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how does one "prove" this algebraic rule...?
an nth index, radicand "a" to nth power = "a"
i understand that (for example) given the number 4 squared under a radicand, that is equal to simply the number 4 because the # of the index (2) matches the exponent (2)
I also understand that this works for any other number, so long as the exponent and index are the same......
What i need to know is how does one PROVE this rule? Please no number examples, i have plenty of those. I need to know how to actually prove the rule using "nth" rather than an actual numerical index/exponent, and "a" under the radicand. This is very tricky to me and although i know the rule to be true i have no idea how to "prove" it.
thank you in advance!!! :)
3 Answers
- BenLv 79 years agoFavorite Answer
This is one of those parts of math that falls into the vague territory between a proof and a definition. What you want to show is that the given definition for a^(1/n) "makes sense."
To phrase this more rigorously, we want to show that this definition is consistent with the other rules we have for exponents. Specifically, we have the following rules for integer powers of a:
a^m * a^n = a^(m+n)
(a^m)^n = a^(mn)
a^1 = a
a^0 = 1 (if a≠0)
All of these can be proved for positive integer powers m and n using the definition of a power of a, that is, multiplying a by itself however many times.
Now, we want to figure out what should happen when you take a^(1/n), if n is an integer. Well, our second rule above says that
(a^(1/n))^n = a^(n * 1/n) = a^1 = a
So, in order to make sure that the rules we have still work, a^(1/n) needs to be a number whose nth power is a. Hence, it has to be the nth root of a.
hope that helps!
PS: you can do the same thing with the first rule (rather than the second) to show that
a^-m = 1/(a^m)
when a≠0
- TychaBraheLv 79 years ago
For n <> 0 and x > 0, the nth root of x is equal to the positive value rendered by x^(1/n)
See, for example sqrt(4) = 2, but 4^(1/2) can be evaluated as either 2 or -2, so we have to define that we are talking only about the positive result.
So
[x^(1/n)]^n = x^[(1/n) * n] = x^1 = x
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