College Algebra Exam Review 18

College Algebra Exam Review 18 - a D .q 1 C q 2 /d C r 2 ;...

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28 1. ALGEBRAIC THEMES the natural number M D p 1 p 2 ±±± p k C 1 . M is not divisible by any of the primes p 1 ;p 2 ;:::;p k , but M is a product of prime numbers, ac- cording to the previous proposition. If p is any prime dividing M , then f p 1 ;p 2 ;:::;p k ;p g is a collection of k C 1 (distinct) prime numbers. n The fundamental fact about divisibility is the following familiar result (division with remainder): It is always possible to divide an integer a by any divisor d ² 1 , to get a quotient q and a remainder r , with the remain- der being strictly smaller than the divisor. I would like to emphasize how the proof of this fact is related to the algorithm for long division which you learned in school. How does this algorithm work (for a a positive number)? In case a < d , we just put q D 0 and r D a . If a ² d , first we guess an approximation q 1 for the quotient such that 0 < q 1 d ³ a , and compute r 1 D a ´ q 1 d , which is less than a . Now we have a D q 1 d C r 1 ; 0 ³ r 1 < a: If r 1 < d , we are finished; we put q D q 1 and r D r 1 . Otherwise, we try to divide r 1 by d , guessing a q 2 such that 0 < q 2 d < r 1 , and compute r 2 D r 1 ´ q 2 d . Now we have a D q 1 d C r 1 ; r 1 D q 2 d C r 2 ; 0 ³ r 2 < r 1 < a: Combining the two equations, we get
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Unformatted text preview: a D .q 1 C q 2 /d C r 2 ; r 2 &lt; r 1 &lt; a: If r 2 &lt; d , then we put q D q 1 C q 2 and r D r 2 , and we are done. Oth-erwise, 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 ; r n &lt; &lt; r 2 &lt; r 1 &lt; a: Eventually, it must be that r n &lt; d , because the r n are nonnegative and strictly decreasing. (Can you justify this obvious conclusion by ref-erence 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 refer-ence to the principle of mathematical induction, constitutes a proof of the...
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