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 .