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integer [2013/09/03 00:39]
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 [(0,0)] = [(1,1)] = \dots = [(k,k)] $+  * $ 0 \equiv [\langle0,0\rangle] = [\langle1,1\rangle] = \dots = [\langlek,k\rangle] $
  
-  * $ 1 \equiv [(1,0)] = [(2,1)] = \dots = [(k+1,k)] $+  * $ 1 \equiv [\langle1,0\rangle] = [\langle2,1\rangle] = \dots = [\langlek+1,k\rangle] $
  
-  * $ -1 \equiv [(0,1)] = [(1,2)] = \dots = [(k,k+1)] $+  * $ -1 \equiv [\langle0,1\rangle] = [\langle1,2\rangle] = \dots = [\langlek,k+1\rangle] $
  
-  * $ 2 \equiv [(2,0)] = [(3,1)] = \dots = [(k+2,k)] $+  * $ 2 \equiv [\langle2,0\rangle] = [\langle3,1\rangle] = \dots = [\langlek+2,k\rangle] $
  
-  * $ -2 \equiv [(0,2)] = [(1,3)] = \dots = [(k,k+2)] $+  * $ -2 \equiv [\langle0,2\rangle] = [\langle1,3\rangle] = \dots = [\langlek,k+2\rangle] $
  
-  * $ 3 \equiv [(0,3)] = \dots $+  * $ 3 \equiv [\langle0,3\rangle] = \dots $
  
 where $k$ is any natural number. where $k$ is any natural number.
  
 === Theorems === === Theorems ===
-$-[\langle a,​b\rangle]$ is the additive inverse of $[\langle a,​b\rangle]$ and can be computed as+The integer ​$-[\langle a,​b\rangle]$ is the additive inverse of $[\langle a,​b\rangle]$ and can be computed as
  
 $-[\langle a,​b\rangle]=[\langle b,​a\rangle]$ $-[\langle a,​b\rangle]=[\langle b,​a\rangle]$
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