Question

True or False: Let A={0,1} and N be the set of natural numbers.Then the mapping f:N→A defined by f(2n-1)=0,f(2n)=1,n∈N,is onto.

Answer 1

Answer:

True

Step-by-step explanation:

By the definition of f,

f(13) = 0, since 13 = 2n-1 for n = 7

and

f(28) = 1, since 2 = 2n for n = 14

So all values in A do occur as values of f.

That means f is onto.

Answer 2

Answer:

Mark me as brainliest and thank me if the answer is useful.

Step-by-step explanation:

Given f:N→A such that

f(2n−1)=0 and f(2n)=1

So, image of all even natural numbers is 1 and image of all odd natural numbers is 0.

Thus, distinct elements of N has same image in A.

Hence, f is many-one.

Since, R(f)={0,1}

which is equal to co-domain A.

Hence, f is onto.