Suppose f:N⟶N is a function defined by f(x)=a(x+b), where a,b∈N. If f is a bijective function, then find the value of a+b
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Answer:
Step-by-step explanation:
(n)={
2
n+1
,if n is odd
2
n
,if n is even
} for all n∈N
f:N→N is defined as
It can be observed that:
f(1)=
2
1+1
=1 and f(2)=
2
2
=1
∴f(1)=f(2), where 1
=2
∴f is not one-one.
Consider a natural number (n) in co-domain NCase I: n is odd
∴n=2r+1 for some r∈N.
Then, there exists 4r+1∈N such that f(4r+1)=
2
4r+1+1
=2r+1
Case II: n is even
∴n=2r for some r∈N.
Then, there exists 4r∈N such that f(4r)=
2
4r
=2r
∴f is onto.
Hence, f is not a bijective function
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