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# ReviewContent-QuestionsWithAnswers - Review Some Highlights...

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Review - Some Highlights of Math 135 1. NOT ( x, P ( x )) is equivalent to ... ( x, NOT P ( x )) 2. NOT ( x, P ( X )) is equivalent to ... ( x, NOT P ( x )) 3. Contrapositive Law: P Q is equivalent to... NOT Q NOT P . 4. Describe Proof by Contradiction for proving P . Assume that P is false and shows that leads to a contradiction. 5. Describe Proof by Mathematical Induction 1. Prove statement is true for base case (usually n = 1 ) 2. Assume statement is true for n = k , for some k P , k 1 3. Prove statement is true for n = k + 1 6. Describe Strong Induction Proof 1. Prove statement is true for base cases 2. Assume statement is true for n = 1 , 2 , ..., k 3. Prove statement is true for n = k + 1 7. How would you decide when to use strong induction vs mathematical induc- tion? math for statements n positive integers. If there is recursion involved, or your hypothesis requires the statement to be true for 2 or more previous statements, use strong induction. 8. Definition of b | a b | a if there exists y Z such that a = by . 9. Definition of gcd( a, b ) gcd( a, b ) is the largest positive integer dividing both of a and b . 10. GCD Characterization Theorem 2.24: If d is a non-negative common divisor of the integers a and b and there exist integers x and y such that ax + by = d , then d = gcd( a, b ) . 11. Prop 2.28: If a, b, c Z with c | ab and gcd( a, c ) = 1 then c | b . 1

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12. Prop 2.29 An integer d is gcd( a, b ) if and only if i. d 0 . ii. d divides both a and b . iii. Any divisor of both a and b divides d . 13. Linear Diophantine Equation Thm 2.31 i. The linear Diophantine equation ax + by = c has a solution if and only if gcd( a, b ) | c . ii. If gcd( a, b ) = d = 0 and x = x 0 , y = y 0 is one particular solution to ax + by = c then the complete solution is x = x 0 + n b d , y = y 0 - n a d n Z .
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