assignment2_sol

# assignment2_sol - MATH 135 Assignment 2 Solutions Hand-In...

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MATH 135 Fall 2005 Assignment 2 Solutions Hand-In Problems Exercise 1-64: Use the Contrapositive Proof Method to prove that ( S T = ) AND ( S T = T ) = S = . Solution: We want to prove the contrapositive of the statement. That is ( S 6 = ) = ( S T 6 = ) OR ( S T 6 = T ) . (We have used the fact that NOT( P AND Q ) is equivalent to (NOT P ) OR (NOT Q ).) It then makes sense to use Proof Method 1.56 (for proving statements of the form P = ( Q OR R )).) So we now want to prove the statement (( S 6 = ) AND ( S T = ) = ( S T 6 = T ) . Proving this will be equivalent to proving our original statement. Therefore, we assume that S 6 = and also that S T = . Because S 6 = then there is an x S . It follows that x / T (since S T = ). Because x S T , then S T 6 = T . We have shown ( S 6 = ) AND NOT ( S T 6 = ) = ( S T 6 = T ) . This is equivalent to the statement ( S 6 = ) = ( S T 6 = ) OR ( S T 6 = T ) . Thus by the Contrapositive Law we have proven the original statement. Exercise 1-68: Prove or give a counterexample to each statement. ( S T ) U = S ( T U ), for any sets S , T , and U . Solution 1: A speciﬁc counterexample would be if S = { 1 } , T = { 1 , 3 } and U = { 2 } . Then, ( S T ) U = ( { 1 } ∩ { 1 , 3 } ) ∪ { 2 } = { 1 } ∪ { 2 } = { 1 , 2 } . Also, S ( T U ) = { 1 } ∩ ( { 1 , 3 } ∪ { 2 } ) = { 1 } ∩ { 1 , 2 , 3 } = { 1 } . Thus, in this example ( S T ) U 6 = S ( T U ). Solution 2: The statement is false. To see this notice that for any set A , A ∩ ∅ = and A ∪ ∅ = A . Let

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## This homework help was uploaded on 04/18/2008 for the course LINEAR ALG MATH135 taught by Professor Vanderburgh during the Spring '08 term at Waterloo.

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assignment2_sol - MATH 135 Assignment 2 Solutions Hand-In...

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