126 9 Number SystemsTheorem 9.3.4
Va; G R, x y 0, Vn G N 3 a Mrogtte y G R, y X 0 9 yn = x.Proof. [Existence]:
Given x X 0, n G N. Let £ = {t G R : i X 0 and tn < x).
Claim 1: E ^ 0
Let £ = ^y x 0, £ -< 1, t -< x; 0 -< t -< 1 => i" -< t
(0 -< i -< 1 => 0 -< t^2 •< t < 1 => ... => 0 -< T -< i < 1).
Also we have, t < x ^> tn < x\ therefore, £ = ^j G E,
Claim 2: E^1 is bounded above
If 1 + x is an upper bound of E.
If not, 3t€E3tyl + x. In particular, t X 1 (because a; X 0) =>
f X /, X 1 + x X• x; therefore, t ^ E: Contradiction!
y = sup E G R because R has lub property.
y X 0, because (x X 0).
Claim 3: yn = x
If not, then either yn -< x or x < yn.
We know the following:
Let () ^ a < b. Then, bn - an = (b - a){bn~l + bn~^2 a + ••• + a'^1 "^1 ) =>
(*) : bn -an -< {b - a)nbn-^1.
0 J/" "< => ^Tff-r >- 0. Find n G R 3 0 -< ^ 1 and 0 -< =#££?.
(): (y + /i)" - (j/)" -< hn(y + h)n -< hn(y + l)""^1 -< x - yn
Therefore, (y + h)n -<x=>y + h(zE. But y + h X y => y is not an upper
bound of E, Contradiction!
ii) x -< yn. Let k = v, ~Xi >~ 0 and x -< y [because yn — x -< ny"~l).
Claim: y — k is an upper bound of E.
Suppose not, 3t G E 3 t X y — k X 0.
Then, tn X (y - k)n => -tn -< ~{y - k)n => yn - tn •< yn - (y ~ k)n
(): yn -{y-'k)n -< knyn~x = yn-x => yn-tn -< yn~x => tn y x => t <£ E,
Contradiction!
Therefore, y — k is an upper bound of E.
However, y is lub of E, Contradiction!
[ Uniqueness]:
Suppose y y 0, y' y 0 arc two positive roots 3 y ^ y' and yn = x — (?/)"•
Without loss of generality, we may assume that , y' y y y 0, (because y ^
?/) =^ 2/" ^ (?/)"> Contradiction! Thus, y is unique. O
Definition 9.3.5 Real numbers which are not rational are called irrational
numbers.
Example 9.3.6 \/2 is an irrational number.
Corollary 9.3.7 Let a. y 0, 6x0 and n&N. Then, (ab)^1 /" = al/nbl/n.