Question: Let (R, +, ) be a ring, with a R. Define 0a = z, la = a, and (n + 1)a = na +

Let (R, +, •) be a ring, with a ∈ R. Define 0a = z, la = a, and (n + 1)a = na + a, for all n ∈ Z+. (Here we are multiplying elements of R by elements of Z, so we have yet another operation that is different from the multiplications in either of Z or R.) For n > 0, we define (-n)a = n(-a), so, for example, (-3)a = 3 (-a) = 2(-a) + (-a) = [(-a) + (-a)] + (-a) - [-(a + a) ] + (-a) = -[(a + a) + a] = - [2a + a] = -(3a).
For all d, b ∈ R, and all m, n ∈ Z, prove that
(a) ma + nd = (m + n)a
(b) m(nd) = (mn)a
(c) n(d + b) = nd + nb
(d) n(db) = (nd)b = d(nb)
(e) (md)(nb) = (mn)(db) = (nd)(mb)

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