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[SPAM] [obm-l] Res: [obm-l] RES: [obm-l] Res: [obm-l] demonstração: pequeno teorema de FERMAT
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- Subject: [SPAM] [obm-l] Res: [obm-l] RES: [obm-l] Res: [obm-l] demonstração: pequeno teorema de FERMAT
- From: Rodrigo Cientista <rodrigocientista@xxxxxxxxxxxx>
- Date: Mon, 26 Nov 2007 16:54:53 -0800 (PST)
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--0-2054223372-1196124893=:2781
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Por indu=E7=E3o, =E9 simples!!=0A=0ASabemos que n^p =3D=3D n mop p para alg=
um n(n=3D1, por exemplo), queremos saber se =E9 v=E1lido para todo n.=0A=0A=
expandindo, (n+1)^p =3D n^p + C_p,1*a^p-1 + ... + C_p,k*a^p-k + ... + 1=0A=
=0Aobs*** C_x,y =3D combina=E7=E3o de x e y=0A=0AComo p divide C_p,k (pois =
o numerador =E9 p! =3D p(p-1)(p-2)...), segue (n+1)^p =3D=3D n^p + 1 mod p=
=0A=0AMas por hip=F3tese de indu=E7=E3o, j=E1 estava provado que n^p =3D=3D=
n mop p oq implica n^p +1 =3D=3D n + 1 mop p=0A=0AAssim, (n+1)^p =3D=3D n =
+ 1 mod p, provando Fermat por indu=E7=E3o sobre n=0A=0ARealmente, essa =E9=
a prova mais simples, mas n=E3o =E9 minha=0A=0A-----Mensagem original-----=
=0ADe: owner-obm-l@xxxxxxxxxxxxxx [mailto:owner-obm-l@xxxxxxxxxxxxxx]Em nom=
e de Rodrigo Cientista=0AEnviada em: segunda-feira, 26 de novembro de 2007 =
13:41=0APara: obm-l@xxxxxxxxxxxxxx=0AAssunto: [obm-l] Res: [obm-l] demonstr=
a=E7=E3o: pequeno teorema de FERMAT=0A=0A=0ASalhab, realmente houve uma fal=
ha=0A =0Ao teorema diz que n^p =3D=3Dn mod p, o que n=E3o sabemos...=0A =0A=
seja x um resto qualquer da divis=E3o de n por p, tal que n =3D=3D x mod p=
=0A =0Aseja um k qualquer tal que x-k =3D 1 (chamarei de r) e n-k =3D w, as=
sim n =3D=3D x mop p =E9 equivalente a n - k =3D=3D x - k mop p que pode se=
r reescrito como w =3D=3D r mod p =0A =0Aw =3D=3D r mod p implica w^p =3D=
=3D r^p mod p =0A =0Aw^p -w =3D=3D r^p - r =3D=3D 0 mod p, assim w^p =3D=3D=
w =3D=3D 1 mod p (oq s=F3 demonstra o teorema quando w deixa resto 1 na di=
vis=E3o por p, tentei provar por indu=E7=E3o para w+1, mas n=E3o saiu...)=
=0A=0A=0A =0A----- Mensagem original ----=0ADe: Marcelo Salhab Brogliato <m=
sbrogli@xxxxxxxxx>=0APara: obm-l@xxxxxxxxxxxxxx=0AEnviadas: S=E1bado, 24 de=
Novembro de 2007 20:16:58=0AAssunto: Re: [obm-l] demonstra=E7=E3o: pequeno=
teorema de FERMAT=0A=0AOl=E1 Rodrigo,=0A=0An=E3o entendi essa passagem: x^=
p - x =3D=3D n^p - n =3D=3D 0 mod p ...=0Ade onde veio o 0?=0A=0Aabra=E7os,=
=0ASalhab=0A=0A=0A=0AOn Nov 24, 2007 6:01 PM, Rodrigo Cientista < rodrigoci=
entista@xxxxxxxxxxxx> wrote:=0A=0AEm primeiro lugar ol=E1 a todos, sou novo=
na lista, e gostaria de saber se uma demonstra=E7=E3o que dei para o peque=
no teorema de fermat est=E1 equivocada ou n=E3o, conforme segue: =0A=0Ao te=
orema diz que n^p =3D=3Dn mod p, o que n=E3o sabemos...=0A=0Aescreverei n =
=3D=3D x mod p, assim n =3D=3D x mod p implica n^p =3D=3D x^p mod p (das pr=
opriedades de congru=EAncia)=0A=0An^p =3D=3D x^p mod p equivale a x^p =3D=
=3D n^p mod p (das propriedades de congru=EAncia) =0A=0Ase n =3D=3D x mod p=
e x^p =3D=3D n^p mod p ent=E3o n + x^p =3D=3D x+ n^p mod p (das propriedad=
es de congru=EAncia)=0A=0Aassim x^p - x =3D=3D n^p - n =3D=3D 0 mod p impli=
ca n^p =3D=3D n mod p como quer=EDamos demonstrar=0A=0A=0A Abra sua con=
ta no Yahoo! Mail, o =FAnico sem limite de espa=E7o para armazenamento! =0A=
http://br.mail.yahoo.com/=0A=0A=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=0AInstru=E7=F5es para entrar na lista, sair da =
lista e usar a lista em =0Ahttp://www.mat.puc-rio.br/~obmlistas/obm-l.html=
=0A=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=0A=
=0A=0A=0A=0A=0A=0A=0A=0AAbra sua conta no Yahoo! Mail, o =FAnico sem limite=
de espa=E7o para armazenamento!=0A=0A=0A Abra sua conta no Yahoo! Mai=
l, o =FAnico sem limite de espa=E7o para armazenamento!=0Ahttp://br.mail.ya=
hoo.com/
--0-2054223372-1196124893=:2781
Content-Type: text/html; charset=iso-8859-1
Content-Transfer-Encoding: quoted-printable
<html><head><style type=3D"text/css"><!-- DIV {margin:0px;} --></style></he=
ad><body><div style=3D"font-family:times new roman, new york, times, serif;=
font-size:12pt"><DIV style=3D"FONT-SIZE: 12pt; FONT-FAMILY: times new roman=
, new york, times, serif"><BR>=0A<DIV style=3D"FONT-SIZE: 12pt; FONT-FAMILY=
: times new roman, new york, times, serif">=0A<DIV><FONT face=3DArial color=
=3D#0000ff size=3D2>Por indu=E7=E3o, =E9 simples!!</FONT></DIV>=0A<DIV><FON=
T face=3DArial color=3D#0000ff size=3D2></FONT> </DIV>=0A<DIV><FONT fa=
ce=3DArial color=3D#0000ff size=3D2>Sabemos que n^p =3D=3D n mop p para alg=
um n(n=3D1, por exemplo), queremos saber se =E9 v=E1lido para todo n.</FONT=
></DIV>=0A<DIV><FONT face=3DArial color=3D#0000ff size=3D2></FONT> </D=
IV>=0A<DIV><FONT face=3DArial color=3D#0000ff size=3D2>expandindo, (n+1)^p =
=3D n^p + C_p,1*a^p-1 + ... + C_p,k*a^p-k + ... + 1</FONT></DIV>=0A<DIV><FO=
NT face=3DArial color=3D#0000ff size=3D2></FONT> </DIV>=0A<DIV><FONT f=
ace=3DArial color=3D#0000ff size=3D2>obs*** C_x,y =3D combina=E7=E3o de x e=
y</FONT></DIV>=0A<DIV><FONT face=3DArial color=3D#0000ff size=3D2></FONT>&=
nbsp;</DIV>=0A<DIV><FONT face=3DArial color=3D#0000ff size=3D2>Como p divid=
e C_p,k (pois o numerador =E9 p! =3D p(p-1)(p-2)...), segue (n+1)^p =3D=3D =
n^p + 1 mod p</FONT></DIV>=0A<DIV><FONT face=3DArial color=3D#0000ff size=
=3D2></FONT> </DIV>=0A<DIV><FONT face=3DArial color=3D#0000ff size=3D2=
>Mas por hip=F3tese de indu=E7=E3o, j=E1 estava provado que n^p =3D=3D n mo=
p p oq implica n^p +1 =3D=3D n + 1 mop p</FONT></DIV>=0A<DIV><FONT face=3DA=
rial color=3D#0000ff size=3D2></FONT> </DIV>=0A<DIV><FONT face=3DArial=
color=3D#0000ff size=3D2>Assim, (n+1)^p =3D=3D n + 1 mod p, provando Ferma=
t por indu=E7=E3o sobre n</FONT></DIV>=0A<DIV><FONT face=3DArial color=3D#0=
000ff size=3D2></FONT> </DIV>=0A<DIV><FONT face=3DArial color=3D#0000f=
f size=3D2>Realmente, essa =E9 a prova mais simples, mas n=E3o =E9 minha</F=
ONT></DIV>=0A<DIV> </DIV>=0A<DIV><FONT face=3DTahoma size=3D2>-----Men=
sagem original-----<BR><B>De:</B> owner-obm-l@xxxxxxxxxxxxxx [mailto:owner-=
obm-l@xxxxxxxxxxxxxx]<B>Em nome de </B>Rodrigo Cientista<BR><B>Enviada em:<=
/B> segunda-feira, 26 de novembro de 2007 13:41<BR><B>Para:</B> obm-l@xxxxx=
uc-rio.br<BR><B>Assunto:</B> [obm-l] Res: [obm-l] demonstra=E7=E3o: pequeno=
teorema de FERMAT<BR><BR></DIV></FONT>=0A<BLOCKQUOTE>=0A<DIV style=3D"FONT=
-SIZE: 12pt; FONT-FAMILY: times new roman, new york, times, serif">=0A<DIV =
style=3D"FONT-SIZE: 12pt; FONT-FAMILY: times new roman, new york, times, se=
rif">Salhab, realmente houve uma falha</DIV>=0A<DIV style=3D"FONT-SIZE: 12p=
t; FONT-FAMILY: times new roman, new york, times, serif"> </DIV>=0A<DI=
V style=3D"FONT-SIZE: 12pt; FONT-FAMILY: times new roman, new york, times, =
serif">o teorema diz que n^p =3D=3Dn mod p, o que n=E3o sabemos...</DIV>=0A=
<DIV style=3D"FONT-SIZE: 12pt; FONT-FAMILY: times new roman, new york, time=
s, serif"> </DIV>=0A<DIV style=3D"FONT-SIZE: 12pt; FONT-FAMILY: times =
new roman, new york, times, serif">seja x um resto qualquer da divis=
=E3o de n por p, tal que n =3D=3D x mod p</DIV>=0A<DIV style=3D"FONT-SIZE: =
12pt; FONT-FAMILY: times new roman, new york, times, serif"> </DIV>=0A=
<DIV style=3D"FONT-SIZE: 12pt; FONT-FAMILY: times new roman, new york, time=
s, serif">seja um k qualquer tal que x-k =3D 1 (chamarei de r) e n-k =
=3D w, assim n =3D=3D x mop p =E9 equivalente a n - k =3D=3D x - k mop p qu=
e pode ser reescrito como w =3D=3D r mod p </DIV>=0A<DIV style=3D"FONT-SIZE=
: 12pt; FONT-FAMILY: times new roman, new york, times, serif"> </DIV>=
=0A<P>w =3D=3D r mod p implica w^p =3D=3D r^p mod p </P>=0A<P> </P>=0A=
<P>w^p -w =3D=3D r^p - r =3D=3D 0 mod p, assim w^p =3D=3D w =3D=3D 1 mod p =
(oq s=F3 demonstra o teorema quando w deixa resto 1 na divis=E3o por p, ten=
tei provar por indu=E7=E3o para w+1, mas n=E3o saiu...)</P>=0A<DIV style=3D=
"FONT-SIZE: 12pt; FONT-FAMILY: times new roman, new york, times, serif"><BR=
><BR> </DIV>=0A<DIV style=3D"FONT-SIZE: 12pt; FONT-FAMILY: times new r=
oman, new york, times, serif">----- Mensagem original ----<BR>De: Marcelo S=
alhab Brogliato <msbrogli@xxxxxxxxx><BR>Para: obm-l@xxxxxxxxxxxxxx<BR=
>Enviadas: S=E1bado, 24 de Novembro de 2007 20:16:58<BR>Assunto: Re: [obm-l=
] demonstra=E7=E3o: pequeno teorema de FERMAT<BR><BR>Ol=E1 Rodrigo,<BR><BR>=
n=E3o entendi essa passagem: x^p - x =3D=3D n^p - n =3D=3D 0 mod p ...<BR>d=
e onde veio o 0?<BR><BR>abra=E7os,<BR>Salhab<BR><BR><BR>=0A<DIV class=3Dgma=
il_quote>On Nov 24, 2007 6:01 PM, Rodrigo Cientista <<A href=3D"mailto:r=
odrigocientista@xxxxxxxxxxxx" target=3D_blank rel=3Dnofollow ymailto=3D"mai=
lto:rodrigocientista@xxxxxxxxxxxx"> rodrigocientista@xxxxxxxxxxxx</A>> w=
rote:<BR>=0A<BLOCKQUOTE class=3Dgmail_quote style=3D"PADDING-LEFT: 1ex; MAR=
GIN: 0pt 0pt 0pt 0.8ex; BORDER-LEFT: rgb(204,204,204) 1px solid">Em primeir=
o lugar ol=E1 a todos, sou novo na lista, e gostaria de saber se uma demons=
tra=E7=E3o que dei para o pequeno teorema de fermat est=E1 equivocada ou n=
=E3o, conforme segue: <BR><BR>o teorema diz que n^p =3D=3Dn mod p, o que n=
=E3o sabemos...<BR><BR>escreverei n =3D=3D x mod p, assim n =3D=3D x mod p =
implica n^p =3D=3D x^p mod p (das propriedades de congru=EAncia)<BR><BR>n^p=
=3D=3D x^p mod p equivale a x^p =3D=3D n^p mod p (das propriedades de cong=
ru=EAncia) <BR><BR>se n =3D=3D x mod p e x^p =3D=3D n^p mod p ent=E3o n + x=
^p =3D=3D x+ n^p mod p (das propriedades de congru=EAncia)<BR><BR>assim x^p=
- x =3D=3D n^p - n =3D=3D 0 mod p implica n^p =3D=3D n mod p como quer=EDa=
mos demonstrar<BR><BR><BR> Abra sua conta no Yahoo! Mail=
, o =FAnico sem limite de espa=E7o para armazenamento! <BR><A href=3D"http:=
//br.mail.yahoo.com/" target=3D_blank
rel=3Dnofollow>http://br.mail.yahoo.com/</A><BR><BR>=3D=3D=3D=3D=3D=3D=3D=
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=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D<BR>Instru=E7=F5es para ent=
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=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D<BR></BLOCKQUOTE></DIV><BR></DIV>=
=0A<DIV style=3D"FONT-SIZE: 12pt; FONT-FAMILY: times new roman, new york, t=
imes, serif"><BR></DIV></DIV><BR>=0A<HR SIZE=3D1>=0AAbra sua conta no <A hr=
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pa=E7o para armazenamento! </BLOCKQUOTE></DIV><BR></DIV></div><br>=0A=0A=0A=
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imite de espa=E7o para armazenamento! =0A</body></html>
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