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