Show that f:N=N given be f(x)={x+1, if x is odd x-1, if x is even } is bijectve - Maths - Relations and Functions

Question

Show that f:N=N given be f(x)={x+1, if x is odd
x-1, if x is even } is bijectve

ramandeep.singh · Accepted Answer

$f (x) = {\begin{matrix} x - 1, i f x i s e v e n \\ x + 1, i f x i s o d d \end{matrix}}$

Let m, n $\in$ N.

If n and m are odd, then,

$f (n) = f (m) \Rightarrow n + 1 = m + 1 \Rightarrow n = m$

If n and m are even, then,

$f (n) = f (m) \Rightarrow n - 1 = m - 1 \Rightarrow n = m$

Thus, we have f(n) = f(m) ⇒ n = m in both the cases.

If n is odd and m is even, then f(n) = n + 1 is even and f(m) = m - 1 is odd. Therefore, n $\neq$ m ⇒ f(n) $\neq$ f(m). Similarly, if n is even and m is odd, then, n $\neq$ m ⇒ f(n) $\neq$ f(m).

Hence, f is injective.

Let n $\in$ N.

If n is an odd natural number, there exists an even number n - 1 $\in$ N, such that

f (n - 1) = n - 1 + 1 = n

If n is an even natural number, there exists an odd number n + 1 $\in$ N, such that

f (n + 1) = n + 1 - 1 = n

Also, f(1) = 0.

Thus, every element of N has its pre-image in N. So, f is surjective (onto function).

Hence, f is bijective.

Show that f:N=N given be f(x)={x+1, if x is odd x-1, if x is even } is bijectve .

Show that f:N=N given be f(x)={x+1, if x is odd

x-1, if x is even } is bijectve .