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What is an intuitive explanation for the rule that $(a^b)^c = a^{bc}$. I'm trying to wrap my head around it, but I can't really do it.

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Maybe ask yourself why $(a^3)^7=a^{21}$. – André Nicolas Sep 20 '11 at 5:02
This probably doesn't help the OP, but if you let $a$, $b$, and $c$ be sets, then the posted equation is essentially currying. – Rahul Sep 20 '11 at 5:29
up vote 16 down vote accepted

I will assume that $b$ and $c$ are positive integers and that $a$ is any "number" (it doesn't really matter much what $a$ is...).

Suppose I have $b \times c$ copies of the number $a$. I can arrange them into a $b \times c$ rectangular array: i.e., with $b$ rows and $c$ columns. When I multiply all $b \times c$ of these $a$'s together, I get $a^{bc}$.

On the other hand, suppose I look at just one column of the array. In this column I have $b$ $a$'s, so the product of all the entries in a column is $a^b$. But now I have $c$ columns altogether, so the product of all the entries is obtained by multiplying the common product of all the entries in a given column by itself $c$ times, or $(a^b)^c$.

Thus $a^{bc} = (a^b)^c$.

If you want to justify this identity when $b$ and $c$ are other things besides positive integers -- e.g. real numbers, or infinite cardinals -- that's another matter: please ask.

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I like your explanation here, together with Zev's it broke through to me. I'm afraid I won't be able to understand your explanation for reals and infinite cardinals though, but I'll bite anyways :). Should I post a new question? – Matt Munson Sep 20 '11 at 5:50
@Matt: I'm glad to help. Indeed my answer is similar to Zev's but expressed a bit more "geometrically". As for your other question: sure, you could post a followup asking about the case of real numbers and/or cardinal numbers. There would be a certain logic in not asking this question until you actually want/need to know the answer, but for such a basic and important question as this, other people would certainly be interested. So either wait until later or post it now, whichever you like. – Pete L. Clark Sep 20 '11 at 6:24

When $b$ and $c$ are positive integers, we can make the following argument: $$(a^b)^c=\underbrace{(a^b)\times(a^b)\times\cdots\times(a^b)}_{c\text{ times}}=$$ $$\underbrace{\left(\underbrace{a\times a\times\cdots \times a}_{b\text{ times}}\right)\times\left(\underbrace{a\times a\times\cdots \times a}_{b\text{ times}}\right)\times\cdots\times \left(\underbrace{a\times a\times\cdots \times a}_{b\text{ times}}\right)}_{c\text{ times}}=$$ $$\underbrace{a\times a\times \cdots\times a}_{bc\text{ times}}=a^{bc}$$

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+1 for excellent formula typesetting skills – Kris Van Bael Sep 29 '11 at 21:33

If you want a physical, intuitive understanding, grab a slide rule. As you'll be manipulating the logarithmic forms of the exponents, it should become clear quite quickly.

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It’s not a rule in the sense of a theorem, but of a definition. (In early mathematics training, even into calculus, there is a tendency, for administrative/teaching convenience, to conflate theorems and definitions, calling all such things “rules”.) It is simply a matter of the notation being chosen such that it is what we might call “fortuitous”. Let’s consider three other such instances.

  1. x – y = x + (-y), as you can verify from numerical examples, but it is really a matter of definition: that is the fortuitous way in which x – y is DEFINED.

  2. x/y = x(1/y), for y not equal to 0, as you can verify from numerical examples, but it is really a matter of definition: that is the fortuitious way in which x/y is DEFINED.

  3. a raised to the power b = exp(b*log(a)), for positive a, as you can verify from numerical examples, but it is really a matter of definition: based on extensive numerical experience, that is the fortuitous way in which mathematicians decided to DEFINE a to the power b.

Now to your question. For positive a and arbitrary b and c (or arbitrary non-zero a and integer b and c), the expression a to the power of (bc) is DEFINED, for notational convenience, to be (a to the power of b) to the power of c. The fact that it is only a notational convenience can be seen if a is negative and not both b and c are integer . Presumably you would agree with the statement that in any (valid) mathematical expression, we should be able to reduce whatever fractions occur to lowest terms without affecting the value of the expression. But that does not happen if a is negative, say, -1, b is 2, and c = 1/4: ((-1 raised to the power 2))raised to the power (1/4) is 1, and this should equal (-1 raised to the (2/4) power), which equals (-1 raised to the 1/2 power), which is the imaginary unit, not unity.

So, the short answer to your question is: It is simply a fortuitous definition, based on extensive numerical experience. Your question arose simply from the fact that you were lacking a sufficient dose of that experience. It’s not enough to just think about it: you have to grok it.

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I disagree with this answer. At some point we define $a^d$ when $d$ is any positive integer, for instance recursively: $a^0 = 1$, for $d \geq 1$, $a^d = a \cdot a^{d-1}$. Anyway, once we have made this definition, we do not need or get to make a new definition in the case $d = b \cdot c$. The result in question really is a proposition not a definition. (Of course it follows from certain definitions, as everything in mathematics does...) – Pete L. Clark Sep 21 '11 at 3:15
@Pete L. Clark: That recursive definition gets discarded when the exp-log definition is finally embraced, much like a caterpillar shedding its cocoon to become a butterfly. – Mike Jones Sep 21 '11 at 10:09
I had another comment where I explained that this argument makes sense for any element of a set endowed with a binary operation, in which context we generally do not have exponentials and logarithms...but I deleted that comment, perhaps because I felt there was insufficient evidence the OP was thinking in this way. (The point though is that the definition I gave is not "discarded": it is used in many parts of mathematics.) But anyway, I said for instance: I don't know of any reasonable definition of exponentiation for which the fact in question is true by definition. Do you? – Pete L. Clark Sep 22 '11 at 4:52
It is similar to the situation with the factorial function. One can offer a strong plausibility argument why 0! = 1, but ultimately it is a matter of defintion. – Mike Jones Sep 24 '11 at 21:30
In the situation of the factorial function $0! = 1$ defines the left hand side: i.e., the symbol $0!$ has not been given any prior meaning. But if we have already defined $x^a$ for positive integers $a$, then the identity $x^{ab} = (x^a)^b$ is a relation between two quantities which have already been defined, so it cannot be true by definition. In any case, I asked you for a reference to support your point of view. If you cannot or do not provide one, I don't see the point in continuing this discussion. – Pete L. Clark Sep 25 '11 at 17:36

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