a) .
k ∈ Z | 16/2k+1 ∈ Z .
ca 16/2k+1 ∈ Z , trebuie ca 2k+1 ∈ D₁₆ .
D₁₆ = {-16 , -8 , -4 , -2 , -1 , 1 , 2 , 4 , 8 , 16 . } .
calculăm 2k + 1 = -16
2k = -17 , k = -17/2 .
2k+1 = -8 .
2k = -9 , k = -9/2 .
2k+1 = -4
2k = -5 , k = -5/2 .
2k + 1 = -2
2k =-3 , k = -3/2 .
2k+1 = -1
2k = -2 , k = -1 .
2k+1 = 1
2k = 0 , k = 0 .
2k+1 = 2
2k = 1 , k = 1/2 .
2k+1 = 4
2k = 3 , k = 3/2 .
2k+ 1 = 8
2k = 7 , k = 7/2 .
2k + 1 = 16
2k = 15 , k = 15/2 . => k ∈ { -1 , 0 } (doar aceste 2 nr ∈ Z 0
. b) n∈Z | 6/n + 2 ∈ Z .
ca 6/n+2 ∈Z , trebuie ca n+2 ∈D₆
D₆={-6,-3,-2,-1 ,1,2,3,6} .
n+2 = -6
n = -8 .
n+2 = -3
n = -5 .
n+2 = -2
n = -4 .
n+2 = -1
n = -3 .
n+ 2 = 1
n = -1 .
n+2 = 2
n = 0 .
n+2 = 3
n = 1 .
n+2 = 6
n = 4 . => n ∈ { -8,-5,-4,-3,-1,0,1,4 } . ( toate ∈ Z ) .
c).n ∈ Z , | 8/3-n∈ Z .
ca 3 8/3-n ∈ Z , trebuie ca 3-n ∈ D₈
D₈ = { -8 , -4 , -2 , -1 , 1 , 2 , 4 , 8 } .
3-n = -8
-n = -11 , n = 11.
3-n = -4
n=7.
3-n = -2
n = 5 .
3-n = -1
n = 4 .
3-n = 1
n = 2 .
3-n = 2
n = 1 .
3-n = 4
n = -1
3-n = 8
n = -5 . => n ∈ { -5 , -1 , 1 , 2 , 4 , 5 , 7 , 11 }
