Question

law of indicies proof

Answer 1

Theorem A: a
n+m = a
na
m
Theorem B: (a
n
)
m = a
nm
Theorem C: (ab)
n = a
n
b
n
Theorem D: a
b
n
=
a
n
b
n
Theorem E: a
n
am
= a
n−m
Theorem F: a
0 = 1
Theorem G: a
−n =
1
a
n
Theorem H: a
n/m =
m√
a
n
Like all theorems, these do not come out of nowhere. They come from a definition and logical deduction.
Let us start, then, with a definition. What is exponentiation, anyway? What do we mean when we
write a number with a superscript? I contend that we mean1
something like repeated multiplication. When
we write, for example, 54
, we mean this simply as a more convenient way of writing 5 · 5 · 5 · 5 (or, even
more simply, 625). Formally, let’s define exponentiation in the following way2
:
a
n ≡ a · a · a · · · a | {z }
n times
(The triple-equals-sign here is used to show that this is a definition—that this equation is not the conse-
quence of some earlier theorem or axiom, but that with it we’re defining what we mean by a
n
.) Let us
prove these theorems.
Theorem A: a
n+m = a
na
m
Proof: We’ll start with the left side of the equation, apply the definition of exponentiation, do some
algebra, and eventually end up with the right side.
a
n+m = a · a · a · · · a | {z }
n+m times
(by the definition of exponen-
tiation)
= (a · a · a · · · a | {z }
n times
)(a · a · a · · · a | {z }
m times
) (because multiplication is as-
sociative)
= a
n
· a
m (applying the definition of ex-
ponentiation again)
(done! this stylized A is my
end-of-proof symbol.)
Theorem B: (a
n
)
m = a
nm
Proof: To prove this, we’ll need to apply the definition of exponentiation—twice. (Well, actually three
times, but the last time doesn’t count.)
1For the most part. If you’re a mathematician, we could have a lengthy conversation about everything I say on this sheet,
but for our purposes this definition is sufficient. If you disagree (and this is in an imaginary conversation between myself and
a mathematician), I will say but this: do you believe we should teach students Dedekind cuts before we let them utter the
words, “real number”?
2Note that we haven’t put any specifications on what a and n can and/or should be—must they be integers? real numbers?
complex numbers? The way our definition works, what with the “n times” business, it must only hold for n being a positive
integer, but these theorems hold for n being any real number. What we’ll see in our derivations is that, even though we start
by considering n only as a positive integer, we can extend our idea of “exponentiation” in a very logical way such that it holds
for any real number. But this is in a footnote for a reason.


(a
n
)
m = (a · a · a · · · a | {z }
n times
)
m (by definition)
= (a · a · a · · · a | {z }
n times
) · (a · a · a · · · a | {z }
n times
)· · ·(a · a · a · · · a | {z }
n times
)
| {z }
m times
(by definition, again)
= a · a · a · · · a | {z }
n*m times
(multiplication)
= a
nm (finish by re-applying the def-
inition in reverse)
Theorem C: (ab)
n = a
n
b
n
Proof: See a pattern? We’ll apply the definition of exponentiation, do some algebra, and eventually get
what we want.
(ab)
n = ab · ab · · · ab
| {z }
n times
(by definition)
= (a · a · · · a | {z }
n times
) · (b · b · · · b
| {z }
n times
) (multiplication is commuta-
tive —we can rearrange)
= a
n
b
n
(definition)
Theorem D:
a
b
n
=
a
n
b
n
Proof: How do you think this should go?
Theorem E:
a
n
am
= a
n−m.
Proof: This is where things start to get interesting. The clear first step is to write out the entire quantity:
a
n
am
=
n times
z }| {
a · a · · · a
a · a · · · a | {z }
m times
But where we go from here will depend on the relative size of n and m. n could be bigger than m, it could
be smaller, or the two could be equal.
Let’s start with the first case—that of n > m. In that case, both the top and the bottom will have at
least m of the a’s, and the top will have more— it’ll have n − m more (because m + (n − m) = n).


Hope it is helpful to u