at k=0,f(1)=1=f(2)=1⇒f is not one-one 2−k covers all integers {1,0,−1,...} k+1 covers {1,2,3,...} ⇒f(n) covers all integers ⇒f(n) is onto but not one-one g(n)={
3+2n,‌
n≥0
−2n,‌
n<0
Notice that 1∉ range of g(x) as 3+2n≠1,n≥0 and −2n≠1,n<0 ⇒g(n) is not onto g(f(n))={
3+2f(n),‌
f(n)≥0
−2f(n),‌
f(n)<0
={
3+2(‌
n+1
2
),‌
n=2k+1
3+2(‌
4−n
2
),‌
n=2k
={
n+4
n=2k+1
7−n,‌
n=2k
. g(f(n)) is always odd ⇒ not onto at g(f(1))=5=g(f(2))⇒0 not one-one f(g(n))={