hwk1 - Solutions Math 567 Homework #1 September 11th, 2009...

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Unformatted text preview: Solutions Math 567 Homework #1 September 11th, 2009 19/2 Show that any integer of the form 6 k + 5 is also of the form 3 j + 2, but not con- versely. Suppose that a = 6 k + 5 for any integer k . Then using the division algorithm, we may write 6 = 3 2 and 5 = 3 1 + 2. Using these substitutions, we may write a = 6 k + 5 = ( 3 2 ) k + 3 1 + 2, which we may then factor as a = 3 ( 2 k + 1 ) + 2. Then for a given k , we find that j = 2 k + 1 yields that a may be written as a = 3 j + 2. To see that the converse does not hold, we make take b = 3 j + 2, where j = 2 n for some integer n . Then b = 3 j + 2 = 3 ( 2 n ) + 2 = 6 n + 2. By the uniqueness of the quotient and remainder in the division algorithm, we know that this decomposition for b is unique, and we therefore cannot find some other m for which b = 6 m + 5. 19/3 Use the Division Algorithm to establish the following: (a) The square of any integer is either of the form 3 k or 3 k + 1. (b) The cube of any integer has one of the firms: 9 k , 9 k + 1, or 9 k + 8. (c) The fourth power of any integer is either of the form 5 k or 5 k + 1. (a) Using the division algorithm, we know that we may write any integer as 3 j , 3 j + 1, or 3 j + 2 for some j Z . Now ( 3 j ) 2 = 9 j 2 = 3 k for k = 3 j 2 . Similarly, ( 3 j + 1 ) 2 = 9 j 2 + 6 j + 1 = 3 ( 3 j 2 + 2 j ) + 1 = 3 k + 1. Finally, we compute ( 3 j + 2 ) 2 = 9 j 2 + 12 j + 4 = 3 ( 3 j 2 + 6 j + 1 ) + 1, and this is of the form 3 k + 1 for k = 3 j 2 + 6 j + 1. (b) For part (b), we claim that it suffices to show that r 3 is congruent to 0,1, or 8 modulo 9 for 0 r < 8. Using the division algorithm, we know that if a is any integer, we can write a = 9 q + r where 0 r < 9. When we compute a 3 = ( 9 q + r ) 3 , we notice that if we expand out ( 9 q + r ) 3 , all of the terms of this expansion are divisible by 9 except for the r 3 term. This fact is true by the binomial theorem. Therefore, a 3 = ( 9 q + r ) 3 = 9 j + r 3 for some integer j , and if we can show that r 3 is either 0,1, or 8 modulo 9 for 0 r < 9, the result will follow. We will present this argument again in less detail in future proofs. We therefore compute: 0 3 = 0, 1 3 = 1, and 2 3 = 8, so all integers of the form 9 q , 9 q + 1, and 9 q + 2 have the desired form. 3 3 = 27, which is congruent to 0 mod- ulo 9. We write 3 3 ( mod 9 ) to mean 3 3 is congruent to 0 modulo 9. Likewise, 4 3 = 64 1 ( mod 9 ) and 5 2 = 125 8 ( mod 9 ) , so integers of the form 9 q + 3, 9 q + 4, and 9 q + 5 all have the required property. Finally, 6 3 = 216 ( mod 9 ) , 7 3 = 343 1 ( mod 9 ) , and 8 3 = 512 8 ( mod 9 ) ....
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This note was uploaded on 06/12/2011 for the course MATH 103 taught by Professor Wouters during the Spring '08 term at Wisc Oshkosh.

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hwk1 - Solutions Math 567 Homework #1 September 11th, 2009...

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