Let N be the set of natural numbers and two functions f and g be defined as f, g: N → N such that: f(n) = ⎧⎪⎪⎨⎪⎪⎩n+12 if n is odd; n² if n is even and g(n) = n −⌊n/2⌋n. The fog is:

f(x) = ⎧⎪⎪⎨⎪⎪⎩n+12 if n is odd; n² if n is even
g(n) = n −⌊n/2⌋
Let us find fog(n)
fog(n) = f(g(n)) = f(n −⌊n/2⌋)
If n is even, let n = 2k
Then g(n) = 2k − k = k
f(g(n)) = f(k)
If k is odd, f(k) = k + 12
If k is even, f(k) = k²
If n is odd, let n = 2k + 1
Then g(n) = 2k + 1 − k = k + 1
f(g(n)) = f(k+1)
If k is even, k+1 is odd, f(k+1) = k + 1 + 12 = k+13
If k is odd, k+1 is even, f(k+1) = (k+1)²
Thus fog(n) is neither one-one nor onto