Question #ea4c3

1 Answer
Nov 6, 2015

By writing out a few terms for the function z, we see that
#z={(1,1),(2,2),(3,8),(4,13),(5,8),(6,13),(7,8), 8,13),........}#

By definition, a function #f:X->Y# is onto (surjective) #iff AAy in Y, EE X in X# such that #f(x)=y#.

So in this case for example, #EE 4 in ZZ# such that #x(x) != 4 AA x in ZZ#.
#therefore z# is not onto.

By definition, a function #f:X->Y# is 1 -1 (injective) #iff f(x_1)=f(x_2)=>x_1=x_2#.

So in this case for example, #3!=5# but #z(3)=z(5)=8#.

#therefore z# is not 1 - 1.