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

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In summary, the conversation discusses the equation a^-n = a^0-n = (a^0)/(a^n) = 1/a^n and the process of proving it. The individual speaking is a beginner and is trying to understand the concept of exponentiation for negative and zero powers. They mention a proof they made and ask for feedback. The conversation then delves into the definition of exponentiation and its extension to rational numbers. The concept of isomorphism is mentioned and its relation to exponentiation is discussed.
  • #1
okunyg
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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
 
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  • #2
you assume that a^b a^c = a^(b+c).
 
  • #3
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:
  • #4
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.
 
  • #5
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
 
  • #6
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 .
 
  • #7
in other words: isomorphism ...
 
  • #8
What does isomorphism have to do with anything? :confused:
 
  • #9
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)
 

1. What is the definition of a^-n?

The definition of a^-n is the reciprocal of a^n, or in other words, the number that when multiplied by a^n results in 1.

2. How would you prove that a^-n = 1/a^n?

To prove that a^-n = 1/a^n, we can use the definition of negative exponents, which states that a^-n = 1/a^n. We can also use algebraic manipulation to show that a^-n = 1/a^n. For example, we can rewrite a^-n as 1/a^n and then use the property of exponents that states a^-n = 1/a^n.

3. Can you provide an example to illustrate the relationship between a^-n and 1/a^n?

Yes, for example, if a = 2 and n = 3, then a^-n = 1/2^3 = 1/8. Similarly, 1/a^n = 1/2^3 = 1/8. Therefore, we can see that a^-n = 1/a^n.

4. How does the value of a affect the relationship between a^-n and 1/a^n?

The value of a does not affect the relationship between a^-n and 1/a^n. This relationship holds true for any value of a, as long as a is not equal to 0. This is because the definition of negative exponents and the property of exponents hold true for any base number.

5. Can the relationship between a^-n and 1/a^n be applied to any type of number, such as fractions or decimals?

Yes, the relationship between a^-n and 1/a^n can be applied to any type of number, including fractions, decimals, and even negative numbers. As long as the base number (a) is not equal to 0, the relationship holds true. This is because the definition of negative exponents and the property of exponents apply to all types of numbers, not just whole numbers.

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