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arithmetic_structure_of_real_numbers [2013/09/08 14:40]
nikolaj
arithmetic_structure_of_real_numbers [2013/09/08 14:41]
nikolaj
Line 3: Line 3:
 | @#FFBB00: $\langle \mathbb R,​+_\mathbb{R},​\cdot_\mathbb{R} \rangle$ | | @#FFBB00: $\langle \mathbb R,​+_\mathbb{R},​\cdot_\mathbb{R} \rangle$ |
  
-| @#55EE55: $ r +_\mathbb{R} s = \{q+_\mathbb{Q}p\ |\ q\in r\land p\in s\} $ | +| @#55EE55: $ r +_\mathbb{R} s = \{q+_\mathbb{Q}p\ |\ (q\in r)\land (p\in s)\} $ | 
-| @#55EE55: $ r -_\mathbb{R} s = \{q-_\mathbb{Q}p\ |\ q\in r\land p\in \mathbb Q\setminus s\} $ | +| @#55EE55: $ r -_\mathbb{R} s = \{q-_\mathbb{Q}p\ |\ (q\in r)\land (p\in \mathbb Q\setminus s)\} $ | 
-| @#55EE55: $ -_{\mathbb R}r = \{q-_\mathbb{Q}p\ |\ q<​0\land ​p\in \mathbb Q\setminus r\} $ |+| @#55EE55: $ -_{\mathbb R}r = \{q-_\mathbb{Q}p\ |\ (p\in \mathbb Q\setminus r)\land (q<0)\} $ |
  
-| @#55EE55: $ r\ge 0\land s\ge 0\implies r\cdot_\mathbb{R}s = \{q\cdot_\mathbb{Q}p\ |\ q\in r\land p\in s\land q,p\ge 0\}\cup\{q\ |\ q\in\mathbb Q\land q<​0\} ​ $ |+| @#55EE55: $ r\ge 0\land s\ge 0\implies r\cdot_\mathbb{R}s = \{q\cdot_\mathbb{Q}p\ |\ (q\in r)\land (p\in s)\land (q,p\ge 0)\}\cup\{q\ |\ (q\in\mathbb Q)\land (q<0)\}  $ |
 | @#55EE55: $ r\ge 0\land s <  0\implies r\cdot_\mathbb{R}s = -(r\cdot_\mathbb{R}(-s)) ​ $ | | @#55EE55: $ r\ge 0\land s <  0\implies r\cdot_\mathbb{R}s = -(r\cdot_\mathbb{R}(-s)) ​ $ |
 | @#55EE55: $ r  < 0\land s\ge 0\implies r\cdot_\mathbb{R}s = -((-r)\cdot_\mathbb{R}s) ​ $ | | @#55EE55: $ r  < 0\land s\ge 0\implies r\cdot_\mathbb{R}s = -((-r)\cdot_\mathbb{R}s) ​ $ |
 | @#55EE55: $ r  < 0\land s <  0\implies r\cdot_\mathbb{R}s = (-r)\cdot_\mathbb{R}(-s) ​ $ | | @#55EE55: $ r  < 0\land s <  0\implies r\cdot_\mathbb{R}s = (-r)\cdot_\mathbb{R}(-s) ​ $ |
  
-| @#55EE55: $ r\ge 0\land s >  0\implies r/​_\mathbb{R}s = \{q/​_\mathbb{Q}p\ |\ q\in r\land p\in \mathbb Q\setminus s\} $ |+| @#55EE55: $ r\ge 0\land s >  0\implies r/​_\mathbb{R}s = \{q/​_\mathbb{Q}p\ |\ (q\in r)\land (p\in \mathbb Q\setminus s)\} $ |
 | @#55EE55: $ r\ge 0\land s <  0\implies r/​_\mathbb{R}s = -(r/​_\mathbb{R}(-s)) ​ $ | | @#55EE55: $ r\ge 0\land s <  0\implies r/​_\mathbb{R}s = -(r/​_\mathbb{R}(-s)) ​ $ |
 | @#55EE55: $ r  < 0\land s > 0\implies r/​_\mathbb{R}s = -((-r)/​_\mathbb{R}s) ​ $ | | @#55EE55: $ r  < 0\land s > 0\implies r/​_\mathbb{R}s = -((-r)/​_\mathbb{R}s) ​ $ |
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