Proving Set Equality: Discrete Mathematics Quiz Insights
School
Florida State University**We aren't endorsed by this school
Course
MAD 2104
Subject
Mathematics
Date
Dec 11, 2024
Pages
2
Uploaded by LieutenantNightingaleMaster785
Quiz 4 AnswersDiscrete Mathematics I1. LetA,B, andCbe sets. Prove that∩right-distributes over⊕, i.e.,(A⊕B)∩C= (A∩C)⊕(B∩C).If you do this by translating into formal logic, prove the logical equivalence you useto verify the set equality. If you do this by a verbal argument, show separately thatLHS⊂RHS and RHS⊂LHS.Proof.LHS⊂RHS: Letx∈LHS = (A⊕B)∩C. Thenx∈A⊕Bandx∈C. Sincex∈A⊕B, either (i)x∈Aandx /∈B, or (ii)x /∈Aandx∈B. We consider thesetwo cases in turn.(i) Herex∈A,x /∈B, andx∈C. This implies thatx∈A∩Candx /∈B∩Csox∈(A∩C)⊕(B∩C).(ii) Herex /∈A,x∈B, andx∈C. This implies thatx /∈A∩Candx∈B∩Csox∈(A∩C)⊕(B∩C).Either way we conclude thatx∈(A∩C)⊕(B∩C) = RHS, hence LHS⊂RHS.RHS⊂LHS: Letx∈RHS = (A∩C)⊕(B∩C). Then either (i)x∈A∩Candx /∈B∩C, or (ii)x /∈A∩Candx∈B∩C. We consider these two cases in turn.(i) Here we get fromx∈A∩Cthatx∈Aandx∈C. Sincex /∈B∩Cbutx∈C,we conclude thatx /∈B(otherwise it would contradictx /∈B∩C). It follows thatx∈A⊕Band since we also havex∈C, we getx∈(A⊕B)∩C.(ii) This is argued exactly as (i) is argued withAandBswitched.Either way we conclude thatx∈(A⊕B)∩C= LHS, hence RHS⊂LHS.Since LHS⊂RHS and RHS⊂LHS, these two sets are the same: LHS = RHS.Note:The student may translate into formal logic withp:x∈A,q:x∈B,andr:x∈C. Then the fact that the set equality is true follows from the logicalequivalence(p⊕q)∧r⇐⇒(p∧r)⊕(q∧r).If the student uses this method to verify the set equality, this logical equivalenceshould be proved (not just referenced). 2 points for correct translation to this equiv-alence, 3 points for verification of the equivalence.
2. LetA= 2Z, the set of even integers,B=N, the set of positive integers, andC={n∈Z:-5≤n≤5}={-5,-4,-3,-2,-1,0,1,2,3,4,5}.Identify thefollowing sets and illustrate the set equality of problem 1.A= 2Z={. . . ,-4,-2,0,2,4, . . .}C={-5,-4,-3,-2,-1,0,1,2,3,4,5}B=N={1,2,3,4,5, . . .}A⊕B={. . . ,-4,-2,0,1,3, . . .}={negative even integers}∪{0}∪{positive odd integers}.A∩C={-4,-2,0,2,4}.B∩C={1,2,3,4,5}.LHS = (A⊕B)∩C={-4,-2,0,1,3,5}.RHS = (A∩C)⊕(B∩C) ={-4,-2,0,1,3,5}.