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# hmw1s - Now if A is bounded below then(1 − A ={− x x...

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From Rudin, Chapter 1. Exercise 1 If s and r ± = 0 are rational then so are s + r, r, 1 /r and sr (since the rationals form a ﬁeld). So if r is rational and x is real, then x + r rational implies ( x + r ) r = x is rational. An irrational number is just a non- rational real number, so conversely if x is irrational then x + t must be irrational. Similarly if rx is rational then so is ( xr ) /r = x ;thu si f x is irrational then so is rx. Exercise 3 [(a)] If x ± =0 then x 1 exists and if xy = xz then 1 1 y =( x x ) y = x 1 ( xy )= x 1 ( xz )=( x x ) z = z using ﬁrst (M5) then (M2), (M3), the given condition, (M3) and (M5). [(b)] Is (a) with z =1 . [(c)] Multiply by x 1 so x 1 = x 1 ( xy )=( x 1 x ) y =1 y = y using associativity and deﬁnition of inverse. 1 · [(d)] The identity for x 1 =1 /x, x · x 1 gives by commutativity x x =1 which means 1 / (1 /x )= x by the uniqueness of inverses. Exercise 5 If A is a set of real numbers which is bounded below then inf A is by deﬁnition a lower bound, i.e. inf A a for all a A andi fin f A b for any other lower
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Unformatted text preview: Now if A is bounded below then (1) − A = {− x ; x ∈ A } is bounded above. Indeed if b ≤ x for all x ∈ A then − b ≥ − x for all x ∈ A which means − b ≥ y for all y ∈ − A. Now, if sup( − A ) is the least upper bound of − A it follows that − sup( − A ) is a lower bound for A since x ∈ A = ⇒ − x ∈ − A = ⇒ sup( − A ) ≥ − x = ⇒ − sup( − A ) ≤ x. As noted above, if b is any lower bound for A then − b is an upper bound for − A so − b ≥ sup( − A ) and b ≤ − sup( − A ) . This is the deﬁnition of inf A so inf A = − sup( − A ) . 1...
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## This note was uploaded on 12/16/2011 for the course STAT 5446 taught by Professor Frade during the Fall '09 term at FSU.

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