Unformatted text preview: Why is âˆ’m < r âˆ’ s < m?
From the prefatory material, Apply the Division Algorithm to
obtain
a = km + r ,
b = ï¿¿m + s , where 0 â‰¤ r < m where 0 â‰¤ s < m Subtracting the second equation from the ï¬rst gives
a âˆ’ b = (k âˆ’ ï¿¿) m + ( r âˆ’ s ) , where âˆ’ m < r âˆ’ s < m What is the biggest that r âˆ’ s could be?
ï¿¿ When r is as big as possible (i.e. when r = m âˆ’ 1 and s is as
small as possible (i.e. when s = 0.)
ï¿¿ Then r âˆ’ s = m âˆ’ 1 âˆ’ 0 = m âˆ’ 1, so r âˆ’ s < m.
What is the smallest that r âˆ’ s could be?
ï¿¿ When r is as small as possible (i.e. when r = 0 and s is as
large as possible (i.e. when s = m âˆ’ 1.)
ï¿¿ Then r âˆ’ s = 0 âˆ’ (m âˆ’ 1) = âˆ’(m âˆ’ 1), so âˆ’m < r âˆ’ s .
Combining these gives âˆ’m < r âˆ’ s < m.
Math 135: Sections 06 and 08 Lecture 11: Congruence Why is r = s ? In Step 2, we have m  (r âˆ’ s ).
From the Prefatory material, we have âˆ’m < r âˆ’ s < m.
How can we conclude the r âˆ’ s = 0?
Since m  (r âˆ’ s ), thus r âˆ’ s = km for some integer k .
But since âˆ’m < r âˆ’ s < m, we must have âˆ’m < km < m. The
only integer value of k that works is k = 0. Thus r âˆ’ s = 0m = 0,
so r = s . Math 135: Sections 06 and 08 Lecture 11: Congruence ...
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This note was uploaded on 10/27/2011 for the course MATH 135 taught by Professor Andrewchilds during the Fall '08 term at Waterloo.
 Fall '08
 ANDREWCHILDS
 Addition, Division

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