Editing Integer (section)

=== Equivalence classes of ordered pairs ===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of [[natural number]]s. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]

In modern set-theoretic mathematics, a more abstract construction<ref>Ivorra Castillo: ''Álgebra''</ref><ref>{{Cite book |last1=Kramer |first1=Jürg |title=From Natural Numbers to Quaternions |last2=von Pippich |first2=Anna-Maria |publisher=Springer Cham |year=2017 |isbn=978-3-319-69427-6 |edition=1st |location=Switzerland |pages=78–81 |language=en |doi=10.1007/978-3-319-69429-0}}</ref> allowing one to define arithmetical operations without any case distinction is often used instead.<ref>{{cite book |title=Learning to Teach Number: A Handbook for Students and Teachers in the Primary School |series=The Stanley Thornes Teaching Primary Maths Series |first=Len |last=Frobisher |publisher=Nelson Thornes |year=1999 |isbn=978-0-7487-3515-0 |page=126 |url=https://books.google.com/books?id=KwJQIt4jQHUC&pg=PA126 |access-date=2016-02-15 |archive-url=https://web.archive.org/web/20161208121843/https://books.google.com/books?id=KwJQIt4jQHUC&pg=PA126 |archive-date=2016-12-08 |url-status=live}}.</ref> The integers can thus be formally constructed as the [[equivalence class]]es of [[ordered pair]]s of [[natural number]]s {{math|(''a'',''b'')}}.<ref name="Campbell-1970-p83">{{cite book |author=Campbell, Howard E. |title=The structure of arithmetic |publisher=Appleton-Century-Crofts |year=1970 |isbn=978-0-390-16895-5 |page=[https://archive.org/details/structureofarith00camp/page/83 83] |url-access=registration |url=https://archive.org/details/structureofarith00camp/page/83 }}</ref>

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}.<ref name="Campbell-1970-p83"/> To confirm our expectation that {{nowrap|1 − 2}} and {{nowrap|4 − 5}} denote the same number, we define an [[equivalence relation]] {{math|~}} on these pairs with the following rule:
:<math>(a,b)\sim(c,d) </math>
precisely when
:<math>a+d=b+c </math>.

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers;<ref name="Campbell-1970-p83"/> by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)]+[(c,d)]:=[(a+c,b+d)]</math>.
:<math>[(a,b)]\cdot[(c,d)]:=[(ac+bd,ad+bc)]</math>.

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)]:=[(b,a)]</math>.

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)]-[(c,d)]:=[(a+d,b+c)]</math>.

The standard ordering on the integers is given by:
:<math>[(a,b)]<[(c,d)]</math> [[if and only if]] <math>a+d<b+c</math>.

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are [[embedding|embedded]] into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since −0&nbsp;=&nbsp;0.

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases}a-b,&\mbox{if }a\ge b\\-(b-a),&\mbox{if }a<b\end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar [[group representation|representation]] of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}0&=[(0,0)]&=[(1,1)]&=\cdots& &=[(k,k)]\\1&=[(1,0)]&=[(2,1)]&=\cdots&&=[(k+1,k)]\\-1&=[(0,1)]&=[(1,2)]&=\cdots&&=[(k,k+1)]\\2&=[(2,0)]&=[(3,1)]&=\cdots&&=[(k+2,k)]\\-2&=[(0,2)]&= [(1,3)]&=\cdots&&=[(k,k+2)]\end{align}</math>