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

  • Context: High School 
  • Thread starter Thread starter okunyg
  • Start date Start date
Click For Summary

Discussion Overview

The discussion centers around the proof of the equation a^-n = 1/a^n, exploring the definitions and implications of exponentiation, particularly when dealing with negative integers. Participants share their thoughts on the validity of proposed proofs and the underlying assumptions related to exponentiation.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • One participant presents a proof attempt using the equation a^-n = a^0-n = (a^0)/(a^n) = 1/a^n, questioning its validity.
  • Another participant points out that the proof relies on the assumption that a^b a^c = a^(b+c), which may not be universally accepted without prior definitions.
  • A participant raises a concern about the meaning of exponentiation when n is a negative integer, suggesting that if exponentiation is defined through repeated multiplication, the concept of multiplying by itself a negative number of times is problematic.
  • Further clarification is sought regarding the assumptions made in the discussion, indicating a lack of understanding of the foundational concepts.
  • One participant suggests that by manipulating the equation a^-n = 1/a^n through multiplication by a^m (where m > n), the equation can be made to make sense under certain conditions.
  • Another participant introduces the concept of isomorphism, suggesting a relationship between the operations of addition and multiplication in the context of exponentiation.
  • There is confusion regarding the relevance of isomorphism to the original proof discussion, with participants seeking clarification on its application.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the proof attempts and the assumptions underlying exponentiation. There is no consensus on the definitions or the implications of negative exponents, leading to an unresolved discussion.

Contextual Notes

The discussion highlights limitations in the definitions of exponentiation, particularly regarding negative integers, and the assumptions that participants bring to the topic. The exploration of isomorphism introduces additional complexity without clear resolution.

okunyg
Messages
17
Reaction score
0
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
 
Mathematics news on Phys.org
you assume that a^b a^c = a^(b+c).
 
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?
 
Last edited:
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.
 
okunyg said:
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
 
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 .
 
in other words: isomorphism ...
 
What does isomorphism have to do with anything? :confused:
 
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)
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 55 ·
2
Replies
55
Views
7K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 69 ·
3
Replies
69
Views
7K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K