5^-1 w Z19 (reszta z dzielenia przez 19)
19 = 5*3 + 4 //Bo szukamy 5. 5 zmieści się 3 razy w 19, i zabraknie nam 4.. więc je dodajemy by L = P
5 = 4*1 + 1 //Bo szukamy rozbicia dla 4'ki w Z5 (to czego szukaliśmy poprzednio). Doszliśmy do 1 - elementu neutralnego więc nie szukamy dalej
5 = 4*1 + 1 -> 1 = 5 - 4
19 = 5*3 + 4 -> 4 = 19 - 5*3. // W ciele Z19 19 = 0, więc:
4 = 0 - 5*3
A teraz podstawiamy:
1 = 5 - 4 // Podstawiamy teraz pod 4 = 0 - 5*3
1 = 5 -(- 5*3) = 5 + 5*3 = 5*4
1 = 5*4 //Po prostu uproszczone.
Z wzoru a^-1 * a = a^0 = 1
więc skoro nasze a to 5 to mamy:
a * 4 = 1
Czyli widzimy, że a^-1 = 4
czyli 5^-1 = 4
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