idiomas románicos y teoría de números

-kalitx

-jjeko

-jjeku

-dokitx

-dukitx

-itx [o] -itxins

-kali

-jjoika

-jjuika

-doki

-duki

-i [o] -ins

pantalones:

un pantalokitx curti-dokitx.

un pantalokitx llargui-dokitx.

libros de matemáticas:

funciokitx expansivi-dokitx.

funciokitx contractivi-dokitx.

hroñi qui hroñi, [ sea quien sea ]

sere-proika benvenuti-prom.

hroñi qui no hroñi, [ sea quien no sea ]

sere-proika malvenuti-prom.

Mi luze-jjoika del idiom-kali sere-proika,

para hroñi qui hroñi.

yo amare-proika,

a hroñi qui hroñi.

Mi luze-jjoika del idiom-kali no sere-proika,

para hroñi qui no hroñi.

yo odiare-proika,

a hroñi qui no hroñi.

Teorema:

Si [Ax][Am][ ( x >] 0 & m >] 1 ) ==> ...

... Si ( d_{x}[f(x)] = m & f(x) [< k ) ==> [x] [< k ].

Si [Ax][Am][ ( x >] 0 & m >] 1 ) ==> ...

... Si ( d_{x}[f(x)] = m & f(x) [< k ) ==> ]x[ [< k ].

Demostració:

[x] [< m·[x] [< mx = f(x) [< k

]x[ [< m·]x[ [< mx = f(x) [< k

Teorema:

Si [Ax][Am][ ( x >] 1 & m >] 1 ) ==> ...

... Si ( d_{x...x}^{n}[f(x)] = n!·m & f(x) [< k ) ==> [x] [< k ].

Si [Ax][Am][ ( x >] 1 & m >] 1 ) ==> ...

... Si ( d_{x...x}^{n}[f(x)] = n!·m & f(x) [< k ) ==> ]x[ [< k ].

Demostració:

[x] [< m·[x] [< mx [< mx^{n} = f(x) [< k

]x[ [< m·]x[ [< mx [< mx^{n} = f(x) [< k

cuanto tú fachere-po-mitxli ayuni-jjeko,

te perfumare-po-mitxli la boki-jjeko,

bebento-sam red-bull-kalitx.

cuanto tú pasare-po-mitxli hambri-jjeko,

no te perfumare-po-mitxli la boki-jjeko,

no bebento-sam red-bull-kalitx.

mecániki-jjeko cuántiki-dokitx de Gaugi-kalitx.

mecániki-jjeko cuántiki-dokitx de Des-Gaugi-kalitx.

Si [Am_{k}][ m_{k} = 1 & n = ( p_{1} )^{m_{1}}·...·( p_{n} )^{m_{n}} ] ==> ...

... h(n) = (-1)^{n}

Si [Em_{k}][ m_{k} >] 2 & n = ( p_{1} )^{m_{1}}·...·( p_{n} )^{m_{n}} ] ==> h(n) = 0

h(1) = 0

Teorema:

h(n)+h( 2·( n!/(n+(-2))! ) )+...+h( k·( n!/(n+(-k))! ) ) = h(n)

Demostració:

kj || kj+1 || ... || kj+(k+(-1))

H(n+m) = h(mcd{n,m})

Teorema:

H( 2^{n+1}+(-1) ) = 0

Demostració:

h( 2^{n}+(2^{n}+(-1)) ) = h(mcd{2^{n},2^{n}+(-1)}) = h(1) = 0

mcd{n,n+1} = mcd{n,1} = 1

Teorema:

Siguin p,q€N & mcd{p,q} = 1.

Si ( k€P & n = kp ) ==> H(n+kq) = (-1)

Si ( k€P & n = kp ) ==> [Am][ m >] 1 ==> H(k^{m}n+k^{m+1}q) = 0 ]

Demostració:

H(n+kq) = h(mcd{kp,kq}) = h(k·mcd{p,q}) = h(k) = (-1)

H(k^{m}n+k^{m+1}q) = h(mcd{k^{m+1}p,k^{m+1}q}) = ...

... h(k^{m+1}·mcd{p,q}) = h(k^{m+1}) = 0