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Hunus
 3 years ago
Number Theory problem
Hunus
 3 years ago
Number Theory problem

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Hunus
 3 years ago
Best ResponseYou've already chosen the best response.0The division algorithm states 'For any positive integers a and b there exists a unique pair (q, r) of nonnegative integers such that b = aq + r, r < a.' For the uniqueness, assume that b = aq′ + r′, where q′ and r′ are also nonnegative integers satisfying 0 ≤ r′ < a. Then aq+r = aq′+r′, implying a(q−q′) = r′−r, and so ar′ −r. Hence r′−r ≥ a or r′ −r = 0. Because 0 ≤ r, r′ < a yields r′ − r < a, we are left with r′ − r = 0, implying r′ = r and, consequently, q′ = q.

Hunus
 3 years ago
Best ResponseYou've already chosen the best response.0My question lies with the statement 'Hence r′−r ≥ a or r′ −r = 0' Why does r′−r ≥ a imply that r′ −r = 0? Is it because r must be less than a and r'  r being greater than a is invalid so it must be zero?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2a(q−q′) = r′−r consider both q's and r's to be integer, if \( q' \neq q \), you would get \( k = q  q' \ge 0 \) but \( k a = rr'\) means rr' is some integer multiple of a .. so it follows.

Hunus
 3 years ago
Best ResponseYou've already chosen the best response.0I don't quite understand how that implies that r  r' = 0

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2since both r and r' are assumed to be positive integers which is less than 'a' .. their difference can't be greater than 'a' .. can they?

Hunus
 3 years ago
Best ResponseYou've already chosen the best response.0Ahh. Thank you :) I understand now
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