**Unformatted text preview: **Now assume statement is true for n. 1/(1*2) + 1/(2*3) + 1/(3*4) + … + 1/((n+1)*(n+2)) = 1/(1*2) + 1/(2*3) + … + 1/(n*(n+1)) + 1/((n+1)*(n+2)) = n/(n+1) + 1/((n+1)*(n+2)) = (n 2 + 2n + 1)/(n+1)(n+2)= (n+1)/(n+2) = (n+1)/(n+1+1). 5. Given R = {(1,2), (2,3)} ⊆ {1, 2, 3} × {1, 2, 3}. a. Find the smallest set B such that R ∪ B is an equivalence relation on {1, 2, 3}. B={(1,1),(2,2),(3,3),(2,1),(3,2),(1,3),(3,1)} b. Find R o R (the composition of R with R). R o R={(1,3)} c. Find R o R o R. R o R o R = ∅ d. Find B such that R ∪ B is a function from {1, 2, 3} to {1, 2, 3} which is not 1-1. B = {(3,2)} or B = {(3,3)} e. Find the smallest set B such that R-1 ∪ B is an equivalence relation on {1, 2, 3}. B={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3),(3,1)}...

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- Spring '08
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- Equivalence relation, Binary relation, Alfa-Beta Vassilopoulos, smallest set, R B