Let N be set of natural numbers, then the function f: mathbbN → mathbbN, defined by f(x) = …, is

The function takes a natural number and:

• If $n$n is odd: $f(n) = \frac{n+1}{2}$f(n)=2n+1​
• If $n$n is even: $f(n) = \frac{n}{2}$f(n)=2n​

Examples:

$$f(1) = \frac{1+1}{2} = 1$$f(1)=21+1​=1

$$f(2) = \frac{2}{2} = 1$$f(2)=22​=1

$$f(3) = \frac{3+1}{2} = 2$$f(3)=23+1​=2

$$f(4) = \frac{4}{2} = 2$$f(4)=24​=2

$$f(5) = \frac{5+1}{2} = 3$$f(5)=25+1​=3

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To check if the function is injective, consider whether different inputs can produce the same output.

$$f(1) = 1$$f(1)=1

$$f(2) = 1$$f(2)=1

Since $f(1) = f(2)$f(1)=f(2) but $1 \neq 2$1=2, the function is not injective.

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To check if the function is surjective, determine whether every natural number $k$k can be reached as an output.

For any natural number $k$k, let $n = 2k$n=2k.

Then $f(2k) = \frac{2k}{2} = k$f(2k)=22k​=k

Examples:

For $k = 1$k=1: $f(2) = 1$f(2)=1

For $k = 5$k=5: $f(10) = 5$f(10)=5

For $k = 100$k=100: $f(200) = 100$f(200)=100

Every natural number is reachable as an output, so the function is surjective.

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The function is surjective but not injective.