Practicing Success
If aN = {an : n ∈ N} and bN ∩ cN = dN, where a, b, c ∈ N and b, c are coprime, then |
b = cd c = bd d = bc None of these |
d = bc |
We have bN = {bn : n ∈ N}, cN = {cn : n ∈ N} and dN = {dn : n ∈ N} We have bN ∩ cN = dN ∴ d = d . 1 ∈ bN ∩ cN ⇒ d = bn1 and d = cn2 where n1, n2 ∈ N ⇒ b/d and c/d ⇒ bc/d, because b and c are coprime. Also bc ∈ bN and bc = cb ∈ cN ∴ bc ∈ bN ∩ cN or bc ∈ dN ⇒ bc = dn3 for n3 ∈ N ⇒ d/bc ∴ bc = d Hence (3) is the correct answer. |