# Differences

This shows you the differences between two versions of the page.

 integer [2013/09/03 00:39]nikolaj integer [2013/09/03 00:40]nikolaj Both sides previous revision Previous revision 2013/09/03 00:40 nikolaj 2013/09/03 00:40 nikolaj 2013/09/03 00:39 nikolaj 2013/09/03 00:39 nikolaj 2013/09/03 00:15 nikolaj 2013/09/02 23:54 nikolaj 2013/09/02 23:53 nikolaj 2013/09/02 23:53 nikolaj 2013/09/02 23:53 nikolaj 2013/09/02 23:50 nikolaj 2013/09/02 23:50 nikolaj 2013/09/02 23:49 nikolaj 2013/09/02 23:49 nikolaj 2013/09/02 23:47 nikolaj 2013/09/02 23:45 nikolaj 2013/09/02 23:45 nikolaj 2013/08/31 16:50 nikolaj 2013/08/12 15:22 nikolaj 2013/08/06 01:45 nikolaj 2013/08/06 01:33 nikolaj 2013/05/20 18:54 nikolaj 2013/05/19 23:15 nikolaj 2013/05/19 23:15 nikolaj 2013/05/19 23:15 nikolaj created Next revision Previous revision 2013/09/03 00:40 nikolaj 2013/09/03 00:40 nikolaj 2013/09/03 00:39 nikolaj 2013/09/03 00:39 nikolaj 2013/09/03 00:15 nikolaj 2013/09/02 23:54 nikolaj 2013/09/02 23:53 nikolaj 2013/09/02 23:53 nikolaj 2013/09/02 23:53 nikolaj 2013/09/02 23:50 nikolaj 2013/09/02 23:50 nikolaj 2013/09/02 23:49 nikolaj 2013/09/02 23:49 nikolaj 2013/09/02 23:47 nikolaj 2013/09/02 23:45 nikolaj 2013/09/02 23:45 nikolaj 2013/08/31 16:50 nikolaj 2013/08/12 15:22 nikolaj 2013/08/06 01:45 nikolaj 2013/08/06 01:33 nikolaj 2013/05/20 18:54 nikolaj 2013/05/19 23:15 nikolaj 2013/05/19 23:15 nikolaj 2013/05/19 23:15 nikolaj created 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 - * $0 \equiv [(0,0)] = [(1,1)] = \dots = [(k,k)]$ + * $0 \equiv [\langle0,0\rangle] = [\langle1,1\rangle] = \dots = [\langle ​k,k\rangle]$ - * $1 \equiv [(1,0)] = [(2,1)] = \dots = [(k+1,k)]$ + * $1 \equiv [\langle1,0\rangle] = [\langle2,1\rangle] = \dots = [\langle ​k+1,k\rangle]$ - * $-1 \equiv [(0,1)] = [(1,2)] = \dots = [(k,k+1)]$ + * $-1 \equiv [\langle0,1\rangle] = [\langle1,2\rangle] = \dots = [\langle ​k,k+1\rangle]$ - * $2 \equiv [(2,0)] = [(3,1)] = \dots = [(k+2,k)]$ + * $2 \equiv [\langle2,0\rangle] = [\langle3,1\rangle] = \dots = [\langle ​k+2,k\rangle]$ - * $-2 \equiv [(0,2)] = [(1,3)] = \dots = [(k,k+2)]$ + * $-2 \equiv [\langle0,2\rangle] = [\langle1,3\rangle] = \dots = [\langle ​k,k+2\rangle]$ - * $3 \equiv [(0,3)] = \dots$ + * $3 \equiv [\langle0,3\rangle] = \dots$ where $k$ is any natural number. where $k$ is any natural number.