For all $ a, b \text{ and } c \in \mathbb{R}$ and $a>1$, Prove that $a^b\cdot a^c=a^{b+c}$

I have come across this question and its bugging me. Its a basic property that we learn in HS and I was hoping someone can enlighten me

  • $\begingroup$ How do you define this when $b$ and $c$ are not rational? $\endgroup$ – Javier Jul 4 '13 at 2:17
  • $\begingroup$ I have no clue @JavierBadia $\endgroup$ – math101 Jul 4 '13 at 2:19
  • $\begingroup$ How would you prove this if b and c were rational? $\endgroup$ – math101 Jul 4 '13 at 2:19
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    $\begingroup$ The details depend very much on how the exponential function is defined. You will almost always see $a^x$ defined as $\exp(x\ln a)$. Here we are using $\exp(t)$ to mean $e^t$. But the exponential function $\exp$ itself may be defined by a power series, or as the solution to a DE, or as the inverse of $\ln$, where $\ln$ is defined as being a certain integral. $\endgroup$ – André Nicolas Jul 4 '13 at 2:21
  • $\begingroup$ Well can you give me a proof based on one of these definitions of the exponential function? $\endgroup$ – math101 Jul 4 '13 at 2:35

On way of doing this is to define $a^x$ for real $x$ using the least upper bound property of the real numbers.

$$ a^x = \text{ l.u.b of } \lbrace a^t \mid t< x, t\in \mathbb{Q} \rbrace $$

This means we consider the values for all rational powers which are less than $x$ and define the result of raising to the $x$ power as the smallest real number that is greater than or equal to the elements of this set.

To prove the product rule then we can first look at the meaning of $a^xa^y$ when $x,y\in \mathbb{R}$. Consider the following set,

$$ B= \lbrace a^r a^t \mid r < x, t < y, r \in \mathbb{Q}, t \in \mathbb{Q} \rbrace $$

We can conclude the following,

$$ a^r < a^x \text{ by the definition of } a^x.$$ $$ a^t < a^y \text{ by the definition of } a^y.$$ $$ \text{ Therefore } a^r a^t < a^x a^y$$

I am going to leave the proof that $a^xa^y$ is the l.u.b of $B$ to the thoughtful reader.

We can now look at $a^{x+y}$, this is by definition

$$ a^{x+y} = \text{l.u.b of } \lbrace a^s \mid s < x+ y, s \in \mathbb{Q} \rbrace = \text{ l.u.b of } \lbrace a^{r+t} \mid r < x, t < y, r \in \mathbb{Q}, t \in \mathbb{Q} \rbrace. $$

Using our knowledge of the power rule for rational number (which is a different proof) we notice that this new set is really just $B$. Therefore $a^{x+y}$ is just the l.u.b of $B$. A set cannot have two distinct least upper bounds therefore $a^{x+y}=a^xa^y$.

Notice that there is no need to appeal to the exponential function (which is far easier) to establish the existence and properties of the real powers. All that is necessary is the defining property of the real numbers which distinguishes them from the rationals (the least upper bound property).

For rational powers we have to take a different approach. We first define rational powers using $n$th roots. If $m,n \in \mathbb{Z}$ then define,

$$ a^{m/n} \equiv \sqrt[n]{a^m} .$$

Now suppose we multiply two expressions with different rational powers,

$$ a^{m/n} a^{p/q} = \sqrt[n]{a^m} \sqrt[q]{a^p} .$$

We will label the left hand side with the variable $K$ giving us,

$$ K = \sqrt[n]{a^m} \sqrt[q]{a^p} .$$

Taking the $n\cdot q$ power of both sides (remember $n$ and $q$ are integers so this is well defined) we get,

$$ K^{nq} = \left(\sqrt[n]{a^m} \sqrt[q]{a^p}\right)^{nq}$$

$$ K^{nq} = \left(\sqrt[n]{a^m} \right)^{nq} \left( \sqrt[q]{a^p}\right)^{nq}$$

$$ K^{nq} = \left[ \left( \sqrt[n]{a^m} \right)^n\right]^{q} \left[ \left(\sqrt[q]{a^p} \right)^q \right]^{n}$$

$$ K^{nq} = \left(a^m \right)^{q} \left( a^p\right)^{n}$$

$$ K^{nq} = a^{mq} a^{pn} $$

$$ K^{nq} = a^{mq+pn} $$

Take note that in every step above I only either used the rules of exponents for integers or the definition of $n$'th roots.

Now we will remove the power from $K$ using radicals.

$$ K = \sqrt[nq]{a^{mq+pn}} $$

Recalling the definition of rational powers we rewrite the right hand side as,

$$ K = a^{\frac{mq+pn}{nq}} = a^{\frac{m}{n} + \frac{p}{q}}$$

Replacing $K$ with its value we have,

$$a^{m/n} a^{p/q} = a^{\frac{m}{n} + \frac{p}{q}}$$

and therefore we have established the product rule for exponents.

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    $\begingroup$ I added the proof for rational powers as well. I learned most of these details solving a homework problem out of Rudin's "Introduction to Mathematical Analysis". $\endgroup$ – Spencer Jul 4 '13 at 3:04
  • $\begingroup$ Wow, Thanks. I was just wondering what property you used when $K^{nq}=a^{mq}*a^{np}$ turned into $K^{n+q}=a^{mq+np}$ Were you not using the exact property we were trying to prove? $\endgroup$ – math101 Jul 4 '13 at 3:16
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    $\begingroup$ I was worried it would appear that way. Remember that $mq$ and $np$ are integers. The property we proved was for rational numbers. The proof for addition of integer powers is actually very simple and is just a matter of considering integer powers as repeated multiplication. For instance $x^2 x^3 = (xx)(xxx) = (xxxxx) = x^5$ The order in which my proofs are presented are actually in the reverse of the order you would do them logically. First you would prove the addition rule for integers, then rationals, then reals. $\endgroup$ – Spencer Jul 4 '13 at 3:27
  • $\begingroup$ I also just realized reading your post that I made a typo. The second $K^{n+q}$ should be $K^{nq}$ I will fix that. $\endgroup$ – Spencer Jul 4 '13 at 3:31
  • $\begingroup$ Ohhh ok everything makes sense now. Thanks for that mini proof of addition of integer powers. Everything is slowly starting to click in my brain. Thanks :) $\endgroup$ – math101 Jul 4 '13 at 3:37

We give a formal argument. It will not be entirely easy.

Let's assume that we have defined the exponential function $\exp(t)=e^t$ in one of the many ways it can be done, and proved that its derivative is $e^t$. We define $a^x$ as $\exp(x\ln a)$.

For simplicity, write $k$ instead of $\ln a$. We want to prove that $\exp(ku)\exp(kv)=\exp(k(u+v)) $.

Keep $u$ fixed, and let $$f(v)=\frac{\exp(k(u+v))}{\exp(kv)}.\tag{1}$$ Differentiate with respect to $v$. We get by the Quotient Rule and the Chain Rule that $f'(v)=0$. (If you need details here, they can be supplied.)

So $f(v)$ is a constant function. Which constant?

Set $v=0$. We find that our constant is $\exp(ku)$. Now (1) gives us the result.

  • $\begingroup$ Thanks alot. Love this proof $\endgroup$ – math101 Jul 4 '13 at 3:21

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