Show that the function `f: N->N` given by, `f(n)=n-(-1)^n` for all `n in N` is a bijection.

Options

A. one-one and into
B. one-one and onto
C. many-one and into
D. many-one and onto

Answer

A

Explanation

We have,
`f(n)=n(-1)^(n)={{:(,n-1,"if n is even"),(,n+1,"if n is odd"):}`
Injectivity Let n, m be any two even natural number. Then,
`f(n)=f(m) Rightarrow n-1=m-1Rightarrow m`
If n, m are any two odd natural numbers. Then,
`f(n)=f(m) Rightarrow n+1=m+1 Rightarrow n=m`
Thus, in both the cases, we have
`f(n)=f(m)Rightarrow n=m`
If n is even and m is odd, then `n ne m`. Also, f(n) is odd f(m) is even. So, `f(n)nef(m)`
`"Thus", n ne m Rightarrow f(n) ne f(m)`
So, f is an injective map.
Surjectivity Let n be an arbitary natural number.
If n is odd natural number, then there exists an even natural number n+1 such that
`f(n+1)=n+1-1=n`
If n is an even natural number, then there exists an odd natural number (n-1) such that
`f(n-1)=n-1+1=n`
Thus, every `n in N` has its pre-image in N`.
So, `f: N to N is a surjection`.
Hence, `f:N to N` is a bijection.