Question

$Inteiros\ dados\ pela\ classe\ de\ equivalencia\ :\\\\Adic\~ao\ em\ Z\ :(a,b)+(c,d)=(a+c,b+d)\\Multiplicac\~ao\ em\ Z\ :\ (a,b).(c,d)=(ac+bd,ad+bc)\\\\\\Represente\ \boxed{(\ 2\ )\ .\ (-1)\ }\ :$


$a)\ (2,1).(0,0)=(2.0+0.1,0.1+2.1)=(0,2)\\\\b)\ (2,0).(0,1)=(1.0+2.1,2.0+1.0)=(2,0)\\\\c)\ (2,0).(0,1)=(2.1+0.1,2.0+0.0)=(2,0)\\\\d)\ (0,0).(2,1)=(2.0+0.1,0.1+2.1)=(0,2)\\\\e)\ (2,0).(0,1)=(2.0+0.1,2.1+0.0)=(0,2)\\\\\\obs:\ Todos\ est\~ao\ sublinhados\ (-)\ em\ cima\ dos\ pares.$

Answer 1

(2).(-1)


O termo que forma o par ordenado é:

(2) = (2,0)

(-1) = (0,1)


Resolvendo essa multiplicação pelo padrão dado pelo exercício teremos:

(2).(-1)

(2,0).(0,1) = (2.0 + 0.1, 2.1 + 0.0) = (0,2)


Letra e


Bons estudos!