cs235-hw1-sol

cs235-hw1-sol - CS 235: Algebraic Algorithms Assignment 1...

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CS 235: Algebraic Algorithms Assignment 1 Solutions 1 Modular Multiplication 1.1 Proof by induction (10 points) Let n and a be given. Base case: b=0. Then a · 0 = 0, so the claim reduces to 0 = ( a mod n ) · (0 mod n ) (mod n ), which is trivial. Induction step: Assume for our induction hypothesis that the claim is true for b - 1. We must show it for b as well. All equalities are modulo n . ab = a ( b - 1) + a = [ a ( b - 1) mod n ] + ( a mod n ) (By the rule for addition) = ( a mod n )( b - 1 mod n ) + ( a mod n ) (By the IH) = ( a mod n )[( b - 1 mod n ) + 1] Now if ( b - 1) mod n is less than n - 1, then [( b - 1 mod n )+1] = ( b mod n ) and we are done. Otherwise, b mod n = 0, so the right side of the equation is zero, and the left side is a multiple of n , which is also 0 mod n . This completes the proof for b 0. For b < 0, note that we can repeat the preceding exactly, with an induction hypothesis of b + 1, decreasing b with each step and flipping signs throughout the argument. 1.2
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cs235-hw1-sol - CS 235: Algebraic Algorithms Assignment 1...

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