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integer [2013/09/03 00:39] nikolaj |
integer [2013/09/03 00:40] nikolaj |
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| 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 = [\langle k,k\rangle] $ |
| - | * $ 1 \equiv [(1,0)] = [(2,1)] = \dots = [(k+1,k)] $ | + | * $ 1 \equiv [\langle1,0\rangle] = [\langle2,1\rangle] = \dots = [\langle k+1,k\rangle] $ |
| - | * $ -1 \equiv [(0,1)] = [(1,2)] = \dots = [(k,k+1)] $ | + | * $ -1 \equiv [\langle0,1\rangle] = [\langle1,2\rangle] = \dots = [\langle k,k+1\rangle] $ |
| - | * $ 2 \equiv [(2,0)] = [(3,1)] = \dots = [(k+2,k)] $ | + | * $ 2 \equiv [\langle2,0\rangle] = [\langle3,1\rangle] = \dots = [\langle k+2,k\rangle] $ |
| - | * $ -2 \equiv [(0,2)] = [(1,3)] = \dots = [(k,k+2)] $ | + | * $ -2 \equiv [\langle0,2\rangle] = [\langle1,3\rangle] = \dots = [\langle k,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. | ||
