Math135Lecture11Additional

Math135Lecture11Additional - Why is −m < r − s...

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Unformatted text preview: Why is −m < r − s < m? From the prefatory material, Apply the Division Algorithm to obtain a = km + r , b = ￿m + s , where 0 ≤ r < m where 0 ≤ s < m Subtracting the second equation from the first gives a − b = (k − ￿) m + ( r − s ) , where − m < r − s < m What is the biggest that r − s could be? ￿ When r is as big as possible (i.e. when r = m − 1 and s is as small as possible (i.e. when s = 0.) ￿ Then r − s = m − 1 − 0 = m − 1, so r − s < m. What is the smallest that r − s could be? ￿ When r is as small as possible (i.e. when r = 0 and s is as large as possible (i.e. when s = m − 1.) ￿ Then r − s = 0 − (m − 1) = −(m − 1), so −m < r − s . Combining these gives −m < r − s < m. Math 135: Sections 06 and 08 Lecture 11: Congruence Why is r = s ? In Step 2, we have m | (r − s ). From the Prefatory material, we have −m < r − s < m. How can we conclude the r − s = 0? Since m | (r − s ), thus r − s = km for some integer k . But since −m < r − s < m, we must have −m < km < m. The only integer value of k that works is k = 0. Thus r − s = 0m = 0, so r = s . Math 135: Sections 06 and 08 Lecture 11: Congruence ...
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