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cs70_sp09_mt1_sol

# cs70_sp09_mt1_sol - CS70 Discrete Mathematics and...

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CS70 Discrete Mathematics and Probability Theory, Spring 2009 Midterm 1 Solutions Note: These solutions are not necessarily model answers. Rather, they are designed to be tutorial in nature, and sometimes contain more explanation (occasionally much more) than an ideal solution. Also, bear in mind that there may be more than one correct solution. The maximum total number of points available is 60. 2 points are deducted for a wrong answer in Q. 1 1. [Logic] 2pts (a) Valid: We can see this by rewriting the implication and applying De Morgan’s laws. n ( ¬ Q ( n ) P ( n )) n ( Q ( n ) P ( n )) n ¬ ( ¬ P ( n ) ∧ ¬ Q ( n ) ¬ [ n ( ¬ P ( n ) ∧ ¬ Q ( n ))] 2pts (b) Invalid: We can see this by looking at the right hand side of the equivalence. Note that the logical expression ( P ∨ ¬ P ) is always going to be true. Therefore, we see that the expression n [( P ( n ) ∨ ¬ P ( n )) ( Q ( n ) ∨ ¬ Q ( n ))] is always true. However, it is not true that the left hand side expression n ( P ( n ) Q ( n )) is always true, and so the equivalence is invalid. 2pts (c) Valid: We can see this by rewriting the implication and applying De Morgan’s laws. m n [ R ( m, n ) ⇒ ¬ ( l [ R ( m + l, n ) R ( n, m + l )])] m n [ ¬ R ( m, n ) ∨ ¬ ( l [ R ( m + l, n ) R ( n, m + l )])] m n [ ¬ R ( m, n ) ∨ ∃ l ¬ ( R ( m + l, n ) R ( n, m + l ))] m n [ ¬ R ( m, n ) ∨ ∃ l ( ¬ R ( m + l, n ) ∧ ¬ R ( n, m + l ))] 2pts (d) Invalid: We can see this just by coming up with a simple counterexample. In order for the equivalence to be valid, it needs to be valid for all possible R ( m, n ) . If we can find an proposition R ( m, n ) such that the equivalence does not hold, then it must be invalid. Consider the proposition R ( m, n ) = ( m = n 2 ) . We are looking at the equivalence n m R ( m, n ) ≡ ∃ m n R ( n, m ) Now, it is true that n, m ( m = n 2 ) - we can just pick m = n 2 which is a natural number, and so the left hand side of the equivalence is true with this particular choice of R ( m, n ) . However, it is not true that m n ( n = m 2 ) . This should be obvious to see that all of the natural numbers n are not all the square of the same natural number m . Note: This is obviously not the only possible counterexample that we can use. For example another counterexample, R ( m, n ) = ( m > n ) would be valid as well. 2. [Induction] Prove by induction that the sum of the interior angles of a convex polygon with n vertices is exactly ( n - 2) π . Proof : (a) Base case: For n = 3 we have a triangle. From elementary geometry we know that the sum of the 3pts interior angles of a triangle is (3 - 2) π = π .

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(b) Inductive Step: The inductive hypothesis is that for all polygons with up to n edges, we have that the 7pts sum of their interior angles is ( n - 2) π . Using this inductive hypothesis we will show this holds for polygons with n + 1 edges. To prove this we will reduce the case of an n + 1 polygon (polygon with n + 1 edges) to the n polygon case. The idea of the proof is to show that any n + 1 polygon can be represented as an n polygon and a triangle, s.t. adding the sum of the interior angles of the triangle and the n
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