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I need to prove the exponent identity $(a^b)^c = a^{bc}$, where $a,b,c \in \mathbb{Z}$.

Apparently this proof is elementary/trivial, but I can't think of how to prove it. I need it as a lemma for another problem. Can anyone please point me in the right direction or show me why this is true?

Martin Sleziak
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Newb
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    See here. – 2012ssohn Jan 15 '14 at 02:07
  • @2012ssohn that proof uses $e\ln(a)$, which strikes me as an illegitimate trick. I would like to see this proven in more elementary terms. – Newb Jan 15 '14 at 02:08
  • What's your definition of exponentiation, and what type of objects are $a, b, c$? – Joshua P. Swanson Jan 15 '14 at 02:10
  • How legitimate that trick is depends on how you define exponentiation. What exactly does $2^\pi$ mean? – Henry Swanson Jan 15 '14 at 02:11
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    @JSwanson $a,b,c \in \mathbb{Z}$. In this context, exponentiation is repeated multiplication. – Newb Jan 15 '14 at 02:13
  • $\mathbb{Q}^\times$ is a group under multiplication, and for general groups, $(g^a)^b = g^{ab}$, where here exponentiation is "repeated multiplication" with some fiddling with non-positive exponents. The proof is by the generalized associative law: expand both sides and you'll see they're the same up to parenthesization, again with some fiddling for negatives. Proving generalized associativity is really the meat of a fully rigorous proof along these lines; most people just assume it as obvious. – Joshua P. Swanson Jan 15 '14 at 02:15
  • I think this is a legitimate question. @Newb , Please tell us how you define exponentiation so we have somewhere to start. – Christian Chapman Jan 15 '14 at 02:18
  • @enthdegree see my comment from five minutes earlier: exponentiation is defined as repeated muliplication, e.g. $a^3 = a \cdot a \cdot a$, as is typical in integer exponentiation. – Newb Jan 15 '14 at 02:19
  • @Newb, if it is repeated multiplication, then the domain for $a, b, c$ must be whole numbers and not integers: repeated multiplication for negative powers? – Parth Thakkar Jan 15 '14 at 13:48

2 Answers2

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Presumably by definition you know that $a^n \cdot a = a^{n+1}$.

Use this to prove that $a^n \cdot a^m = a^{n+m}$.

Then you have $$(a^n)^k = \underbrace{a^n \cdot a^n \cdot \cdots \cdot a^n}_{k\ \text{times}} = a^{\overbrace{n+n+\cdots+n}^{k\ \text{times}}} = a^{nk}$$

There are some details to be filled in, e.g. what happens when $k$ is negative, but this should help you get the idea.

Clive Newstead
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Let's prove it, first, for $c\ge0$. Then the assertion is obvious for $c=0$. Assume it holds for $c\ge0$: $$ (a^b)^{c+1}=(a^b)^c\cdot a^b=a^{bc}\cdot a^b=a^{bc+b}=a^{b(c+1)} $$ so, by induction the assertion is proved for $c\ge0$. If $c<0$, then, setting $c=-d$, we have $$ (a^b)^c=\frac{1}{(a^b)^d}=\frac{1}{a^{bd}}=\frac{1}{a^{-bc}}=a^{bc} $$

Fill in the details.

egreg
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