Soln 2 - MACM 101 Discrete Mathematics I Exercises on...

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MACM 101 — Discrete Mathematics I Exercises on Predicates and Quantifiers. Due: Wednesday, October 13th (at the beginning of the class) 1. Use predicates and quantifiers to express this statement “There is a customer in this coffee shop who has tried every type of cookies produced by one of the suppliers.” Solution: x [ CustomerInThisCoffeeShop ( x ) o [( CookieType ( o ) ∧∃ y ( CookieSupplier ( y ) Produces ( y,o ))) Tried ( x,o )]] In practice we use P , Q , R , etc., but I kept long predicate names so that you can understand better. Also, there are several other logically equivalent solutions. Many people wrote much simpler statements whee they simply used a predicate to denote say “x tried every type of cookies produced by one of the suppliers”. That was OK too, but not as good as the solution above. In general, you should try to extract as much logical information as possible from a sentence. 2. Determine the truth value of each of these statements if the universe of each variable consists of (i) rational numbers, (ii) all integers. Justify your answer. (a) x y z ( z = x + y 1000 ) ; (b) z x y ( z = x + y 1000 ) ; (c) x y ( y > 0 → ∃ z ( y x = z )) . Solution: (a) x y z ( z = x + y 1000 ); 1
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(i) True for rationals. No matter which x and y we take, the rational number x + y 1000 ) does exist. (ii) False for integers. Take x = 1, y = 2. Then x + y 1000 ) is not an integer. (b) z x y ( z = x + y 1000 ); (i, ii) False for rationals and for integers. There is no rational (or integer) number z which simultaneously equals to all possible sums of two arbitrary rational (or integer) numbers, divided by 1000! If such a number existed, call it a , then a would be simultaneously equal to say 2 (if x = y = 1000) and 1 (if x = y = 500), and thus 2 would be equal to 1, which is impossible.
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Soln 2 - MACM 101 Discrete Mathematics I Exercises on...

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