How would you proove a^-n = 1/a^n?

[okunyg]

I can't find many proofs of how the equation in the thread title is correct. How would you proove the equation?

I tried to make one, but I don't know if it's worth mentioning. What do you think about my "proof"?

a^-n = a^0-n = (a^0)/(a^n) = 1/a^n

 

[mathwonk]

you assume that a^b a^c = a^(b+c).

 

[okunyg]

I'm sorry, could you elaborate a little? Do I assume? Do you assume? What does that equation have to do with this?

You might notice that I'm a beginner.


Edit: Oh, I believe I know what you mean: I did not proove anything, I just wrote something obvious that someone else has figured out? Kind of like not thinking on my own?

 

[JohnDuck]

The problem is this: what does xⁿ mean when n is a negative integer? If you have defined exponentiation in terms of repeated multiplication, then what does it mean to multiply something by itself -3 or 0 times? Once you have defined exponentiation for these (and other) values, there is nothing really to prove.

 

[matt grime]

I'm sorry, could you elaborate a little? Do I assume? Do you assume? What does that equation have to do with this?

You assume that you know what x^2, x^3 means etc. Then you notice that (x^a)^b = x^(ab), and other such rules, that make you realize that it is reasonable to extend thee definition from just the positive whole numbers to all rational numbers via x^{1/n} as the n'th root of x, and x^-n as 1/x^n

 

[lalbatros]

Because by multiplying both side of the equation by a^m, with m>n, it makes sense.

Doing so on this equation:

a^-n = 1/a^n

leads to

a^(m-n) = a^m/a^n

which makes sense (already) for m>=n .

 

[lalbatros]

in other words: isomorphism ...

 

[Office_Shredder]

What does isomorphism have to do with anything?

 

[lalbatros]

well, summing exponents is similar to making products
there is an isomorphism between:

- the set of positive numbers equipped with multiplication
- the set of all numbers equipped with the addition

(sorry for the lose speaking, this is very old for me)