1}a^{x_2}-a^{r_1+r_2} \right |<\epsilon\left ( |a^{r_1}|+|a^{x_2}| \right )<\epsilon\left ( a^{x_1}+\epsilon+|a^{x_2}| \right )$$ Right? – Nameless Sep 11 '18 at 7:06
• I would say $M\epsilon+a^{x_1}\epsilon$, where $M$ is an upper bound for $a^{r_2}$. – Nicolas FRANCOIS Sep 11 '18 at 10:00
• That is the same, because when $|r_1-x_1|<\delta_1$ we have: $$|a^{x_1}-a^{r_1}|<\epsilon \Rightarrow |a^{r_1}|=a^{r_1}<a^{x_1}+\epsilon$$ and so $M=a^{x_1}+\epsilon$. No? – Nameless Sep 11 '18 at 10:33