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hmwk_6_sol (1)

# hmwk_6_sol (1) - Discrete Structures Assignment 6 Due at...

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Discrete Structures Sept 26 2011 Assignment 6: Due at beginning of class Wednesday, Oct 12 Prof. Hopcroft *******Note: this homework is subject to the style guide on the website. Points will be deducted for homeworks not following the guidelines.******** *******Please print out and staple the grade sheet to the back of your homework!****** 1. If, for a given a there exists x such that ax 1 mod m does that imply that a and m are relatively prime? Prove why or why not. Proof. This implies that x is a ’s inverse in mod m . We have a theorem, that says if a and m are relatively prime, then x exists (Euclid’s Extended Algorithm). Hence, we have the intuition that we should try to prove a and m are relatively prime. But note, we can not directly use Euclid’s Algorithm, as it assumes what we want to prove. (Another way to phrase it, is we want to prove the reverse direction.) There are a few proofs that work, I present the most straight forward one: Using the given let us manipulate it a little bit: ax 1 mod m ax - km = 1 for some integer k ax = 1 + km Now, let us look at the gcd gcd( ax, m ) = gcd( km + 1 , m ) . It may seem obvious from here, but if we add a little bit of argumentation, our proof will be mathe- matically sound. Let d be the gcd( km + 1 , m ). Because d divides m and d divides km + 1, this implies that d divides 1. Which, can only be the case if d = 1. Hence, we have shown gcd( ax, m ) = 1, which implies no number that divides m divides either a or x . Hence, there is no number (other than 1) that divides both m and a .

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hmwk_6_sol (1) - Discrete Structures Assignment 6 Due at...

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