arXiv:0805.3303v1 [math.GM] 21 May 2008 e od n phrases and words Key em 1.1 Lemma edfieteset the define We facts Auxiliary 1 ht( that Proof above. from bounded and empty not is i,Mnzls Caldas. Manizales, bia, sa pe on for bound upper an is es pe on Axiom Bound Upper Least xssalatuprbudfor bound upper least a exists h es pe on of bound upper least The B ento 1.1 Definition A ti ayt hwta if that show to easy is It if o any For . . x ∗ t eateto ahmtc,Uiesddd ads Departm Caldas, de Universidad Mathematics, of Department and ≤ a . x 1) + y E rvn h xsec fthe of existence the Proving nti ae epoeb nuto on induction by prove we paper this In o any for r oiienmesfrwhich for numbers positive are sntepy eas 0 because empty, not is > δ n a > ups that Suppose emyfind may we 0 Let t o,if Now, . R ∈ fra ubr sanmrcodrdfil nwihtefollowi the which in field ordered numeric a as numbers real of email n E E htis, that , eapstv nee and integer positive a be : . n a [email protected] : hro,speu,latuprbudaxiom. bound upper least supremum, root, th E a an has a > t > a suiu n ti eoe ysup by denoted is it and unique is E t 0 x ∈ If : . then 1 + and sa pe on for bound upper an is n E hro,te hsro suiu.Ti olw rmtefc t fact the from follows This unique. is root this then root, th ∈ 6 ∅ uhthat such n = E E sapstv nee.Teset The integer. positive a is nteohrhn,i ses oso yidcinon induction by show to easy is it hand, other the On . E x = t laoSalas Alvaro n n ⊆ { = > Abstract t R n y ≥ x ( > a htaypstv elnme has number real positive any that n sannepystbuddfo bv,te there then above, from bounded set empty non a is a − then , 1 0 1) + t < δ | 0 t n esythat say We . n n a < n fMteais nvria ainld Colom- de Nacional Universidad Mathematics, of ent x a > E ≤ hro yinduction. by root th ∗ = . } x . y n then and . The . E If . n x hro of root th x sa is / t sup = ∈ E. n hro of root th E hsmasthat means This a hn: then , n sdntdby denoted is hroot. th gaimholds: axiom ng a if x n = a √ n a 1 + hat . a . n 2 Existence of nth root of a positive number
Theorem 2.1 For any positive integer n 1 and for any a> 0 there exists √n a> 0. ≥ n We first prove by induction that for any positive integer n and for any b> 0 there exists 2√b. This is true for n = 1. Indeed, according to Lemma 1.1, the set E = t > 0 t2 < a has a least upper { | } bound x = sup E 0. We claim that x2 = a. Indeed, let 0 <ε< 1 and define δ = ε/(2x + 1). ≥ Clearly, 0 < δ ε< 1. There exists an element t E such that x δ < t x. We have ≤ ∈ − ≤ x2 < (t + δ)2 = t2 +2tδ + δ2 < a + (2t + 1)δ a + ε. ≤ Since the inequality x2 < a + ε holds for any ε (0, 1), we conclude that x2 a. ∈ ≤ On the other hand, x + δ / E, so ∈ a (x + δ)2 < x2 + (2x + 1)δ = x2 + ε. ≤ Since the inequality x2 + ε > a holds for any ε (0, 1), we conclude that x2 a. We have proved ∈ ≥ the equality x2 = a. In particular, x = √a> 0. k Suppose that for some positive integer k the number 2√c is defined for every c > 0. Let b be a k positive number. By inductive hypothesis, for c = √b there exists y > 0 such that y2 = √b. From this it follows that 2 k k+1 b = √b = y2·2 = y2
k+1 and then y = 2 √b, that is, our Lemma is also true for n = k + 1. Finally, we shall prove that for any positive integer n 2 and for any a> 0 there exists √n a> 0. We ≥ proceed by induction on n. Our theorem is true for n = 2 since any positive number a has a square root. Suppose that n−√1 c exists for any c> 0 and for some n 3. Let a> 0 and define the set ≥ E = t 0 tn < a . { ≥ | } By Lemma 1.1 this set has a supremum, say x = sup E 0. Define m =2n n. Then m + n =2n ≥ − n and m is a positive integer, since 2n > n. There exists a positive y for which y2 = ym+n = axm.
Observe that m x yn = y−mym+n = y−maxm = a (2.1) y We claim that xn = a. Indeed, suppose that xn < a. Then xm+n < axm = ym+n. This implies that x
2