Hunus
  • Hunus
Number Theory problem
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Hunus
  • Hunus
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.
Hunus
  • Hunus
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?
experimentX
  • experimentX
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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Hunus
  • Hunus
I don't quite understand how that implies that |r - r'| = 0
experimentX
  • experimentX
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?
Hunus
  • Hunus
Ahh. Thank you :) I understand now
experimentX
  • experimentX
yw

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