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and drag on elements in order hese statements in order to produce a proof that the integers a and b are congruent modulo m if and only Instructions Then there is an integer k such that km = a – b, so that a = b + k Now, we perform a direct proof of the other direction, beginning with the assumption that there Hence, m divides a – b, so that a = b (mod m). We perform a direct proof in both directions, first proving sufficiency and th

and drag on elements in order hese statements in order to produce a proof that the integers a and b are congruent modulo m if and only Instructions Then there is an integer k such that km = a – b, so that a = b + k Now, we perform a direct proof of the other direction, beginning with the assumption that there Hence, m divides a – b, so that a = b (mod m). We perform a direct proof in both directions, first proving sufficiency and th

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Ordering Question Click and drag on elements in order Put these statements in order to produce a proof that the integers a and b are congruent modulo m if and only if there is an integer k such that a = b + km. i Instructions Then there is an integer k such that km = a - b, so that a = b + km. Now, we perform a direct proof of the other direction, beginning with the assumption that there is an integer k such that a = b + km. Hence, m divides a - b, so that a = b (mod m). We perform a direct proof in both directions, first proving sufficiency and then necessity. If there is an integer k such that a = b + km, then km = a -b. If a = b (mod m) then mI (a – b).
Transcribed Image Text:Ordering Question Click and drag on elements in order Put these statements in order to produce a proof that the integers a and b are congruent modulo m if and only if there is an integer k such that a = b + km. i Instructions Then there is an integer k such that km = a - b, so that a = b + km. Now, we perform a direct proof of the other direction, beginning with the assumption that there is an integer k such that a = b + km. Hence, m divides a - b, so that a = b (mod m). We perform a direct proof in both directions, first proving sufficiency and then necessity. If there is an integer k such that a = b + km, then km = a -b. If a = b (mod m) then mI (a – b).
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