Consider f : N → N, g : N → N and h : N → R defined as f (x) = 2 x, g(y) = 3 y + 4 and h(z) = sin z ∀ x, y and z in N. Show that h o (g o f) = (h o g) o f.
X= {1, 2, 3}, Y = {a, b, c}
f : X → Y is one-one and onto such that
f(1) = a f(2) = b, f(3) = c
Let g : Y ∴ Y X be a function such that g (a) = 1, g (b) = 2, g (c) = 3
Now (g o f) (1) = g (f (1)) = g(a) = 1
(g o f) (2) = g (f (2)) = g (b) = 2
(g o f)(3) =g(f(3)) = g (c) = 3 ∴ g o f = 1x
Again (f o g) (a) = f(g (a)) = f(1) = a (f o g) (b) = f(g (b)) = f(2) = b
(f o g) (c) = f(g (c)) = f (3) = c
∴ f o g = IY.
Let f, g and h be function from R to R. Show that (f + g) o h = f o h + g o h
(f . g) o h = (f o h) . (g o h)