A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing

This Question is Closed

Hunus
 2 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
 2 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
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.0I don't quite understand how that implies that r  r' = 0

experimentX
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.0Ahh. Thank you :) I understand now
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.