Question

$Classes\ de\ equivalencia\ : \\\\Adic\~ao\em\ Z\ : (a,b)+(c,d)=(a+c,b+d)\\Multiplicac\~ao\ :\ (a,b).(c,d)=(ac+bd,ad+bc)\\\\\\Represente\ o\ resultado\ de\ :\boxed{(-4).(3)+(-2)}$

$a)\ (0,4).(0,3)+(0,2)=(0,-14)\\\\b)\ (0,-4).(3,0)+(0,-2)=(0,-14)\\\\c)\ (0,4).(3,0)+(0,2)=(0+2,0+12)=(0,14)\\\\d)\ (0,4).(3,0)+(0,2)=(0+0,12+2)=(0,14)\\\\e)\ (4,0).(3,0)+(0,-2)=(0,14)\\\\obs:\ (todos\ os\ pares\ dados\ t\~ao\ sublinhados\ (-)\ em\ cima\ deles\ )$

Answer 1

Olá Optimistic, bom dia!

 Pensei no seguinte: verificar as opções de acordo com as operações indicadas no enunciado!

  Segue,

$\\ \bullet \quad \textbf{adi\c{c}\~ao:} \ \qquad \qquad \mathsf{(\overline{a}, \overline{b}) + (\overline{c}, \overline{d}) = (\overline{a + c}, \overline{b + d})} \\\\ \bullet \quad \textbf{multiplica\c{c}\~ao:} \ \, \ \mathsf{(\overline{a}, \overline{b}) \cdot (\overline{c}, \overline{d}) = (\overline{ac} + \overline{bd}, \overline{ad} + \overline{bc})}$


Item a):

$\\ \mathsf{(\overline{0}, \overline{4}) \cdot (\overline{0}, \overline{3}) + (\overline{0}, \overline{2})=} \\\\ \mathsf{(\overline{0} \cdot \overline{0} + \overline{4} \cdot \overline{3}, \overline{0} \cdot \overline{3} + \overline{4} \cdot \overline{0}) + (\overline{0}, \overline{2}) =} \\\\ \mathsf{(\overline{0 \cdot 0} + \overline{4 \cdot 3}, \overline{0 \cdot 3} + \overline{4 \cdot 0}) + (\overline{0}, \overline{2}) =} \\\\ \mathsf{(\overline{0} + \overline{12}, \overline{0} + \overline{0}) + (\overline{0}, \overline{2}) =} \\\\ \mathsf{(\overline{0 + 12}, \overline{0 + 0}) + (\overline{0}, \overline{2}) =} \\\\ \mathsf{(\overline{12}, \overline{0}) + (\overline{0}, \overline{2}) =} \\\\ \mathsf{(\overline{12 + 0}, \overline{0 + 2}) =} \\\\ \mathsf{(\overline{12}, \overline{2})}$


Item b)

$\\ \mathsf{(\overline{0}, \overline{- 4}) \cdot (\overline{3}, \overline{0}) + (\overline{0}, \overline{- 2})=} \\\\ \mathsf{(\overline{0}, \overline{4}) \cdot (\overline{3}, \overline{0}) + (\overline{0}, \overline{2})=} \\\\ \mathsf{(\overline{0} \cdot \overline{3} + \overline{4} \cdot \overline{0}, \overline{0} \cdot \overline{0} + \overline{4} \cdot \overline{3}) + (\overline{0}, \overline{2}) =} \\\\ \mathsf{(\overline{0 \cdot 3} + \overline{4 \cdot 0}, \overline{0 \cdot 0} + \overline{4 \cdot 3}) + (\overline{0}, \overline{2}) =} \\\\ \mathsf{(\overline{0} + \overline{0}, \overline{0} + \overline{12}) + (\overline{0}, \overline{2}) =} \\\\ \mathsf{(\overline{0 + 0}, \overline{0 + 12}) + (\overline{0}, \overline{2}) =} \\\\ \mathsf{(\overline{0}, \overline{12}) + (\overline{0}, \overline{2}) =} \\\\ \mathsf{(\overline{0 + 0}, \overline{12 + 2}) =} \\\\ \mathsf{(\overline{0}, \overline{14})}$


Item c):

$\\ \mathsf{(\overline{0}, \overline{4}) \cdot (\overline{3}, \overline{0}) + (\overline{0}, \overline{2})=} \\\\ \mathsf{(\overline{0 \cdot 3} + \overline{4 \cdot 0}, \overline{0 \cdot 0} + \overline{4 \cdot 3}) + (\overline{0}, \overline{2}) =} \\\\ \mathsf{(\overline{0 + 0}, \overline{0 + 12}) + (\overline{0}, \overline{2}) =} \\\\ \mathsf{(\overline{0}, \overline{12}) + (\overline{0}, \overline{2}) =} \\\\ \mathsf{(\overline{0 + 0}, \overline{12 + 2}) =} \\\\ \mathsf{(\overline{0}, \overline{14})}$


Item d): análogo ao item anterior.


Item e):

$\\ \mathsf{(\overline{4}, \overline{0}) \cdot (\overline{3}, \overline{0}) + (\overline{0}, \overline{2})=} \\\\ \mathsf{(\overline{4 \cdot 3} + \overline{0 \cdot 0}, \overline{4 \cdot 0} + \overline{0 \cdot 3}) + (\overline{0}, \overline{2}) =} \\\\ \mathsf{(\overline{12 + 0}, \overline{0 + 0}) + (\overline{0}, \overline{2}) =} \\\\ \mathsf{(\overline{12}, \overline{0}) + (\overline{0}, \overline{2}) =} \\\\ \mathsf{(\overline{12 + 0}, \overline{0 + 2}) =} \\\\ \mathsf{(\overline{12}, \overline{2})}$


 Obs1.: talvez tenha ocorrido um erro de digitação no item c), afinal (0 + 2, 0 + 12) = (2, 12).

 Obs.: há duas opções correctas (a meu ver): "c" e "d". Mas, o item c) apresenta um erro de soma no desenvolvimento. Então, o item correcto é a opção "d"!!