hw1 - Math 113 Homework 1 selected solutions 2 In the...

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Unformatted text preview: Math 113 Homework # 1, selected solutions 2. In the following, ( a,b ), ( c,d ), etc. will always denote pairs of integers where the second integer is assumed to be nonzero. (a) Reflexive: ( a,b ) ∼ ( a,b ) because ab = ba since multiplication of integers is commutative. Symmetric: If ( a,b ) ∼ ( c,d ), i.e. ad = bc , then since multiplica- tion of integers is commutative, cb = da , so ( c,d ) ∼ ( a,b ). Transitive: Suppose ( a,b ) ∼ ( c,d ) and ( c,d ) ∼ ( e,f ), i.e. ad = bc and cf = de . We need to show that ( a,b ) ∼ ( e,f ), i.e. af = be . Multiplying the two equations we know by f and b respectively, we get adf = bcf and bcf = bde , so adf = bde . Since d is assumed nonzero, it follows that af = be . (b) Suppose that ( a,b ) ∼ ( a ,b ) and ( c,d ) ∼ ( c ,d ). To prove that our definition of addition is well-defined, we must check that ( ad + bc,bd ) ∼ ( a d + b c ,b d ) , i.e. ( ad + bc )( b d ) = ( bd )( a d + b c ), or multiplying out, adb d + bcb d = bda d + bdb c . Since ( a,b ) ∼ ( a ,b ) we can replace ab by a b in the first term on the left, and since ( c,d ) ∼ ( c ,d ) we can replace cd by c d in the second term on the left. Then everything cancels, so the equation we are trying to prove is true....
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This note was uploaded on 01/19/2010 for the course MATH 113 taught by Professor Ogus during the Fall '08 term at Berkeley.

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hw1 - Math 113 Homework 1 selected solutions 2 In the...

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