Sec.%2011%20Ordered%20Fields

Sec.%2011%20Ordered%20Fields - Math 350 Advanced Calculus I...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 350 Advanced Calculus I Fall 2009 Homework Solutions Section 11 ORDERED FIELDS 11.3 Let x , y , and z be real numbers. Prove the following. (a) ( x ) = x Proof: The element ( x ) is by deFnition the additive inverse of x . To show ( x ) = x , we need to show that x is the additive inverse of x . By Axiom A5, the additive inverse of any element is unique, so to show that x is the additive inverse of x , we only need to show that x behaves like the additive inverse of x . That is, we need to show ( x ) + x = 0 . By Axiom A5, since x is the additive inverse of x , then x + ( x ) = 0 . By Axiom A2 Commutative Law of Addition, x + ( x ) = ( x ) + x. Therefore ( x ) + x = 0 , as required. Thus ( x ) = x . (b) ( x ) · y = ( xy ) and ( x ) · ( y ) = xy Proof: The element ( xy ) is by deFnition the additive inverse of xy . So, to show that ( x ) · y = ( xy ), we need to show that ( x ) · y behaves like the additive inverse of xy . That is, we need to show xy + ( x ) · y = 0 . We have xy + ( x ) · y = yx + y · ( x ) = y [ x + ( x )] = y · 0 = 0 by M2 Commutative Law by DL Distributive Law by A5 Existence of Additive Inverses by Theorem 11.1(b) Alternatively, we can use the result of Theorem 11.1(c). We have ( x ) · y = [( 1) · x ] · y = ( 1) · ( x · y ) = ( xy ) by Theorem 11.1(c) by M3 Associative Law by Theorem 11.1(c) To prove ( x ) · ( y ) = xy , we can use the Frst result. We have ( x ) · ( y ) = [ x · ( y )] = [( y ) · x ] = [ ( yx )] = [ ( xy )] = xy by the Frst part of (b) by M2 Commutative Law by the Frst part of (b) by M2 Commutative Law by part (a) 12
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Math 350 Homework Solutions page 13 Section 11 Ordered Fields 11.3 (continued) (e) If x 6 = 0, then x 2 > 0. Proof: Suppose x 6 = 0. We need to show x 2 > 0 . By Axiom O1 Trichotomy Law, exactly one of the relations x = 0, x > 0, or x < 0 holds. Since x 6 = 0 by assumption, it follows that either x > 0 or x < 0. We need to prove x 2 > 0 in both of these cases. Case 1. Suppose x > 0. By Axiom O4, since x > 0, when we multiply both sides of the inequality x > 0 by x , we preserve the inequality, so we obtain x 2 = x · x > x · 0 . By Theorem 11.1(b), x · 0 = 0, so this implies x 2 > 0, as required. Case 2.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 6

Sec.%2011%20Ordered%20Fields - Math 350 Advanced Calculus I...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online