Arrangement:
For n = 1, 2
f(1) = (n+1)/2 = (1+1)/2 = 1 and f(2) = (n)/2 = (2)/2 = 1
f(1) = f(2), yet 1 ≠ 2
f isn’t one-one.
For a characteristic number, “a” in co-space N
In case n is odd
n = 2k + 1 for k ∈ N , then, at that point 4k + 1 ∈ N to such an extent that f(4k+1) = (4k+1+1)/2 = 2k + 1
In case n is even
n= 2k for some k ∈ N to such an extent that f(4k) = 4k/2 = 2k
f is onto
In this manner, f is onto however not bijective capacity.
Solution:
For n = 1, 2
f(1) = (n+1)/2 = (1+1)/2 = 1 and f(2) = (n)/2 = (2)/2 = 1
f(1) = f(2), yet 1 ≠ 2
f isn’t one-one.
For a characteristic number, “a” in co-space N
In case n is odd
n = 2k + 1 for k ∈ N , then, at that point 4k + 1 ∈ N to such an extent that f(4k+1) = (4k+1+1)/2 = 2k + 1
In case n is even
n= 2k for some k ∈ N to such an extent that f(4k) = 4k/2 = 2k
f is onto
In this manner, f is onto however not bijective capacity.