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Unformatted text preview: Introductory Number Theory.Topics for the make up exam 1. Prove: There exists an infinity of prime numbers. 2. True or false: n 2 81 n + 1681 is prime for all n N ? 3. Prove that the product of any three consecutive numbers is divisible by 6. 4. Let b N , b 1. Prove that every a N can be expressed in the form a = d n b n + + d 1 b + d = n k =0 d k b k for a unique n Z , n 0,unique d , . . . , d n { , 1 , . . . , b 1 } , d n = 0. Hint: This is, of course, the representation of a in base b . Uniqueness means that if n k =0 d k b k = m j =0 e j b j , d n = 0 = e n , d k , e j { , 1 , . . . , b 1 } then n = m and d k = e k for 0 k n . One way of proving this is to say that without loss of generality we may assume that m n , and if m &lt; n , set e m +1 = = e n = 0, so we now deal with the same number of digits. It now suces to prove that d k = e k for k = 0 , . . . , n . This automatically implies n = m , because then e n = d n = 0, so we must have had n = m . For a contradiction assume that not all digits are equal; there exists then a last k , 0 k n such that d k = e k . For example, if n &gt; m , this last one will already be d n ; i.e., k = n . If k &lt; k n , then d k = e k . (If k = n , then the set k &lt; k n is empty.) Without loss of generality we may assume d k &gt; e k ; the proof if e k &gt; d k is identical. We now have d k b k + + d 1 b + d = e k b k + + e 1 b + e (since the terms preceding the k term are all equal). If k = 0 we have an immediate contradiction: d = e (since with k = 0 wed have d &gt; e ), so henceforth we assume k 1. Rearranging the equation, ( d k e k ) b k = ( e k 1 d k 1 ) b k 1 + + ( e 1 d 1 ) b + ( e d ) . On the RHS,  e k d k  b 1, thus the RHS of this equation satisfies  ( e k 1 d k 1 ) b k 1 + +( e 1 d 1 ) b +( e d )  ( b 1) ( b k 1 + + b + 1 ) = ( b 1) b k 1 b 1 = b k 1 &lt; b k ....
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This note was uploaded on 07/13/2011 for the course MAS 3202 taught by Professor Staff during the Summer '11 term at FAU.
 Summer '11
 STAFF

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