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Unformatted text preview: 1 The Real Number System The rational numbers are beautiful, but are not big enough for various pur poses, and the set R of real numbers was constructed in the late nineteenth century, as a kind of an envelope of Q . (More later on this.) For a noncon structive approach, one starts with a list of axioms, called the Field axioms , which R satisfies. This will not be suﬃcient to have R , and we will also need order axioms , and the completeness axiom . In Math we like to build things based on axioms, much like in experimental sciences one starts from facts noticed in nature. Field axioms : There exist two binary operations, called addition + and multiplication ∗ , such that the following hold: 1) commutativity x + y = y + x , xy = yx 2) associativity x + ( y + z ) = ( x + y ) + z , x ( yz ) = ( xy ) z 3) distributivity x ( y + z ) = xy + xz 4) Existence of 0 , 1 such that x + 0 = x , 1 · x = x , 5) Existence of negatives: For every x there exists y such that x + y = 0. 6) Existence of reciprocals: For every x ̸ = 0 there exists y such that xy = 1. A lot of properties can be derived from these axioms. Example 1 a + b = a + c ⇒ b = c Proof Ax. 5 ⇒ ∃ y : a + y = 0. Take y + ( a + b ) = y + ( a + c ) and use Ax. 2 ⇒ ( y + a ) + b = ( y + a ) + c ⇒ 0 + b = 0 + c Use Ax. 4 ⇒ b = c . Example 2 ( − 1)( − 1) = 1. Proof . (by direct verification) By definition of negatives, ( − 1) + 1 = 0. Hence 0 = (( − 1) + 1)(( − 1) + 1) = ( − 1)( − 1) + ( − 1)(1) + (1)( − 1) + (1)(1) , where we have used the distributive property. By definition of the multi plicative unit 1, ( a )(1) = (1)( a ) = a for any a . Taking a to be − 1 and 1, 1 we see that ( − 1)(1) = (1)( − 1) = − 1 and (1)(1) = 1. Thus we get from the equation above, 0 = ( − 1)( − 1) − 1 − 1 + 1 = ( − 1)( − 1) − 1 , which implies, as asserted, that ( − 1)( − 1) is 1. QED....
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 Winter '08
 Borodin,A
 Real Numbers, upper bound, R+

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