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 integer [2013/09/03 00:40]nikolaj integer [2014/03/21 11:11] (current) 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 1: Line 1: ===== Integer ===== ===== Integer ===== - ==== Definition ​==== + ==== Set ==== - | @#FFBB00: $\mathbb Z \equiv \mathbb N\times\mathbb N\ /\ \{\langle \langle a,​b\rangle,​\langle m,​n\rangle\rangle\ |\ a+n = b+m )\}$ | + | @#FFBB00: definiendum ​| @#FFBB00: $\mathbb Z \equiv \mathbb N\times\mathbb N\ /\ \{\langle \langle a,​b\rangle,​\langle m,​n\rangle\rangle\ |\ a+n = b+m )\}$ | with $a,b,n,m\in \mathbb N$. with $a,b,n,m\in \mathbb N$. 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 - * $0 \equiv [\langle0,​0\rangle] = [\langle1,​1\rangle] = \dots = [\langlek,k\rangle]$ + * $0 \equiv [\langle0,​0\rangle] = [\langle1,​1\rangle] = \dots = [\langle k,k\rangle]$ - * $1 \equiv [\langle1,​0\rangle] = [\langle2,​1\rangle] = \dots = [\langlek+1,​k\rangle]$ + * $1 \equiv [\langle1,​0\rangle] = [\langle2,​1\rangle] = \dots = [\langle k+1,​k\rangle]$ - * $-1 \equiv [\langle0,​1\rangle] = [\langle1,​2\rangle] = \dots = [\langlek,​k+1\rangle]$ + * $-1 \equiv [\langle0,​1\rangle] = [\langle1,​2\rangle] = \dots = [\langle k,​k+1\rangle]$ - * $2 \equiv [\langle2,​0\rangle] = [\langle3,​1\rangle] = \dots = [\langlek+2,​k\rangle]$ + * $2 \equiv [\langle2,​0\rangle] = [\langle3,​1\rangle] = \dots = [\langle k+2,​k\rangle]$ - * $-2 \equiv [\langle0,​2\rangle] = [\langle1,​3\rangle] = \dots = [\langlek,​k+2\rangle]$ + * $-2 \equiv [\langle0,​2\rangle] = [\langle1,​3\rangle] = \dots = [\langle k,​k+2\rangle]$ * $3 \equiv [\langle0,​3\rangle] = \dots$ * $3 \equiv [\langle0,​3\rangle] = \dots$ Line 33: Line 33: === Reference === === Reference === Wikipedia: [[http://​en.wikipedia.org/​wiki/​Integer|Integer]] Wikipedia: [[http://​en.wikipedia.org/​wiki/​Integer|Integer]] - ==== Context ​==== + ==== Parents ​==== === Subset of === === Subset of === [[Quotient set]] [[Quotient set]] === Refinement of === === Refinement of === [[Rational number]] [[Rational number]] - === Requirements ​=== + === Context ​=== [[Arithmetic structure of natural numbers]] [[Arithmetic structure of natural numbers]]