0 is not covered in both cases.
A1 ∩ A2 = ∅; however, A 1∪A 2 ≠ Z, as 0 is missing. Therefore, it is not
a partition of the set of all integers.. b. B1 = { n Z: n = 2k, for some integer k } B2 = { n Z: n = 2k + 1, for some integer k } B3 = { n Z: n = 3k, for some integer k }
B1 is the set of all even integers, B2 is the set of all odd integers, and B3
is the set of all integers divisible evenly by 3. E. g. B3 = { 3, 9, 12, 15, 18 …}
This is a partition of Z since B1 U B2 U B3 ¿ Z.
2.( 10 pts) Define f: Z → Z by the rule f( x) = 6x + 1, for all integers x. a. Is f onto?
No. For a function to be onto the codomain must equal the range. A
counter example would be f( x)= y and solve for x. We find that x = y- 1 / 6.
Let y = 0 and x =-1 / 6. This means that, in order to get 0( an integer) as the
resulting value for f( x), we have to input-1 / 6 into the function.-1 / 6 is not
an integer; therefore, the function is not onto. b. Is f one-to-one?
Yes, f is one-to-one. For every x there will be a different f( x) and for every