Number Theory problem

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Number Theory problem

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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The division algorithm states 'For any positive integers a and b there exists a unique pair (q, r) of non-negative 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 a|r′ −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.
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?
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 \( k a = |r-r'|\) means |r-r'| is some integer multiple of a .. so it follows.

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I don't quite understand how that implies that |r - r'| = 0
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?
Ahh. Thank you :) I understand now
yw

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