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CHAPTER II [:] The Real Number System [...] 2. ORDER. "end quote
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O5:       THE RULES OF SIGN

On the left side of the equals sign in the table below,
each instance of   positive   stands indepentantly for any real number that is greater than zero, and
each instance of   negative   stands independantly for any real number that is less than zero.

On the right side of the equals sign in the table below,
positive   means that the result is greater than zero, and
negative   means that the result is less than zero.

The Rules of Sign in Addition and Multiplication
  (1)  
  positive + positive  
  =  
  positive  
  by the order property  
   
   
   
   
   
  (2)  
  negative + negative  
  =  
  negative  
  Let   a   and   b   be any real numbers such that   a0,   and   b0.  
  Then  
  -a0  
  and  
  -b0  
  by O3:
  Add   -a   to both sides of   a0,   and add   -b   to both sides of   b0.
  (Avoid confusion by reading all the inequality symbols from the pointy ends.)
   
   
   
  0(-a)+(-b)  
  by the order property  
  But  
  (-a)+(-b)  
  =  
  -(a+b)  
  by F8  
   
   
   
  a+b0  
  which we get by adding   a+b   to both sides of   0-(a+b)  
   
   
   
   
   
  (3)  
  positive · positive  
  =  
  positive  
  by the order property  
   
   
   
   
   
  (4)  
  positive · negative  
  =  
  negative  
  Let   a   and   b   be any real numbers such that   a0,   and   b0.  
   
  Because  
   
  -b0  
  by part two of the order property  
   
  we have  
   
  a(-b)0  
  by the order property.  
  But  
  a(-b)  
  =  
  a((-1)·b)  
  by F10  
   
   
  =  
  ((-1)·a)b  
  by the associativity and commutativity of multiplication  
   
   
  =  
  (-1)·(ab)  
  by the associativity of multiplication  
   
   
  =  
  -(ab)  
  by F10  
   
  Consequently  
   
  -(ab)0  
  by the chain of equals signs.  
   
  Therefore  
   
  ab0  
  by part two of the order property  
   
   
   
   
   
  (5)  
  negative · negative  
  =  
  positive  
  Let   a   and   b   be any real numbers such that   a0,   and   b0  
  Then  
  -a0  
  and  
  -b0  
  by part two of the order property  
   
   
   
  (-a)·(-b)0  
  by the order property  
  But  
  (-a)·(-b)  
  =  
  ((-1)·a)(-b)  
  by F10  
   
   
  =  
  (-1)·(a(-b))  
  by the associativity of multiplication  
   
   
  =  
  (-1)·(a((-1)·b))  
  by F10  
   
   
  =  
  (-1)·( ((-1)·a)b )  
  by the associativity and commutativity of multiplication  
   
   
  =  
  ( (-1)·((-1)·a) )b  
  by the associativity of multiplication  
   
   
  =  
  ( ((-1)·(-1))a )b  
  by the associativity of multiplication  
   
   
  =  
  ( (-(-1))a )b  
  by F10  
   
   
  =  
  ( (1)a )b  
  by F6  
   
   
  =  
  ( a )b  
  by the property of the neutral element for multiplication  
   
   
  =  
  ab  
  because we no longer need the parentheses  
   
  Consequently  
   
  ab0  
  by the chain of equals signs  

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