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

Let m, n N

If n and m are odd, then,

If n and m are even, then,

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 m ⇒ f(n) f(m). Similarly, if n is even and m is odd, then,  n m ⇒ f(n) f(m).

Hence, f is injective.

Let n N

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

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

If n is an even natural number, there exists an odd number n + 1 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.

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