congruencias y topología y análisis-matemático

Definición:

f(a) = b <==> a =[m]= b

Teorema:

Sea a =[2]= 0 ==>

x^{2}+ax =[2]= p <==> x =[2]= p 

x = 2k & p = 2j

x = 2k+1 & p = 2j+1

Demostración:

a =[2]= 0

f(a) = 0

x+ax =[2]= x^{2}+ax =[2]= p

f(x) = f(x)+f(a)·f(x) = f(x+ax) = p

x =[2]= p

Teorema:

Sea a =[2]= 1 ==>

x^{2}+ax+(-1) =[2]= p <==> x =[2]= p

x = 2k+1 & p = 2j+1

Demostración:

a =[2]= 1

f(a) = 1

ax =[2]= 1+ax+(-1) =[2]= x^{2}+ax+(-1) =[2]= p

f(x) = f(a)·f(x) = f(ax) = p

x =[2]= p

Teorema:

Sea a =[2]= 1 ==>

x^{2}+ax =[2]= p <==> 2x =[2]= p

x = 2k & p = 4j

x = 2k+1 & p = 4j+2

Demostración:

a =[2]= 1

f(a) = 1

x+ax =[2]= x^{2}+ax =[2]= p

f(2x) = f(x)+f(x) =f(x)+f(a)·f(x) = f(x+ax) = p

2x =[2]= p

Anexo:

k = 1 & j = 4

9+3·3 = 18 = 4·4+2

k = 2 & j = 7 

16+4·3 = 28 = 4·7

Teorema:

[ m+(-1) // k ] =[m]= (-1)^{k}

Demostración:

[ m+(-1) // k ] = (1/k!)·(m+(-1))·...·(m+(-k)) =[m]= (1/k!)·(-1)^{k}·k! = (-1)^{k}

Teorema:

Si m = 2k ==> 2^{m+(-1)} =[m]= 0

Si m = 2k+1 ==> 2^{m+(-1)} =[m]= 1

Demostración:

2^{m+(-1)}+(-1) = sum[k = 1]-[m+(-1)][ (1/k!)·(m+(-1))·...·(m+(-k)) ] =[m]= ...

... sum[k = 1]-[m+(-1)][ (-1)^{k}·(1/k!)·k! ] =[m]= sum[k = 1]-[m+(-1)][ (-1)^{k} ] = ( 0 || (-1) )

Definición: [ de índice logarítmico ]

Ind(p) = |p|

Ind(pq) = Ind(p)+Ind(q)

Teorema:

Ind(p^{n}) = Ind(p)+...(n)...+Ind(p) = n·Ind(p) = np

Teorema:

Ind(1) = 0

Demostración:

1 = 1^{0}

Ind(1) = ind(1^{0}) = 0·Ind(1) = 0

Teorema:

( x =[2]= 1 & x =[2^{n+(-1)}]= (-1) ) <==> ( x = 2k+1 & n = 2 ) 

Demostración:

(x+(-1)) = 2k & (y+1) = 2^{n+(-1)}·j

(x+(-1))·(y+1) =[2^{n}]= 0

x = (-1) & y = 1

4 = Ind(x+(-1))+Ind(y+1) = Ind( (x+(-1))·(y+1) ) = Ind(2^{n}) = n·Ind(2) = 2n

Teorema:

( x =[2^{n}]= 1 & x =[2^{n+(-1)}]= (-1) ) <==> ( x = 4k+1 & n = 2 ) 

Demostración:

(x+(-1)) = 2^{n}·k & (y+1) = 2^{n+(-1)}·j

(x+(-1))·(y+1) =[2^{2n+(-1)}]= 0

x = (-3) & y = 1

4+2 = Ind(x+(-1))+Ind(y+1) = Ind( (x+(-1))·(y+1) ) = Ind(2^{2n+(-1)}) = n·Ind(2) = 2·(2n+(-1))

Topología cociente:

< A [&] ¬B , A [ || ] ¬B > € VxV || < ¬A [ || ] B , ¬A [&] B > € VxV

Teorema

< A [&] ¬A , A [ || ] ¬A > = < 0 , E > € VxV

< ¬A [ || ] A , ¬A [&] A > = < E , 0 > € VxV

Teorema:

< A [&] ¬B , A [ || ] ¬B > € VxV

<==>

< ¬A [ || ] B , ¬A [&] B > € VxV

Teorema:

Si < (A [&] ¬B) [ || ] (B [&] ¬C), (A [ || ] ¬B) [&] (B [ || ] ¬C) > € VxV ==> ...

... < A [&] ¬C , A [ || ] ¬C > € VxV

Si < (¬A [ || ] B) [&] (¬B [ || ] C), (¬A [&] B) [ || ] (¬B [&] C) > € VxV ==> ...

... < ¬A [ || ] C , ¬A [&] C > € VxV

Teorema:

¬( < 1,0 > ) = < 1+(-1),1+(-0) > = < 0,1 >

¬( < (1/3),(2/3) > ) = < 1+(-1)·(1/3),1+(-1)·(2/3) > = < (2/3),(1/3) >

Teorema:

int[x = 0]-[1][ e^{x}·cos(x^{(1/2)}) ]d[x] = sum[k = 0]-[oo][ (-1)^{k}·(1/(2k+1)!)·e ]+(-1)

0 [< cos(1) [< cos(x^{(1/2)}) [< 1

Demostración:

x = y^{2} & d[x] = 2y·d[y]

int[ e^{x}·cos(x^{(1/2)}) ]d[x] = int[ 2ye^{y^{2}}·cos(y) ]d[y] = e^{y^{2}} [o(y)o] sin(y)

Teorema:

int[x = 0]-[1][ e^{x}·sin(x^{(1/2)}) ]d[x] = 1+(-1)·sum[k = 0]-[oo][ (-1)^{k}·(1/(2k)!)·e ]

(-1) [< sin(-1) [< sin(x^{(1/2)}) [< 0

Demostración:

x = y^{2} & d[x] = 2y·d[y]

int[ e^{x}·sin(x^{(1/2)}) ]d[x] = int[ 2ye^{y^{2}}·sin(y) ]d[y] = e^{y^{2}} [o(y)o] (-1)·cos(y)

Si conocéis a alguien que es,

el Mal no vos a va a decir que la gente es,

porque es la negación de cuando no conocéis,

o no hay ninguien que sea,

en vuestra vida.