# Algebra Question

#### acm

Two unrelated questions that I need checked:
i) Show that a/b + c/d and an/(bn) + c/d , n is an element of I are equivalent. All denominators are assumed to be nonzero.
ii) Prove that, in a ring R, m.0 = 0 for each m in R.

i) an/(bn) + c/d = {and + (bn)c}/(bn)d = {and + (b)nc}/(b)nd (I is distributive), Hence, a/(b) + c/d = LHS. RHS = a/b + c/d. Hence LHS ~ RHS #.

ii) m.0 = m(n - n) where n is in R
Thus m.0 = mn -mn (R is distributive)
Hence m.0 = 0

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#### HallsofIvy

Homework Helper
Two unrelated questions that I need checked:
i) Show that a/b + c/d and an/(bn) + c/d , n is an element of I are equivalent. All denominators are assumed to be nonzero.
ii) Prove that, in a ring R, m.0 = 0 for each m in R.
Okay, n is an integer but what are a,b,c,d? Are they integers or are they members of some general ring? Isn't it obvious that an/(bn) is equivalent to a/b?

i) an/(bn) + c/d = {and + (bn)c}/(bn)d = {and + (b)nc}/(b)nd (I is distributive), Hence, a/(b) + c/d = LHS. RHS = a/b + c/d. Hence LHS ~ RHS #.

Okay, that's a bit more complicated but works.

ii) m.0 = m(n - n) where n is in R
Thus m.0 = mn -mn (R is distributive)
Hence m.0 = 0
Yes, that's fine. Another proof is mn= m(n+ 0)= mn+ m(0) and then cancel.