índex de matemàtiques

index de etiquetes matemàtiques

números fraccionaris binaris

(0.01)_{b} = (01/100)_{b} = (1/4)
(0.10)_{b} = (10/100)_{b} = (2/4) = (1/2)
(0.11)_{b} = (11/100)_{b} = (3/4)


(0.001)_{b} = (001/1000)_{b} = (1/8)
(0.010)_{b} = (010/1000)_{b} = (2/8) = (1/4)
(0.011)_{b} = (011/1000)_{b} = (3/8)
(0.100)_{b} = (100/1000)_{b} = (4/8) = (1/2)
(0.101)_{b} = (101/1000)_{b} = (5/8)
(0.110)_{b} = (011/1000)_{b} = (6/8) = (3/4)
(0.111)_{b} = (111/1000)_{b} = (7/8)


(100)_{b}·x+(-1)·x=(11)_{b}·x
(0.01...)_{b} = (1/3)
(0.10...)_{b} = (2/3)
(0.11...)_{b} = (3/3)


(1000)_{b}·x+(-1)·x=(111)_{b}·x
(0.001...)_{b} = (1/7)
(0.010...)_{b} = (2/7)
(0.011...)_{b} = (3/7)
(0.100...)_{b} = (4/7)
(0.101...)_{b} = (5/7)
(0.110...)_{b} = (6/7)
(0.111...)_{b} = (7/7)


(0.0011...)_{b} = (3/15) = (1/5)
(0.0110...)_{b} = (6/15) = (2/5)
(0.1001...)_{b} = (9/15) = (3/5)
(0.1100...)_{b} = (12/15) = (4/5)
(0.1111...)_{b} = (15/15) = (5/5)


(0.000111...)_{b} = (7/63) = (1/9)
(0.001110...)_{b} = (14/63) = (2/9)
(0.010101...)_{b} = (21/63) = (3/9) = (1/3)
(0.011100...)_{b} = (28/63) = (4/9)
(0.100011...)_{b} = (35/63) = (5/9)
(0.101010...)_{b} = (42/63) = (6/9) = (2/3)
(0.110001...)_{b} = (49/63) = (7/9)
(0.111000...)_{b} = (56/63) = (8/9)
(0.111111...)_{b} = (63/63) = (9/9)


(0.1)_{b}·(0.0011...)_{b} = (1/10)
(0.1)_{b}·(0.0110...)_{b} = (2/10)
(0.1)_{b}·(0.1001...)_{b} = (3/10)
(0.1)_{b}·(0.1100...)_{b} = (4/10)
(0.1)_{b}·(0.1111...)_{b} = (5/10) = (1/2)
(0.1)_{b}·(1.0011...)_{b} = (6/10)
(0.1)_{b}·(1.0110...)_{b} = (7/10)
(0.1)_{b}·(1.1001...)_{b} = (8/10)
(0.1)_{b}·(1.1100...)_{b} = (9/10)
(0.1)_{b}·(1.1111...)_{b} = (10/10)


(1/10)+1 = (0.1)_{b}·(10.0011...)_{b}
(1/10)+2 = (0.1)_{b}·(100.0011...)_{b}
(1/10)+3 = (0.1)_{b}·(110.0011...)_{b}
(1/10)+4 = (0.1)_{b}·(1000.0011...)_{b}


1+(-1)·(1/10) = (0.1)_{b}·(01.1100...)_{b}
2+(-1)·(1/10) = (0.1)_{b}·(011.1100...)_{b}
3+(-1)·(1/10) = (0.1)_{b}·(101.1100...)_{b}
4+(-1)·(1/10) = (0.1)_{b}·(0111.1100...)_{b}


(0.1)_{b}·(0.01...)_{b} = (1/6)
(0.1)_{b}·(0.10...)_{b} = (2/6) = (1/3)
(0.1)_{b}·(0.11...)_{b} = (3/6) = (1/2)
(0.1)_{b}·(1.01...)_{b} = (4/6) = (2/3)
(0.1)_{b}·(1.10...)_{b} = (5/6)
(0.1)_{b}·(1.11...)_{b} = (6/6)


(0.0001011101...)_{b} = (93/1023) = (1/11)

maxim comú divisor álgebra

mcd{x,y} = x [M] y (intersecció en divisors)
mcm{x,y} = x [W] y (reunió en divisors)


mcd{(1/x),(1/y)} = (1/x) [M] (1/y)
mcm{(1/x),(1/y)} = (1/x) [W] (1/y)


mcd{2,6} = 2 [M] 6 = 2
mcm{2,6} = 2 [W] 6 = 6


mcd{(1/2),(1/6)} = (1/2) [M] (1/6) = (1/6)
mcm{(1/2),(1/6)} = (1/2) [W] (1/6) = (1/2)


2·3=6
(1/6)·3=(1/2)


mcd{mcd{x,y},z} =  mcd{x,mcd{y,z}}
mcm{mcm{x,y},z} =  mcm{x,mcm{y,z}}


mcd{mcd{x,y},z} =  mcd{x,mcd{y,z}}
mcd{mcd{x,y},z} = mcd{x,y} [M] z
mcd{mcd{x,y},z} = (x [M] y) [M] z
mcd{mcd{x,y},z} = x [M] (y [M] z)
mcd{mcd{x,y},z} = x [M] mcd{y,z}
mcd{mcd{x,y},z} =  mcd{x,mcd{y,z}}


mcm{mcm{x,y},z} =  mcm{x,mcm{y,z}}
mcm{mcm{x,y},z} = mcm{x,y} [W] z
mcm{mcm{x,y},z} = (x [W] y) [W] z
mcm{mcm{x,y},z} = x [W] (y [W] z)
mcm{mcm{x,y},z} = x [W] mcm{y,z}
mcm{mcm{x,y},z} =  mcm{x,mcm{y,z}}


mcd{x,y} =  mcd{y,x}
mcm{x,y} =  mcm{y,x}


mcd{x,x} =  x
mcm{x,x} =  x


mcm{mcd{x,y},z}=mcd{mcm{x,z},mcm{y,z}}
mcd{mcm{x,y},z}=mcm{mcd{x,z},mcd{y,z}}


m·k=n <==> m | n <==> mcd{m,n}=m <==> mcm{m,n}=n
(1/n)·k=(1/m) <==> (1/n) | (1/m) <==> mcm{(1/m),(1/n)}=(1/m) <==> mcd{(1/m),(1/n)}=(1/n)

ecuacions diofàntiques enteres

x^{2}+y^{2}=z
[2k+1]^{2}+[(2i)·k]^{2}=[4k+1]


x^{2}+y^{2}=z^{2}
[3k]^{2}+[4k]^{2}=[5k]^{2}


x^{3}+y^{3}+z=s^{2}
[3k+1]^{3}+[3·e^{(pi/3)i}k]^{3}+[3k]=[9k^{2}+6k+1]=[3k+1]^{2}


x^{4}+( y^{2}+(-5)z^{2} )·[k]^{2}=[k]·s^{3}
[3k]^{4}+( [8k]^{2}+(-5)[2k]^{2} )·[k]^{2}=[125k^{4}]=[k]·[5k]^{3}


x^{4}+y^{4}+(-2)z^{2}+3=4s^{3}
[3k+1]^{4}+[3e^{(pi/4)i}k]^{4}+(-2)[3k]^{2}+3=4[27k^{3}+9k^{2}+3k+1]=4[3k+1]^{3}

combinacions racionals

(5/6k) = 1·(1/2k) + 1·(1/3k)
( (2n+1)/(n^{2}+n)k  ) = 1·(1/nk) + 1·(1/(n+1)k)


(3/4k) = 1·(1/2k) + 1·(1/4k)
( (m+1)/(m·n)k  ) = 1·(1/nk) + 1·(1/(m·n)k)

index de una congruencia

ind(x^{n}) =[m]= n·ind(x)


x^{p} =[m]= a^{p}
p·ind(x) =[m]= ind(a)+...(p)...+ind(a)
p·ind(x) =[m]= p·ind(a)
x =[m]= a
(mk+a)^{p} = P(m)+a^{p}


x^{p} =[m]= a^{np}
p·ind(x) =[m]= ind(a^{n})+...(p)...+ind(a^{n})
p·ind(x) =[m]= p·ind(a^{n})
x =[m]= a^{n}
(mk+a^{n})^{p} = P(m)+a^{np}

problemes de enters

sum[k=0-->p]( (q/p)·k ) + (-1)·sum[k=0-->q]( (p/q)·k ) = (1/2)·(q+(-p))


sum[k=0-->p]( (q/p)(q+1)·k^{2} ) + (-1)·sum[k=0-->q]( (p/q)(p+1)·k^{2} ) = ...
...(1/6)·(q+(-p))(p+1)(q+1)


sum[k=0-->p]( (q/p^{2})(q+1)·k^{3} ) + (-1)·sum[k=0-->q]( (p/q^{2})(p+1)·k^{3} ) = ...
...(1/4)·(q+(-p))(p+1)(q+1)




proposició:
si n=2^{k}+(-1) ==> [Ep][ p€N & 2n+1=2^{p}+(-1) ]
demostració
2n+1=2·(2^{k}+(-1))+1=2^{k+1}+(-2)+1
2n+1=2^{k+1}+(-1)=2^{p}+(-1)


proposició:
si n=m^{k}+(-1) ==> [Ep][ p€N & m·n+(m+(-1))=m^{p}+(-1) ]
proposició:
si n=m^{k}+1 ==> [Ep][ p€N & m·n+(-1)(m+(-1))=m^{p}+1 ]


a=mb_{a}+r_{a}
a+(-m)b_{a}=r_{a}


sum[a]( (a+(-m)b_{a})·(1/m(m+(-1))) )=(1/2)


sum[a]( (a+(-m)b_{a})^{2}·(1/m(m+(-1))(2m+(-1))) )=(1/6)


sum[a]( (a+(-m)b_{a})^{3}·(1/m^{2}(m+(-1))^{2}) )=(1/4)

punts enters de una regió

x^{2}+y^{2} [< r^{2}


y [< ( r^{2}+(-1)x^{2} )^{(1/2)}
x [< ( r^{2}+(-1)y^{2} )^{(1/2)}


x+y [< ( r^{2}+(-1)y^{2} )^{(1/2)}+( r^{2}+(-1)x^{2} )^{(1/2)}


si x=y ==>
2y^{2} [< r^{2} <==> y [< (r/2^{(1/2)})
2x^{2} [< r^{2} <==> x [< (r/2^{(1/2)})


si ( x=0 or y=0 ) ==>
y [< r
x [< r


sum( [x+y] ) [<
...1+...
...4·r+...
...4·sum[y=0-->(r/2^{(1/2)})]( [( r^{2}+(-1)y^{2} )^{(1/2)}] )+...
...4·sum[x=0-->(r/2^{(1/2)})]( [( r^{2}+(-1)x^{2} )^{(1/2)}] )+...
...(-4)[(r/2^{(1/2)})]


x > 0 & y > 0 & xy [< n


y [< (n/x)
x [< (n/y)


x+y [< (n/y)+(n/x)

si x=y ==>
y^{2} [< n <==> y [< n^{(1/2)}
x^{2} [< n <==> x [< n^{(1/2)}


sum( [x+y] ) [< ...
...sum[(y > 0)-->n^{(1/2)}]( [(n/y)] )+...
...sum[(x > 0)-->n^{(1/2)}]( [(n/x)] )+...
...(-1)[n^{(1/2)}]

congruencies

teorema:
Si p€P ==> n^{p} =[p]= n
demostració:
(n+1)^{p}=n^{p}+pk+1=(npk+n)+pk+1=(n+1)pk+(n+1)


teorema:
Si p€P ==> n^{p+(-1)} =[p]= 1
demostració:
(n+1)^{p+(-1)}=(n^{p}+pk+1)/(n+1)=(n·n^{p+(-1)}+pk+1)/(n+1)=...
(n+1)^{p+(-1)}=(n(pk+1)+pk+1)/(n+1)=((n+1)pk+(n+1))/(n+1)=pk+1


proposició:
Si p€P ==> 1^{p+(-1)}+...+n^{p+(-1)} =[p]= n


proposició:
Si p€P ==> 1^{p}+...+n^{p} =[p]= ( n(n+1)/2 )


teorema:
ax =[a·m]= ab <==> x =[m]= b
demostració:
ax=a·mt+ab
x=mt+b




3x =[9]= 18 <==> x =[3]= 6 =[3]= 0
2x =[8]= 18 <==> x =[4]= 9 =[4]= 1
3x =[6]= 18 <==> x =[2]= 6 =[2]= 0
2x =[6]= 18 <==> x =[3]= 9 =[3]= 0


teorema:
Si ( p€P & p=2k+1 ) ==> ax^{2} =[p]= a <==> x =[p]= a^{(p+(-1))/2}
demostració:
(a^{(p+(-1)/2)})^{2} = a^{p+(-1)} =[p]= 1
a^{p} =[p]= a


2x^{2} =[5]= 2 <==> x =[5]= 4
3x^{2} =[5]= 3 <==> x =[5]= 9 =[5]= 4
4x^{2} =[5]= 4 <==> x =[5]= 16 =[5]= 1


teorema:
Si ( p€P & p=2k+1 ) ==> ax^{2} =[p]= (-a) <==> x =[p]= a^{(p+(-1))/2}·i
demostració:
(a^{(p+(-1)/2)}·i)^{2} = (-1)a^{p+(-1)} =[p]= (-1)
(-1)a^{p} =[p]= (-a)


2x^{2} =[5]= (-2) <==> x =[5]= 4i
3x^{2} =[5]= (-3) <==> x =[5]= 9i =[5]= 4i
4x^{2} =[5]= (-4) <==> x =[5]= 16i =[5]= i


teorema:
Si ( p€P & p=3k+1 ) ==> ax^{3} =[p]= a <==> x =[p]= a^{(p+(-1))/3}
demostració:
(a^{(p+(-1)/3)})^{3} = a^{p+(-1)} =[p]= 1
a^{p} =[p]= a


2x^{3} =[7]= 2 <==> x =[7]= 4
3x^{3} =[7]= 3 <==> x =[7]= 9 =[7]= 2
4x^{3} =[7]= 4 <==> x =[7]= 16 =[7]= 2


teorema:
Si p€N ==> ...
...x^{2}+2bx =[p^{2}]= (-1)b^{2} <==> x =[p]= (-b)
demostració:
x^{2}+2bx =[p^{2}]= (-1)b^{2}
x^{2}+2bx+b^{2} =[p^{2}]= 0
(x+b)^{2} =[p^{2}]= 0
x+b =[p]= 0
x =[p]= (-b)


x^{2}+4x =[9]= (-4) <==> x =[3]= (-2) =[3]= 1
x^{2}+6x =[9]= (-9) <==> x =[3]= (-3) =[3]= 0
x^{2}+8x =[9]= (-16) <==> x =[3]= (-4) =[3]= (-1) =[3]= 2

problemes de números

s=1+(1/2)+...+(1/n)  ==> ¬( s€N ) or s=1


s = sum[k=1-->n]( (1·...·(k+(-1))(k+1)·...·n)/n! )
sigui m_{k}€N ==>
s = sum[k=1-->n]( m_{k} )=sum[k=1-->n]( (1·...·(k+(-1))(k+1)·...·n)/n! )
sum[k=1-->n]( m_{k} )n!=sum[k=1-->n]( (1·...·(k+(-1))(k+1)·...·n) )
m_{k}=(1/k)


s=1+(1/3)+...+(1/(2n+1))  ==> ¬( s€N ) or s=1


s = sum[k=1-->n]( (1·...·(2(k+(-1))+1)(2(k+1)+1)·...·(2n+1))/(2n+1)!! )
sigui m_{k}€N ==>
s = sum[k=1-->n]( m_{k} )=sum[k=1-->n]( (1·...·(2(k+(-1))+1)(2(k+1)+1)·...·(2n+1))/(2n+1)!! )
sum[k=1-->n]( m_{k} )(2n+1)!!=sum[k=1-->n]( (1·...·(2(k+(-1))+1)(2(k+1)+1)·...·(2n+1)) )
m_{k}=(1/(2k+1))


s=1+(1/2!)+...+(1/n!)  ==> ¬( s€N ) or s=1


s = sum[k=1-->n]( (1·...·(k+(-1))!(k+1)!)·...·n!)/(n!)! )
sigui m_{k}€N ==>
s = sum[k=1-->n]( m_{k} )=sum[k=1-->n]( (1·...·(k+(-1))!(k+1)!)·...·n!)/(n!)! )
sum[k=1-->n]( m_{k} )(n!)!=sum[k=1-->n]( (1·...·(k+(-1))!(k+1)!)·...·n!) )
m_{k}=(1/k!)


s=(p/n)+...(n)...+(p/n)==> s€N


s = ( (p+...(n)...+p)/n )
sigui m_{k}€N ==>
s = sum[k=1-->p]( m_{k} )=( ( p+...(n)...+p )/n )
( m_{1}+...(p)...+m_{p} )=(np/n)
( 1+...(p)...+1 )=p
m_{k}=1


s=(p/n^{q})+...(n)...+(p/n^{q})==> ¬( s€N ) or p=n^{q+(-1)}


s = ( (p+...(n)...+p)/n^{q} )
sigui m_{k}€N ==>
s = sum[k=1-->p]( m_{k} )=( ( p+...(n)...+p )/n^{q} )
( m_{1}+...(p)...+m_{p} )=(np/n^{q})
( 1/n^{q+(-1)}+...(p)...+1/n^{q+(-1)} )n^{q+(-1)}=p
m_{k}=(1/n^{q+(-1)})
( 1/n^{q+(-1)}+...(n^{q+(-1)})...+1/n^{q+(-1)} )n^{q+(-1)}=n^{q+(-1)}

mínim comú múltiple

ab=mcm{a,b}·mcd{a,b}
(mcd{a,b}·p)·(mcd{a,b}·q)=mcm{a,b}·mcd{a,b}


mcm{a,b+c}·mcd{a,b+c}=mcm{a,b}·mcd{a,b}+mcm{a,c}·mcd{a,c}


mcm{a,a+b}·mcd{a,a+b}=a^{2}+mcm{a,b}·mcd{a,b}
mcm{a,a+b}=(a^{2}/mcd{a,b})+mcm{a,b}


mcm{3,6}=mcm{3,3+3}=(9/3)+3=3+3=6
mcm{3,7}=mcm{3,3+4}=9+12=21
mcm{3,8}=mcm{3,3+5}=9+15=24
mcm{3,9}=mcm{3,3+6}=(9/3)+6=3+6=9


mcm{5,10}=mcm{5,5+5}=(25/5)+5=5+5=10
mcm{10,15}=mcm{10,10+5}=(100/5)+10=20+10=30
mcm{2k,2k+1}=4k^{2}+2k=2k(2k+1)
mcm{2k,3k}=mcm{2k,2k+k}=(4k^{2}/k)+2k=4k+2k=6k
mcm{nk,(n+1)k}=mcm{nk,nk+k}=(n^{2}k^{2}/k)+nk=n^{2}k+nk=(n^{2}+n)k=n(n+1)k


(na)·(nb)=mcm{na,nb}·mcd{na,nb}
n^{2}·(ab)=mcm{na,nb}·mcd{na,nb}
n^{2}·mcm{a,b}·mcd{a,b}=mcm{na,nb}·(n·mcd{a,b})
n·mcm{a,b}=mcm{na,nb}


mcm{3,12}=mcm{3,3·4}=3·mcm{1,4}=3·4=12


n=mcm{n,1}·mcd{n,1}
n=mcm{n,1}


mcm{a,a+1}=a^{2}+a=a(a+1)

màxim comú divisor

a=mcd{a,b}·p
b=mcd{a,b}·q
a+b=mcd{a,b}·(q+p)


a=mcd{a,a+b}·k
a+b=mcd{a,a+b}·s
b=mcd{a,a+b}·(s+(-k))


mcd{a,a+b}·(s+(-k))=mcd{a,b}·q
mcd{a,a+b}·s=mcd{a,b}·q+mcd{a,a+b}·k
mcd{a,b}·(q+p)=mcd{a,b}·q+mcd{a,a+b}·k
mcd{a,b}·p=mcd{a,a+b}·k & mcd{p,k}=1
mcd{a,b}=mcd{a,a+b}·(k/p) & ( p=(mk) or  k=(np) )
mcd{a,b}=mcd{a,a+b}·(p/p) & ( (1/m)=(k/p) or  (1/n)=(p/k) ) ==> (n=1  & m=1)


mcd{a,b}=mcd{a,a+b}


mcd{a,a+1}=mcd{a,1}=1








mcd{8,24}=mcd{8,8+16}=mcd{8,16}=mcd{8,8+8}=mcd{8,8}=8
mcd{10,15}=mcd{10,10+5}=mcd{10,5}=mcd{5+5,5}=mcd{5,5}=5


mcd{4k+1,14k+3}=mcd{4k+1,4k+1+10k+2}=mcd{4k+1,10k+2}=...
...mcd{4k+1,6k+1}=mcd{4k+1,2k}=mcd{2k+2k+1,2k}=mcd{2k+1,2k}=1




a=mcd{a,b}·p
b=mcd{a,b}·q
na=mcd{a,b}·np
nb=mcd{a,b}·nq
na=mcd{na,nb}·p
nb=mcd{na,nb}·q
n(a+b)=mcd{na,nb}·(p+q)
n·mcd{a,b}·(p+q)=mcd{na,nb}·(p+q)
n·mcd{a,b}=mcd{na,nb}


descens inductiu:
mcd{(n+1)a,(n+1)b}=mcd{na,nb}+mcd{a,b}
mcd{(n+1)a,(n+1)b}=n·mcd{a,b}+mcd{a,b}
mcd{(n+1)a,(n+1)b}=(n+1)·mcd{a,b}


mcd{3k,6k}=mcd{(2+1)k,(2+1)2k}=mcd{2k,4k}+mcd{k,2k}=2k+k=3k






a=mcd{a,na}·p
na=mcd{a,na}·np
(n+1)a=mcd{a,na}·(n+1)p
(n+1)a=mcd{a,a+na}·(n+1)p
(n+1)a=mcd{a,(n+1)a}·(n+1)p


mcd{6k+1,12k+2}=mcd{6k+1,2·(6k+1)}=6k+1






mcd{a,a^{n}}=a


n+1=2k+1 & n^{2}+1=4k^{2}+1
mcd{2k+1,2k(2k)+1}=1 & k=(1/2)
n+1=2k+2 & n^{2}+1=4k^{2}+4k+2
mcd{2k+2,4k^{2}+4k+2}=2·mcd{k+1,2k^{2}+k+k+1}=...
...2·mcd{k+1,k(2k+1)}=2 & 2k+1=m(1+(1/k)) & k=1



mcd{a,b}=1 & d=a+(-b)
nd=an+n(-b)
(n+1)d=a(n+1)+(n+1)(-b)
mcd{an,nb}=n


6=12+(-6)
6=2·6+6·(-1)
4=16+(-12)
4=4·4+4·(-3)
10=20+(-10)
10=2·10+10·(-1)
24=36+(-12)
12·2=3·12+12·(-1)
12=18+(-6)
6·2=3·6+6·(-1)
12=72+(-60)
12=6·12+12·(-5)
54=72+(-18)
18·3=4·18+18·(-1)
105=120+(-15)
15·7=8·15+15·(-1)




1=3·5+2·(-7)


mcd{2k,3k+1}=mcd{2k,k+1}
si k=2p+1 ==> mcd{2k,k+1}=mcd{4p+2,2p+2}=2·mcd{2p+1,p+1}=2·mcd{p,p+1}=2
si k=2p ==> mcd{2k,k+1}=mcd{4p,2p+1}=1 & 2p+1=4q & p=2q+(-1)(1/2)




a^{2}=(mcd{a,b}·p)^{2}
b^{2}=(mcd{a,b}·q)^{2}
2ab=(mcd{a,b})^{2}·(2pq)
mcd{a^{2},2ab,b^{2}}=(mcd{a,b})^{2}


a^{n} = (mcd{a,b}·p)^{n}
b^{n} = (mcd{a,b}·q)^{n}
[ n // k ]a^{n+(-k)}b^{k} = (mcd{a,b})^{n}·( [ n // k ]p^{n+(-k)}q^{k} )
[ n // n+(-k) ]a^{k}b^{n+(-k)} = (mcd{a,b})^{n}·( [ n // n+(-k) ]p^{k}q^{n+(-k)} )
mcd{a^{n},...,[ n // k ]a^{n+(-k)}b^{k},...,[ n // n+(-k) ]a^{k}b^{n+(-k)},...,b^{n}}=(mcd{a,b})^{n}





ax+by=au_{n}+bv_{n}
ax+by=a(x+(-1)nb)+b(y+na)
ax+by=a(x+(-1)nb)+b(y+na)+(-1)ab+ab
ax+by=a(x+(-1)(n+1)b)+b(y+(n+1)a)
ax+by=au_{n+1}+bv_{n+1}