7/24/2019 rais n-esima de x

1/2

POSITIVE NTH ROOTS

Here we prove, using the least upper bound property ofR

, that positive nth roots exist.The proof is taken from Walter Rudins book Principles of mathematical analysis.

Theorem. Let x > 0 be a positive real number. For every natural number n 1, thereexists a unique positiventh root ofx, which is to sayy >0 R such thatyn =x.

Proof. Givenxwe construct a set Swhose supremum will be the desired element y. Let

S=

t Rt >0, tn x

.

First we claim that Sis nonempty. To see this, considert = x/(x + 1).Sincet 1, it follows thatrn r > x

so r /S, and is an upper bound.By the least upper bound property ofR(which is ultimately equivalent to completeness),

Shas a supremum. Lety= sup(S).

We now claim that yn =x. To prove this, we will derive contradictions from the other twopossibilities, that yn > xandyn < x.Both steps use the following estimate

(1) 0< a < b = bn an

7/24/2019 rais n-esima de x

2/2

and note that

0< k < yn

nyn1 =

y

n< y.

Invoking the formula (1) again with a= y k (which weve just shown to be positive) andb= y, we obtain the estimate

yn (y k)n < k n yn1 =(yn

x) n yn1

nyn1 =yn x.

Cancelling the yn terms and multiplying by1, it follows that x x and yn < x, it follows that yn =x.

2