Unformatted text preview: a D .q 1 C q 2 /d C r 2 ; ³ r 2 < r 1 < a: If r 2 < d , then we put q D q 1 C q 2 and r D r 2 , and we are done. Otherwise, we continue by trying to divide r 2 by d . If the process continues to the n th stage, we obtain partial quotients q 1 ;q 2 ;:::;q n , and successive remainders r 1 ;r 2 ;:::;r n , such that a D .q 1 C q 2 C ±±± C q n /d C r n ; ³ r n < ±±± < r 2 < r 1 < a: Eventually, it must be that ³ r n < d , because the r n are nonnegative and strictly decreasing. (Can you justify this “obvious” conclusion by reference to some version of the principle of mathematical induction?) Then we stop, and put q D q 1 C q 2 C ±±± C q n and r D r n . The procedure for negative a is similar. The description of this algorithm, made a little more formal by reference to the principle of mathematical induction, constitutes a proof of the...
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 Fall '08
 EVERAGE
 Algebra, Division, Prime Numbers, Natural number, Prime number, Divisor

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