solution_hw_3_fall_2011

solution_hw_3_fall_2011 - CSE20 Solutions HW 3 Fall 2011...

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CSE20 - Solutions HW 3 Fall 2011 Problem 1 (8 pts) (a) 1. x . H(x) → B(x) 2.  x . B(x) ¬ H(x) 3. x . H(x) B(x) → E(x) (b) Literally: “For every historical essay x there is a book y that is not expensive and is more interesting than x” Elaborating: “You can always find a non-expensive book that is more interesting than a historical essay” (c) ¬ (  x . [H(x) → y . ( ¬E(y) M(y,x))] )  x . ¬ [H(x) → y . ( ¬E(y) M(y,x))] As a rule (negation of implication), ¬ (p → q) p ¬ q  x . [ H(x) ¬ ( y . ( ¬E(y) M(y,x)) ) ]  x . [ H(x) ( y . ¬ ( ¬E(y) M(y,x)))] applying De Morgan's Law:  x . [ H(x) y . ( E(y) ¬M(y,x))] (translated into english [not required] would be: there is always a book that is either expensive or less interesting than a historical essay) Problem 2 (8 pts) 1. For any a, b, x, if x divides both a and b, then x also divides (a+b) Proof: a/x=c, c integer, by definition of “x divides a”, x|a. Hence a=cx. Equally, b=dx, d integer. Thus for any integer x, a ,b, we can write (a+b) as (cx+dx)= (c+d)x (c+d) is integer since c and d are both integers and we can use the closure property of integers under addition. Thus, since (c+d) is integer, we can write x|a+b
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This note was uploaded on 11/21/2011 for the course CSE 20 taught by Professor Foster during the Spring '08 term at UCSD.

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solution_hw_3_fall_2011 - CSE20 Solutions HW 3 Fall 2011...

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