5 Classes de congruence - Nanopdf
Classes de congruence. Soit m ? N. On appelle classe de congruence de a modulo m l'ensemble de tous les entiers qui sont congrus à a modulo m;
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classe 4 qui contient : 4 9
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La relation de congruence modulo n est une relation d'équivalence. La classe d'équivalence de x est l'ensemble x + nZ souvent noté x s'il n'y a pas.
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Congruence and Congruence Classes - Oklahoma State University
The congruence class of a modulo n denoted [a] n is the set of all integers that are congruent to a modulo n; i e [a] n = fz 2Z ja z = kn for some k 2Zg : Example: In congruence modulo 2 we have [0] 2 = f0; 2; 4; 6;g [1] 2 = f 1; 3; 5; 7;g : Thus the congruence classes of 0 and 1 are respectively the sets of even and odd integers
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3 Congruence Congruences are an important and useful tool for the study of divisibility As we shall see they are also critical in the art of cryptography De nition 3 1If a and b are integers and n>0wewrite a bmodn to mean nj(b ?a) We read this as a is congruent to b modulo (or mod) n
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Existence On cherche une relation de Bezout 7u+ 31v= 1 par l’algorithme d’Euclide étendu + + + a i 4 2 3 u i 31 7 3 1 0 7 p i 0 1 4 9 31 +31 q i 1 0 1 2 7 Ontrouve31 2 7 9 = 1 Modulo31ona31 0doncceladevient7 9 1 (mod 31) Onmultipliepar11 donnant7 9 11 7 99 11 (mod 31)etonréduitmodulo31 parla
Searches related to classe de congruence PDF
3 Si (x;y) est solution de (E) alors d’apr es la question 2 x2 et 2y2 ont le m^eme reste modulo 5 Dans les deux tableaux pr ec edents les seuls cas qui v eri ent ce r esultats sont x 0 [5] et y 0 [5] Donc x et y sont multiples de 5 4 Si (x;y) est solution de (E) alors x = 5k et y = 5k0 avec k et k0deux entiers
LECTURE 11
Congruence and Congruence Classes
Definition11.1.Anequivalence relation~on a setSis a rule or test applicable to pairs of elements ofSsuch that (i)aa ;8a2S(re exive property) (ii)ab)ba(symmetric property) (iii)abandbc)ac(transitive property):You should think of an equivalence relation as a generalization of the notion of equality. Indeed, the usual
notion of equality among the set of integers is an example of an equivalence relation. The next denition
yields another example of an equivalence relation. Definition11.2.Leta;b;n2Zwithn >0. Thenaiscongruent tobmodulon; ab(modn) provided thatndividesab.Example.
175 (mod 6)
The following theorem tells us that the notion of congruence dened above is an equivalence relation on the
set of integers. Theorem11.3.Letnbe a positive integer. For alla;b;c2Z (i)aa(modn) (ii)ab(modn))ba(modn) (iii)ab(modn)andbc(modn))ac(modn):Proof.
(i)aa= 0 andnj0, henceaa(modn). (ii)ab(modn) means thatab=nkfor somek2Z. Therefore,ba=nk=n(k); hence ba(modn). (iii) Ifab(modn) andbc(modn), then ab=nk bc=nk0:Adding these two equations yields
ac=n(k+k0) ; and soac(modn). 3811. CONGRUENCE AND CONGRUENCE CLASSES 39
Theorem11.4.Ifab(modn)andcd(modn), then
(i)a+cb+d(modn) (ii)acbd(modn):Proof.
(i) By the denition of congruence there are integerssandtsuch thatab=snandcd=tn. Therefore, ab+cd=sn+tn=n(s+t) or, addingb+dto both sides of this equation, a+c=b+d+n(s+t):Hence,a+cb+d(modn).
(ii) Using the fact thatbc+bc= 0 we have acbd=ac+ 0bd =ac+ (bc+bc)bd =c(ab) +b(cd) =c(sn) +b(tn) =n(cs+bt) and sonj(acbd). Hence,acbd(modn). Definition11.5.Letaandnbe integers withn >0. Thecongruence class ofamodulon, denoted [a]n, is the set of all integers that are congruent toamodulon; i.e., [a]n=fz2Zjaz=knfor somek2Zg:Example:
In congruence modulo 2 we have
[0]2=f0;2;4;6;g
[1]2=f1;3;5;7;g:
Thus, the congruence classes of 0 and 1 are, respectively, the sets of even and odd integers.In congruence modulo 5 we have
[3]5=f3;35;310;315;g
=f;12;7;2;3;8;13;18;g:Theorem11.6.ac(modn) if and only if[a]n= [c]n.
Proof.
Assumeac(modn). Letb2[a]n. Then by denitionba(modn). By the transitivity property of congruence we then have ab(modn) andac(modn))bc(modn): Sob2[c]n. Thus, any elementbof [a]nis also an element of [c]n. Reversing the roles ofaandcin the argument above we similarly conclude that any element of [c]nis also an element of [a]n. Therefore [a]n= [c]n:11. CONGRUENCE AND CONGRUENCE CLASSES 40
Conversely, suppose [a] = [c]. Sinceaa(modn), by the re exive property of congruence, we havea2[a] and so, since by hypothesis [a] = [c],a2[c]. Hence, ac(modn): Recall that ifAandCare arbitrary sets, there are in general three possibilities: (i)A\C6=;andA=C (ii)A\C6=;andA6=C (iii)A\C=;:In the last case we say that the setsAandCaredisjoint. The following corollary says that for congruence
clases the immediary case (ii) does not exist. Corollary11.7.Two congruence classes modulonare either disjoint or identical.Proof.
If [a]nand [b]nare disjoint there is nothing to prove. Suppose then that [a]n\[b]n6=;. Then there is an
integerbsuch thatb2[a]nandb2[c]n. Soba(modn) andbcmodn). By the symmetry and transitivity properties of congruence we then have ac(modn):Hence [a]n= [c]nby Theorem 2.3.
Corollary11.8.There are exactlyndistinct congruence classes modulon; namely,[0]n,[1]n,[2]n,:::, [n1]n.Proof.
We rst show that no two of 0;1;2;:::;n1 are congruent modulon. To see this, suppose that0s < t < n :
Thentsis a positive integer andts < n. Thus,ndoes not dividetsand sotis not congruent tosmodulon. Since no two of 0;1;2;:::;n1 are congruent, the classes [0]n;[1]n;:::;[n1]nare all distinct,
by Theorem 2.3.To complete the proof we need to show that every congruence class is one of these classes. Leta2Z. By
the Division Algorithm, (11.1)a=nq+r or with 0r < n. The condition onrimpliesr2 f0;1;2;:::;n1g. If we rewrite (11.1) as ar=nq it is clear thatar(modn). Thus, any integerais congruent modulonto somer2 f0;1;2;:::;n1g. Definition11.9.The set of all congruence classes modulonis denotedZn(which is read \Zmodn"). Thus, Z n=f[0];[1];[2];:::;[n1]g:Note that while the set
Z=f:::;3;2;1;0;1;2;3;:::g
has an innite number of elements, the setZnhas onlynelements. Note also that the individual elements ofZnare not integers, but rather innte sets of integers; e.g., [2] =f:::;23n;22n;2n;2;2 +n;2 + 2n;2 + 3n;::::g:11. CONGRUENCE AND CONGRUENCE CLASSES 41
We proved last time that congruence modulonis an equivalence relation; i.e., (i)aa(modn) (ii)ab(modn))ba(modn) (iii)ab(modn) andbc(modn))ac(modn); and that congruence modulonalso is compatible with the addition and multiplication of integersTheorem11.10.Ifab(modn)andcd(modn), then
(i)a+cb+d(modn) (ii)acbd(modn): Definition11.11.Letaandnbe integers withn >0. Thecongruence class ofamodulon, denoted [a], is the set of all integers that are congruent toamodulon; i.e., [a] =fz2Zjaz=knfor somek2Zg:Example:
In congruence modulo 2 we have
[0]2=f0;2;4;6;g
[1]1=f1;3;5;7;g:
Thus, the congruence classes of 0 and 1 are, respectively, the sets of even and odd integers.In congruence modulo 5 we have
[3] =f3;35;310;315;g =f;12;7;2;3;8;13;18;g:Theorem11.12.ac(modn) if and only if[a]n= [c]n.
Proof.
Assumeac(modn). Letb2[a]. Then by denitionba(modn). By the transitivity property of congruence we then have ab(modn) andac(modn))bc(modn): Sob2[c]n. Thus, any elementbof [a]nis also an element of [c]. Reversing the roles ofaandcin the argument above we similarly conclude that any element of [c]nis also an element of [a]n. Therefore [a] = [c]: Conversely, suppose [a] = [c]. Sinceaa(modn), by the re exive property of congruence, we havea2[a] and so, since by hypothesis [a] = [c],a2[c]. Hence, ac(modn): Recall that ifAandCare arbitrary sets, there are in general three possibilities: (i)A\C6=;andA=C (ii)A\C6=;andA6=C (iii)A\C=;:In the last case we say that the setsAandCaredisjoint. The following corollary says that for congruence
clases the immediary case (ii) does not exist. Corollary11.13.Two congruence classes modulonare either disjoint or identical.11. CONGRUENCE AND CONGRUENCE CLASSES 42
Proof.
If [a]nand [b]nare disjoint there is nothing to prove. Suppose then that [a]n\[b]n6=;. Then there is an
integerbsuch thatb2[a]nandb2[c]n. Soba(modn) andbcmodn). By the symmetry and transitivity properties of congruence we then have ac (modn):Hence [a]n= [c]nby Theorem 2.3.
Corollary11.14.There are exactlyndistinct congruence classes modulon; namely, [0], [1], [2],:::,[n-1]
Proof.
We rst show that no two of 0;1;2;:::;n1 are congruent modulon. To see this, suppose that0s < t < n :
Thentsis a positive integer andts < n. Thus,ndoes not dividetsand sotis not congruent tosmodulon. Since no two of 0;1;2;:::;n1 are congruent, the classes [0]n;[1]n;:::;[n1]nare all distinct,
by Theorem 2.3.To complete the proof we need to show that every congruence class is one of these classes. Leta2Z. By
the Division Algorithm, (11.2)a=nq+r or with 0r < n. The condition onrimpliesr2 f0;1;2;:::;n1g. If we rewrite (11.2) as ar=nq it is clear thatar(modn). Thus, any integerais congruent modulonto somer2 f0;1;2;:::;n1g. Definition11.15.The set of all congruence classes modulonis denotedZn(which is read\Zmodn"). Thus, Z n=f[0]n;[1]n;[2]n;:::;[n1]ng:Note that while the set
Z=f:::;3;2;1;0;1;2;3;:::g
has an innite number of elements, the setZnhas onlynelements. Note also that the individual elements ofZnare not integers, but rather innte sets of integers; e.g., [2] n=f:::;23n;22n;2n;2;2 +n;2 + 2n;2 + 3n;::::g:quotesdbs_dbs44.pdfusesText_44[PDF] écart type relatif
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