3.1. If R is a ring and a, b, c, d ∈ R, evaluate (a + b)(c + d).
3.2. Prove that if a, b ∈ R, then (a + b)2 = a2 + ab + ba + b2 where
by x2 we mean xx.
3.1
(a+b)(c+d)=a(c+d)+b(c+d)(a + b)(c + d) = a(c + d) + b(c + d)(a+b)(c+d)=a(c+d)+b(c+d)
by distributive law:
=(ac+ad)+(bc+bd)=ac+ad+bc+bd= (ac + ad) + (bc + bd) = ac + ad + bc + bd=(ac+ad)+(bc+bd)=ac+ad+bc+bd
3.2
(a+b)2=(a+b)(a+b)=a(a+b)+b(a+b)=a2+ab+ba+b2(a + b)^2 = (a + b)(a + b) = a(a + b) + b(a + b)=a^2 + ab + ba + b^2(a+b)2=(a+b)(a+b)=a(a+b)+b(a+b)=a2+ab+ba+b2
Note that if R is not a commutative ring ab≠baab\ne baab=ba
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