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2 books I recommend: how to prove it, and the book of proof, get a real analysis book, and if possible take a course. it took me around 5 years to be able to prove stuff... you need relaxation, fluid thinking, and a breadth of knowledge of facts you can use to prove stuff

we can try your example

we want to show that (a^b)^c = a^bc

let's work on the LHS

by the definition of an exponent, we know that a^b is just a * a * ... * a b times, so we can rewrite it as:

(a_0 * a_1 * ... * a_b)^c

by the same definition, we can multiply the quantity inside the parenthesis by itself c times:

(a_0 * a_1...a_b)_0 * (a_0 * a_1 * ... * a_b)_1 * ... * (a_0 * a_1 * ... * a_b)_c

now, use the fact that a^m * a^n = a^(m+n) to consolidate the parenthesis, since each factor has an exponent of 1 we can use simple counting:

(a^b)_0 * (a^b)_1 * ... * (a^b)_c

Repeat the previous step c times, we end up with

a^(b_0 + b_1 + ... + b_c)

which of course is just

a^(bc)

therefore (a^b)^c = a^(bc)



side note; there are some subtleties here for the rationals and serious difficulties with this method for irrationals. I agree this is a good start for someone thinking about how to prove some simple math ’rules’ though, so please don’t take this as discouragement; rather encouragement to keep opening up minds when discussing math!


Suppose you have three finite sets A, B and C, each with a, b and c elements.

Let A->B be the set of functions from A to B. Then it has b^a elements.

Let AxB the set of couples with the first element in A and the second on B. Then it has ab elements.

So to prove that (a^b)^c = a^(cb) you have to prove that there is a bijection between C->(B->A) and (CxB)->A.

Ever heard of currying and uncurrying?

(BTW, I didn't really use the hypothesis that the sets are finite, the proof is also valid for transfinites if you care about those)




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