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Many-One into Function

15. let a=(0 ...

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15. let a=(0 1) and the set of natural numbers. then the mapping f:N-A defined by f(2n-1)=0,f(2n)=1 for all n belongs to N is 1)one one function 2)many one onto 3)one one onto 4)many one into

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Q. Let A = {0, 1} and N the set of all natural numbers. Then the mapping

f : N \to A

f (2 n - 1) = 0, f (2 n) = 1 \forall n ϵ N

is many-one onto.

A = {0, 1}

N

the set of all natural numbers. Then show that the mapping

f : N \to A

f (2 n - 1) = 0, f (2 n) = 1 \forall n \in N

is many-one onto.

State true or false:

A = {0, 1}

N

the set of all natural numbers. Then the mapping

f : N \to A

f (2 n - 1) = 0, f (2 n) = 1

n \in N

, is many-one onto.

f

be a function on non negative integers defined as follows

f (2 n) = f (f (n))

f (2 n + 1) = f (2 n) + 1 :

f (0) = 0

f (n)

n

.

N

be the set of natural numbers and two functions

f

g

f, g : N \to N

f (n) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{n + 1}{2} & if n is odd \frac{n}{2} & if n is even \end{matrix}

g (n) = n - (- 1)^{n}

f \circ g

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