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Hunus Group TitleBest ResponseYou've already chosen the best response.0
The 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.
 one year ago

Hunus Group TitleBest ResponseYou've already chosen the best response.0
My 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?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
a(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.
 one year ago

Hunus Group TitleBest ResponseYou've already chosen the best response.0
I don't quite understand how that implies that r  r' = 0
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
since 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?
 one year ago

Hunus Group TitleBest ResponseYou've already chosen the best response.0
Ahh. Thank you :) I understand now
 one year ago
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