THEOREMS
Solution: In case we need not find inverse, then we can just show that the functions are one-one & onto. In this case inverse has to be found as well, so lets find gof & gof, & see if they are equal.
Given that Y = { y ∈ N: y = 4x + 3 for some x ∈ N }. y = 4x + 3 or x =( y-3)/ 4 Given, f: N → Y f( x)= y, lets define g: y à N be g( y) = x, or g( y) =( y-3)/ 4
gof( x) = g( f( x)) = g( 4x + 3) =( 4x + 3-3)/ 4 = x---- replacing f( x) with 4x + 3 since f( x) = 4x + 3
fog( y) = f( g( y)) = f( x) = y Since gof = In & fog = In So f is invertible. Function g is the inverse of f. g: y à N be g( y) =( y-3)/ 4
THEOREMS
Theorem 1: If f: X → Y, g: Y → Z and h: Z → S are functions, then ho( gof) =( hog) o f.
Numerical: Consider f: N → N, g: N → N and h: N → R defined as f( x) = 2x, g( y) = 3y + 4 and h( z) = sin z, ∀ x, y and z in N. Show that ho( gof) =( hog) of.
Solution: LHS = ho( gof) = ho( g( fx)) = ho( g( 2x)) = h( 3 * 2x + 4) = h( 6x + 4) = sin( 6x + 4)
RHS =( hog) of = Hog( f( x)) = hog( 2x) = h( g( 2x)) = h( 3 * 2x + 4) = h( 6x + 4) = sin( 6x + 4)
Thus LHS = RHS
Theorem 2: Let f: X → Y and g: Y → Z be two invertible functions. Then gof is also invertible with( gof) – 1 = f – 1 og – 1
Numerical: Consider f: { 1, 2, 3 } → { a, b, c } and g: { a, b, c } → { apple, ball, cat } defined as f( 1) = a, f( 2) = b, f( 3) = c, g( a) = apple, g( b) = ball and g( c) = cat.