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Unformatted text preview: Homework 7 Solutions Section 4.2 4. a) ¡¢£¤¥ ¢£ ¦ ¢ § ¨ © ª § « ¡¢¬¤¥ ¢¬ ¦ § ¨ © ¢ § « ¡ª®¤¥¯ª® ¦ ¨ § ° © ® § « ¡ª¢¤¥ ª¢ ¦ ® § ° © § « b) Assume ¡±¤ for all ¢£¯ ² ±¯ ³ ´ c) ¡´¤ d) ´ ¦ ¨ © ¡´ µ ¨¤ ¦ ¯¨ © ¡¨¶ © «·¤ ¦ ¨¡¶ © ¢¤ © «· for some nonnegative i, j e) We have shown that the statement is true for several basis cases ( ¢£¯ ² ´¯ ² ª¢ ) and that in the inductive steps all other cases are reducible to smaller cases, which are themselves reducible to the basis cases. Therefore by the principle of strong induction, the statement is true for all ´ ¸ ¢£¹ 12. Basis case: n=1: ´ ¦ ª º Inductive step: Assume that for all ¢¯ ² ±¯ ³ ´ , n is representable as a sum of distinct non negative powers of two. This proof is broken into two cases: n is even, and n is odd. n is even: ´ ¦ ª± for some ±¯ ³ ´ . By the inductive hypothesis, ± ¦ ª » ¼ © ª » ½ ©¯¹ ¹ ¹ ©ª » ¾ for m distinct values of a . Since ´ ¦ ª± , ´ ¦ ª » ¼ ¿À © ª » ½ ¿À ©¯¹ ¹ ¹ ©ª » ¾ ¿À . Thus n is representable as a sum of distinct powers of two. n is odd: ´ ¦ ª± © ¢ for some ± ³ ´ . By the inductive hypothesis, ± ¦ ª » ¼ © ª » ½ ©¯¹ ¹ ¹ ©ª » ¾ for m distinct values of a . Since ´ ¦ ª± +1, ´ ¦ ª » ¼ ¿À © ª » ½ ¿À ©¯¹ ¹ ¹ ©ª » ¾ ¿À © ª º . Since ®¯ Á · © ¢ for any nonnegative j, ª º is distinct among the powers of two. Thus n is representable as a sum of distinct powers of two. 32. The proof fails because there is no way to form 5 cents from 3 and 4 cent postage stamps. More specifically, the proof assumes that if we can represent 4 cents that we can either replace a 3cent stamp with a 4cent stamp or replace two 4cents stamps with three 3cent stamps. However, 4 cents is not formed with either a 3cent stamp or two 4cent stamps, so there is a logical flaw in the...
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This note was uploaded on 02/12/2012 for the course EECS 203 taught by Professor Yaoyunshi during the Spring '07 term at University of Michigan.
 Spring '07
 YaoyunShi

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