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07Ma6aMidSol

# 07Ma6aMidSol - Math 6a Midterm Solutions Problem 1 A...

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Math 6a - Midterm Solutions Problem # 1 A commutative ring is a set B with two binary operations + and * called addition and multiplication and two constants 0 and 1 such that: (a) Addition is associative and commutative (b) For all b in B, b + 0 = b = 0 + b (c) For all b in B, there exists an element c in C such that b + c = 0 = c + b ; we usually denote this element c as - b (d) Multiplication is associative and commutative (e) For all b in B, b * 1 = b = 1 * b (f) Multiplication distributes over addition; that is, for all b, c, and d in B, b ( c + d ) = bc + bd and ( c + d ) b = cb + db. Now assume that B is a commutative ring such that, for all b in B, let b = b + c + bc and b c = bc. Define a unary operation on B as follows: for all b in B let b 0 = 1 + b. Prove that B with the operations and is a Boolean algebra with these two operations denoting join and meet respectively (where b 0 is the complement of b ). Solution. We will be using the ring properties, like commutativity, associativity, etc., freely. First, note that we have a b = a + b + ab = b + a + ba = b a a b = ab = ba = b a so both operations are commutative. Also ( a b ) c = ( a + b + ab ) c = a + b + ab + c + ( a + b + ab ) c ( a b ) c = a + b + ab + c + ac + bc + abc = a + b + c + bc + ab + ac + abc ( a b ) c = a + ( b + c + bc ) + a ( b + c + bc ) = a ( b + c + bc ) = a ( b + c + bc ) = a ( b c ) so is associative and ( a b ) c = ( ab ) c = a ( bc ) = a ( b c ) so is associative. Also, consider any b B. Then we have, by the fact that x 2 = x for all x B, b + b = ( b + b ) 2 = ( b + b )( b + b ) = b 2 + b 2 + b 2 + b 2 = b + b + b + b 1

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2 so if we add - b + - b to both sides we get 0 = b + b for all b B. So now, for any a, b B a ( a b ) = a ( ab ) = a + ab + a 2 b = a + ab + ab = a a ( a b ) = a ( a + b + ab ) = a 2 + ab + a 2 b = a + ab + ab = a showing that both operations in B satisfy the absorption property. Thus, since we showed both operations satisfy the commutative, associative, and absorption laws, it follows from a result in Homework 3 that B is a lattice with the operation defined by a b if and only if a b = a. So note that for any a, b, c in B we have a ( b c ) = a ( b + c + bc ) = ab + ac + abc = ab + ac + a 2 bc = ab + ac + ( ab )( ac ) a ( b c ) = ( ab ) ( ac ) = ( a b ) ( a c ) so by a result in Liu Chapter 9 we have that B is a distributive lattice. Further, for any b B we have b 1 = b * 1 = b implying b 1 and 0 b = 0 + b + 0 * b = b implying that 0 b. Thus, B has universal upper bound 1 and universal lower bound 0 . Further, we have, for any b B, b b 0 = b + b 0 + bb 0 = b + (1 + b ) + b (1 + b ) = b + 1 + b + b + b 2 = 1 + b + b + b + b = 1 b b 0 = bb 0 = b (1 + b ) = b + b 2 = b + b = 0 showing that b 0 is the complement of b and that B is a complemented, distributive lattice, and thus a Boolean algebra. Problem # 2 1 7 + 2 7 + 3 7 + 4 7 · · · + n 7 =? Solution. By a method shown in class, there exist functions f k ( x ) for all positive integers k such that, for positive integers x, we have
3 f k ( x ) = x X i =0 i k and for k 1 we have f k ( x ) = k Z x 0 f k - 1 ( y ) dy + c k x, where we can solve for c k by noting that f k (1) = 1 . Then, we also have from a result in class that f 5 ( x ) = 1 6 x 6 + 1 2 x 5 + 5 12 x 4 - 1 12 x 2 .

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07Ma6aMidSol - Math 6a Midterm Solutions Problem 1 A...

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