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Nvmeri integri: constrvctio, signa et ordo

Svmmarivm

Haec lectio nvmeros integros ex natvralibvs constrvit. Qvisqve integer definitvr vt classis parivm nvmerorvm natvralivm (a,b)(a, b), qvae svbtractionem a−ba - b repraesentant etiamsi haec inter natvrales non exsistit; svmma et prodvctvm classivm per repraesentantes definivntvr, et demonstratvr id qvod efficitvr a repraesentantibvs electis non pendere. Natvrales intra integros continentvr, nihil et opposita apparent, et ostenditvr omnem integrvm esse natvralem, nihil avt oppositvm natvralis. Hoc fvndamento demonstrantvr regvla signorvm, defectvs divisorvm nihili, cancellatio mvltiplicationis et proprietates ordinis, et divisio cvm residvo dividendi negativi ad divisionem dividendi natvralis redvcitvr.

Proposita discendi

  1. Integros vt classes parivm nvmerorvm natvralivm constrvere et demonstrare aeqvivalentiam parivm relationem aeqvivalentiae esse, svmmamqve et prodvctvm classivm bene definita esse.
  2. Natvrales, nihil et opposita intra integros agnoscere, et demonstrare omnem integrvm vnvm ex eis exacte esse.
  3. Regvlam signorvm, defectvm divisorvm nihili et cancellationem mvltiplicationis ex proprietatibvs additionis et mvltiplicationis demonstrare.
  4. Ordinem integrorvm definire, eivs cvm operationibvs congrventiam demonstrare et ad divisionem cvm residvo dividendorvm negativorvm adhibere.

Constrvctio nvmerorvm integrorvm

In lectione de generibvs nvmerorvm integri propositi svnt vt copia qvae natvrales, nihil et opposita natvralivm complectitvr, et eorvm proprietates envntiatae svnt. Atqvi proprietates envntiare non est eas demonstrare: qvamdiv non dicitvr qvid sit nvmervs negativvs, regvla signorvm avt congrventia ordinis cvm mvltiplicatione conventiones svnt, non theoremata. Haec lectio integros constrvit solis instrvmentis axiomatvm Peani et operationvm nvmerorvm natvralivm, et omnes has regvlas ex constrvctione dedvcit. Vt in reliqvo cvrsv, natvrales a 11 incipivnt; nihil in hac constrvctione primvm apparet.

Consilivm hoc est. Aeqvatio x+b=ax + b = a solvtionem natvralem habet tantvm cvm b<ab < a, et haec solvtio est differentia a−ba - b. Qvod si ita est, nvmervs negativvs concipi potest vt svbtractio svspensa, descripta per par nvmerorvm natvralivm (a,b)(a, b) qvod eam efficeret. Mvlta paria eandem svbtractionem describvnt; exempli gratia, (5,2)(5, 2), (4,1)(4, 1) et (9,6)(9, 6) nvmervm 33 describvnt, et (2,5)(2, 5) et (1,4)(1, 4) describvnt id qvod nondvm exsistit et erit −3-3. Dvo paria (a,b)(a, b) et (c,d)(c, d) eandem svbtractionem describvnt cvm a−b=c−da - b = c - d; haec condicio svbtractionibvs vtitvr qvae fortasse non exsistvnt, sed addito b+db + d vtriqve membro fit aeqvalitas inter natvrales qvae semper sensvm habet:

a−b=c−d↔a+d=b+ca - b = c - d \leftrightarrow a + d = b + c
Definitio (paria aeqvivalentia). Dvo paria nvmerorvm natvralivm (a,b)(a, b) et (c,d)(c, d) aeqvivalentia svnt, et scribitvr (a,b)∼(c,d)(a, b) \sim (c, d), cvm a+d=b+ca + d = b + c.

In reticvlo qvorvm pvncta svnt paria nvmerorvm natvralivm, cvm aa in axe horizontali et bb in verticali, par (a+1,b+1)(a + 1, b + 1) aeqvivalens est ipsi (a,b)(a, b), qvia a+(b+1)=b+(a+1)a + (b + 1) = b + (a + 1); vnde paria dato pari aeqvivalentia in recta diagonali parallela disponvntvr, in qva vno loco dextrorsvm et vno svrsvm progreditvr. Vt haec relatio ad paria identificanda valeat, debet se gerere vt aeqvalitas: omne par sibi ipsi aeqvivalens est, relatio non pendet ab ordine qvo paria nominantvr, et dvo paria tertio aeqvivalentia inter se aeqvivalentia svnt. Id est, debet esse reflexiva, symmetrica et transitiva: relatio aeqvivalentiae, sensv lectionis de copiis, relationibvs et fvnctionibvs, cvivs exemplvm de relatione qvae integros constrvet hanc constrvctionem praenvntiabat.

LemmaAeqvivalentia parivm

Aeqvivalentia parivm nvmerorvm natvralivm reflexiva, symmetrica et transitiva est: relatio aeqvivalentiae est. Dvae primae proprietates ad commvtativitatem additionis redigvntvr; tertia cancellationem additionis postvlat, demonstratam in problemate solvto 1 lectionis de operationibvs nvmerorvm natvralivm. Figvra paria (5,2)(5, 2), (4,1)(4, 1) et (7,4)(7, 4) seqvitvr.

Demonstratio

  1. a+b=b+aa + \resaltar{b} = b + \resaltar{a}

    Reflexivitas: par (a,b)(a, b) sibi ipsi aeqvivalens est, qvia condicio est commvtativitas additionis.

  2. a+d=b+c→c+b=d+aa + d = b + c \rightarrow \resaltar{c + b} = \resaltar{d + a}

    Symmetria: si (a,b)(a, b) aeqvivalens est ipsi (c,d)(c, d), membris permvtatis et singvlis svmmis commvtatis obtinetvr condicio pro (c,d)(c, d) et (a,b)(a, b).

  3. a+d=b+c,c+f=d+e\dato{h1}{a + d = b + c}, \quad \dato{h2}{c + f = d + e}

    Transitivitas: ponamvs (a,b)(a, b) aeqvivalens esse ipsi (c,d)(c, d), et (c,d)(c, d) ipsi (e,f)(e, f).

  4. (a+d)+(c+f)=(b+c)+(d+e)\resaltar{(a + d)} + \resaltar{(c + f)} = \resaltar{(b + c)} + \resaltar{(d + e)}

    Dvas aeqvalitates membratim addimvs.

  5. (a+f)+(c+d)=(b+e)+(c+d)\resaltar{(a + f)} + (c + d) = \resaltar{(b + e)} + (c + d)

    Per associativitatem et commvtativitatem qvattvor addenda vtrivsqve membri ita aliter colligimvs vt c+dc + d in vtroqve appareat.

  6. a+f=b+e\dato{t}{a + f = b + e}

    Per cancellationem additionis tollitvr c+dc + d: paria (a,b)(a, b) et (e,f)(e, f) aeqvivalentia svnt.

Integri vt classes parivm

Cvm aeqvivalentia parivm relatio aeqvivalentiae sit, valent pro ea generalia lectionis de copiis, relationibvs et fvnctionibvs, qvae repetere non oportet. Classis paris (a,b)(a, b) est copia omnivm parivm nvmerorvm natvralivm ipsi (a,b)(a, b) aeqvivalentivm; in notatione illivs lectionis esset C((a,b))C((a, b)), et hic brevivs scribitvr C(a,b)C(a, b). Qvodqve par classis eivs repraesentans est. In reticvlo classis est copia pvnctorvm rectae diagonali parallelae. Per lemma de classibvs aeqvivalentiae ibi demonstratvm, dvo paria aeqvivalentia svnt si et solvm si eorvm classes aeqvales svnt, et dvae classes aliqvod par commvne habentes congrvvnt; et per theorema de aeqvivalentiis et partitionibvs classes partitionem parivm nvmerorvm natvralivm efficivnt. Ex qvo seqvitvr regvla qva dvae classes comparantvr:

C(a,b)=C(c,d)↔a+d=b+cC(a, b) = C(c, d) \leftrightarrow a + d = b + c
Definitio (nvmeri integri). Nvmervs integer est classis parivm nvmerorvm natvralivm. Copia integrorvm per Z\mathbb{Z} notatvr; est copia qvotiens parivm nvmerorvm natvralivm secvndvm aeqvivalentiam parivm.
Z=(N×N)/∼\mathbb{Z} = \left( \mathbb{N} \times \mathbb{N} \right) / {\sim}

Definitio mira videri potest, qvia nvmervs fit copia infinita parivm. Sed id iam in fractionibvs accidit: 12\frac{1}{2} et 24\frac{2}{4} svnt diversae scriptvrae eivsdem nvmeri, et id qvod definitvr est qvod omnibvs commvne est. Classis C(5,2)C(5, 2) erit 33; classis C(2,5)C(2, 5), −3-3; classis C(1,1)C(1, 1), 00. Anteqvam his nominibvs vtamvr, operationes definiendae svnt et confirmandvm est natvrales re vera intra integros contineri.

Svmma et prodvctvm classivm

Operationes definivntvr imitando qvod fieret si svbtractiones exsisterent. Si (a,b)(a, b) repraesentat a−ba - b et (c,d)(c, d) repraesentat c−dc - d, svmma vtrivsqve svbtractionis esset (a+c)−(b+d)(a + c) - (b + d), et prodvctvm, evolvtvm, (ac+bd)−(ad+bc)(ac + bd) - (ad + bc). Hic calcvlvs non est demonstratio, qvia svbtractionibvs vtitvr qvae nondvm exsistvnt; ad definitiones eligendas inservit, in qvibvs solae svmmae et prodvcta nvmerorvm natvralivm intervenivnt:

C(a,b)+C(c,d)=C(a+c,b+d)C(a, b) + C(c, d) = C(a + c, b + d)
C(a,b)⋅C(c,d)=C(ac+bd,ad+bc)C(a, b) \cdot C(c, d) = C(ac + bd, ad + bc)

Definitio cvm repraesentantibvs compvtat, et eadem classis infinitos habet. Si mvtato repraesentante mvtaretvr qvod efficitvr, definitio nihil definiret: plvres valores eidem operationi daret. Demonstrandvm est igitvr id qvod efficitvr a solis classibvs pendere; tvm operatio bene definita esse dicitvr.

TheoremaSvmma et prodvctvm a repraesentantibvs non pendent

Si paria (a,b)(a, b) et (p,q)(p, q) aeqvivalentia svnt, tvm pro omni pari (c,d)(c, d) paria (a+c,b+d)(a + c, b + d) et (p+c,q+d)(p + c, q + d) aeqvivalentia svnt, itemqve (ac+bd,ad+bc)(ac + bd, ad + bc) et (pc+qd,pd+qc)(pc + qd, pd + qc). Qvam ob rem svmma et prodvctvm dvorvm integrorvm a repraesentantibvs electis non pendent. Satis est singvlos repraesentantes vicissim mvtare: primvm primi addendi vel factoris, deinde secvndi. Figvra repraesentantibvs (3,1)(3, 1) et (4,2)(4, 2) eivsdem classis vtitvr, et pari (1,2)(1, 2).

Demonstratio

  1. a+q=b+p\dato{h}{a + q = b + p}

    Ab hypothesi proficiscimvr: (a,b)(a, b) et (p,q)(p, q) aeqvivalentia svnt.

  2. (a+c)+(q+d)=(a+q)+(c+d)=(b+p)+(c+d)=(b+d)+(p+c)(a + c) + (q + d) = \resaltar{(a + q)} + (c + d) = \resaltar{(b + p)} + (c + d) = (b + d) + (p + c)

    Svmma: per associativitatem et commvtativitatem aliter colligimvs et hypothesim svbstitvimvs; prodit condicio aeqvivalentiae ipsorvm (a+c,b+d)(a + c, b + d) et (p+c,q+d)(p + c, q + d).

  3. (ac+bd)+(pd+qc)=c(a+q)+d(b+p)(ac + bd) + (pd + qc) = \resaltar{c(a + q)} + \resaltar{d(b + p)}

    Prodvctvm: ostendendvm est (ac+bd)+(pd+qc)(ac + bd) + (pd + qc) aeqvale esse ipsi (ad+bc)+(pc+qd)(ad + bc) + (pc + qd). Per distribvtivitatem membrvm privs in dvo prodvcta colligitvr.

  4. c(b+p)+d(a+q)=bc+pc+ad+qd=(ad+bc)+(pc+qd)c\resaltar{(b + p)} + d\resaltar{(a + q)} = bc + pc + ad + qd = (ad + bc) + (pc + qd)

    Hypothesim in dvas parentheses svbstitvimvs et itervm distribvimvs: obtinetvr membrvm posterivs.

  5. (ca+db,cb+da)=(ac+bd,ad+bc)(ca + db, cb + da) = (\resaltar{ac + bd}, \resaltar{ad + bc})

    Ad repraesentantem secvndi addendi vel factoris mvtandvm satis est animadvertere definitiones symmetricas esse: per commvtativitatem nvmerorvm natvralivm par qvod C(c,d)⋅C(a,b)C(c, d) \cdot C(a, b) definit idem est atqve par qvod C(a,b)⋅C(c,d)C(a, b) \cdot C(c, d) definit, et idem in svmma accidit.

  6. C(a,b)=C(p,q)→C(a,b)⋅C(c,d)=C(p,q)⋅C(c,d)C(a, b) = C(p, q) \rightarrow C(a, b) \cdot C(c, d) = C(p, q) \cdot C(c, d)

    Mvtato primvm vno repraesentante, deinde altero, classis eivs qvod efficitvr non mvtatvr: svmma et prodvctvm integrorvm bene definita svnt.

Proprietates additionis et mvltiplicationis

Operationibvs bene definitis, qvaeqve proprietas integrorvm compvtando cvm repraesentantibvs demonstratvr, et qvaeqve proprietatem nvmerorvm natvralivm qva nititvr hereditate accipit. Pro omnibvs integris xx, yy et zz hae proprietates valent:

  1. Commvtativitas et associativitas additionis, a nvmeris natvralibvs singvlis componentibvs acceptae.
  2. Commvtativitas mvltiplicationis, accepta a commvtativitate additionis et mvltiplicationis nvmerorvm natvralivm, vt in gradv 5 demonstrationis svperioris.
  3. Associativitas mvltiplicationis et distribvtivitas, acceptae ab associativitate, commvtativitate et distribvtivitate nvmerorvm natvralivm.
  4. Elementvm nevtrvm additionis: classis C(1,1)C(1, 1), qvae omnia paria (m,m)(m, m) continet.
  5. Oppositvm: pro qvaqve classe C(a,b)C(a, b), classis C(b,a)C(b, a), qvae obtinetvr paribvs ad diagonalem reflexis.
  6. Elementvm nevtrvm mvltiplicationis: classis C(2,1)C(2, 1).

Qvattvor postremae breves calcvli svnt. Pro distribvtivitate vtrvmqve membrvm per definitiones evolvitvr et idem par obtinetvr:

C(a,b)⋅C(c+e,d+f)=C(ac+bd+ae+bf,ad+bc+af+be)=C(ac+bd,ad+bc)+C(ae+bf,af+be)C(a, b) \cdot C(c + e, d + f) = C(ac + bd + ae + bf, ad + bc + af + be) = C(ac + bd, ad + bc) + C(ae + bf, af + be)

Pro associativitate mvltiplicationis dvo modi colligendi C(a,b)⋅C(c,d)⋅C(e,f)C(a, b) \cdot C(c, d) \cdot C(e, f) post distribvtionem idem par svmmarvm qvattvor terminorvm dant:

C(ace+adf+bcf+bde,acf+ade+bce+bdf)C(ace + adf + bcf + bde, acf + ade + bce + bdf)

Pro elementis nevtris et opposito satis est regvlam comparandi classes adhibere: C(a+1,b+1)=C(a,b)C(a + 1, b + 1) = C(a, b), qvia (a+1)+b=(b+1)+a(a + 1) + b = (b + 1) + a; par (2a+b,a+2b)(2a + b, a + 2b), qvod C(a,b)⋅C(2,1)C(a, b) \cdot C(2, 1) definit, aeqvivalens est ipsi (a,b)(a, b), qvia vtrvmqve membrvm condicionis 2a+2b2a + 2b valet; et C(a,b)+C(b,a)=C(a+b,a+b)C(a, b) + C(b, a) = C(a + b, a + b), qvae est classis parivm dvos terminos aeqvales habentivm.

C(a,b)+C(1,1)=C(a,b),C(a,b)⋅C(2,1)=C(a,b),C(a,b)+C(b,a)=C(1,1)C(a, b) + C(1, 1) = C(a, b), \qquad C(a, b) \cdot C(2, 1) = C(a, b), \qquad C(a, b) + C(b, a) = C(1, 1)