Definition of Bijective Function:
A function is said to be bijective if it is both injective (one-to-one) and surjective (onto). In other words, every element in the domain is mapped to a unique element in the codomain, and every element in the codomain has a pre-image in the domain.

Proof for Injectivity:
To check if the function f is injective, we need to show that different inputs map to different outputs.

Case 1: n is odd
Let's assume that f(a) = f(b), where a and b are odd numbers.
Then, (a-1)/2 = (b-1)/2.
Cross multiplying, we get a - 1 = b - 1.
Therefore, a = b.
Hence, f is injective for odd numbers.

Case 2: n is even
Let's assume that f(a) = f(b), where a and b are even numbers.
Then, a/2 = b/2.
Cross multiplying, we get a = b.
Therefore, a = b.
Hence, f is injective for even numbers.

Since f is injective for both odd and even numbers, it is injective overall.

Proof for Surjectivity:
To check if the function f is surjective, we need to show that every element in the codomain has a pre-image in the domain.

Case 1: n is odd
For any odd number n, let's find its pre-image in the domain.
We can rewrite the function as f(n) = (n-1)/2.
Given any y in the codomain, we need to find an odd number n such that f(n) = y.
Solving the equation (n-1)/2 = y, we get n = 2y + 1.
Therefore, every element in the codomain has a pre-image in the domain for odd numbers.

Case 2: n is even
For any even number n, let's find its pre-image in the domain.
We can rewrite the function as f(n) = n/2.
Given any y in the codomain, we need to find an even number n such that f(n) = y.
Solving the equation n/2 = y, we get n = 2y.
Therefore, every element in the codomain has a pre-image in the domain for even numbers.

Since every element in the codomain has a pre-image in the domain, the function f is surjective overall.

Conclusion:
Since the function f is both injective and surjective, it is bijective.