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integer [2013/09/03 00:40]
nikolaj
integer [2013/09/03 00:40]
nikolaj
Line 12: Line 12:
 So if $[\langle a,​b\rangle]$ is the equivalence class of $\langle a,b\rangle$ with respect to the equivalence relation $\{\langle \langle a,​b\rangle,​\langle m,​n\rangle\rangle\ |\ a+n = b+m )\}$, we have So if $[\langle a,​b\rangle]$ is the equivalence class of $\langle a,b\rangle$ with respect to the equivalence relation $\{\langle \langle a,​b\rangle,​\langle m,​n\rangle\rangle\ |\ a+n = b+m )\}$, we have
  
-  * $ 0 \equiv [\langle0,​0\rangle] = [\langle1,​1\rangle] = \dots = [\langlek,k\rangle] $+  * $ 0 \equiv [\langle0,​0\rangle] = [\langle1,​1\rangle] = \dots = [\langle k,k\rangle] $
  
-  * $ 1 \equiv [\langle1,​0\rangle] = [\langle2,​1\rangle] = \dots = [\langlek+1,​k\rangle] $+  * $ 1 \equiv [\langle1,​0\rangle] = [\langle2,​1\rangle] = \dots = [\langle k+1,​k\rangle] $
  
-  * $ -1 \equiv [\langle0,​1\rangle] = [\langle1,​2\rangle] = \dots = [\langlek,​k+1\rangle] $+  * $ -1 \equiv [\langle0,​1\rangle] = [\langle1,​2\rangle] = \dots = [\langle k,​k+1\rangle] $
  
-  * $ 2 \equiv [\langle2,​0\rangle] = [\langle3,​1\rangle] = \dots = [\langlek+2,​k\rangle] $+  * $ 2 \equiv [\langle2,​0\rangle] = [\langle3,​1\rangle] = \dots = [\langle k+2,​k\rangle] $
  
-  * $ -2 \equiv [\langle0,​2\rangle] = [\langle1,​3\rangle] = \dots = [\langlek,​k+2\rangle] $+  * $ -2 \equiv [\langle0,​2\rangle] = [\langle1,​3\rangle] = \dots = [\langle k,​k+2\rangle] $
  
   * $ 3 \equiv [\langle0,​3\rangle] = \dots $   * $ 3 \equiv [\langle0,​3\rangle] = \dots $
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