1. ## Tricky prove question

Assume the result of (a) and prove (b)

(a) Prove that if 0 <= a < b, then a^n < b^n.
(b) Prove that every nonnegative real number x has a unique nonnegative nth root x^1/n.

HINT: The existence of x^1/n can be seen by applying the intermediate-value theorem to the function f (t) = t^n for t >= 0. The uniqueness follows from (a).

Does anyone have idea how to do this?

2. Originally Posted by 450081592
Assume the result of (a) and prove (b)

(a) Prove that if 0 <= a < b, then a^n < b^n.
(b) Prove that every nonnegative real number x has a unique nonnegative nth root x^1/n.

HINT: The existence of x^1/n can be seen by applying the intermediate-value theorem to the function f (t) = t^n for t >= 0. The uniqueness follows from (a).

Does anyone have idea how to do this?

The function f(t) = t^n is continuous on [0,x], for any x > 0 ==> for any value c in (f(0),f(x)) = (0, x^n) there exists some d in (0,x) s.t. f(d) = c ==> for any positive c there exists an expression d^n = c for it (since any c is contained in some interval of the fomr [0,x] for some positive x, because f is increasing monotonically ==> c^(1/n) = d and c has a positive n-th root d.

About uniqueness it follows at once from (a).

Tonio

3. I do not understand the solution, can you be more specific?
what do c and d stand for? we can prove it has a positive n-th root d, but how do know it is a unique one?