MATH 319 Solutions to Assignment 1

 

10.  Given that  prove that .

  Proof:  By the definition of divides a = be and c = df for some integers e and f.  Thus ac=bedf = bdef=(bd)(ef). Thus  by the definition of divides.

 

14. Prove that if n is odd then n2 – 1 is divisible by 8.

   Proof:  An odd integer n is either a 4k+1 or a 4k+3.

(4k+1)2 – 1 = 16k2+ 8k + 1 - 1 =8(2k2 +k), which is a multiple of 8.

(4k+3)2 – 1 = 16k2+ 24k + 9 - 1 =8(2k2 +3k + 1), which is a multiple of 8.

 

21. Prove that if an integer is of the form 6k+5 it is necessarily of the form 3k-1, but not conversely.

Proof:  6k+5 = 6k + 5 + 1 – 1 = 6k + 6 – 1 = 3(2k+2) – 1, a number of the form 3k -1.

    8 = 3(3) – 1 is of the form 3k-1, however, if 8 = 6k+5 then k = ½ which is not an integer. There is a number of the form 3k-1 that is not of the form 6k+5.

 

 

23. Prove that the square of any integer is of the form 3k or 3k+1 but not of the form 3k+2.

   Proof: Every integer is of one of the three forms: 3k  or 3k+1 or 3k+2.

Thus every square is of the form  (3k)2 or (3k+1)2 or (3k+2)2.

(3k)2  = 3(3k2), a number of the form 3k.

(3k+1)2 = 9k2+6k+1=3(3k2+2k)+1, a number of the form 3k+1.

(3k+2)2 = 9k2 + 12k + 4 = 9k2 + 12k + 3 + 1 = 3(3k2 + 4k +1) + 1, a number of the form 3k+1.