Let x 1 2 3 4 5 Let 3 functions be defined as F XX with F w

Let x = {1, 2, 3, 4, 5}. Let 3 functions be defined as:

F: X->X with F with F(1)= 3, F(2) = 2, F(3) = 2, F(4) = 2 F(5) = 5

G: X->X with G(1)=1, G(2)=3, G(3)=4, G(4)=5, G(5)=2

H: X->X with H(1)=2, H(2) = 4, H(3)=1, H(4)=3, H(5)=5

OR

F={(1,3),(2,2),(3,2),(4,2),(5,5)}

G={(1,1),(2,3),(3,4),(4,5),(5,2)}

H={(1,2),(2,4),(3,1),(4,3),(5,5)}

Find the following:

F(g(x))

Solution

Answer:

Let x = {1, 2, 3, 4, 5}. Let 3 functions be defined as:

F: X->X with F with F(1)= 3, F(2) = 2, F(3) = 2, F(4) = 2 F(5) = 5

G: X->X with G(1)=1, G(2)=3, G(3)=4, G(4)=5, G(5)=2

H: X->X with H(1)=2, H(2) = 4, H(3)=1, H(4)=3, H(5)=5

OR

F={(1,3),(2,2),(3,2),(4,2),(5,5)}

G={(1,1),(2,3),(3,4),(4,5),(5,2)}

H={(1,2),(2,4),(3,1),(4,3),(5,5)}

Consider F(g(1))=F(1)=3

F(g(2))=F(3)=2

F(g(3))=F(4)=2

F(g(4))=F(5)=5

F(g(5))=F(2)=2

Now consider G(H(x))

G(H(1))=G(2)=3

G(H(2))=G(4)=5

G(H(3))=G(1)=1

G(H(4))=G(3)=4

G(H(5))=G(5)=2

Consider F(G(H(x)))

F(G(H(1)))=F(G(2))=F(3)=2

F(G(H(2)))=F(G(4))=F(5)=5

F(G(H(3)))=F(G(1))=F(1)=3

F(G(H(4)))=F(G(3))=F(4)=2

F(G(H(5)))=F(G(5))=F(2)=2

Since F={(1,3),(2,2),(3,2),(4,2),(5,5)}

The inverse of F is F-1={(3,1),(2,2),(2,3),2,4),(5,5)}

Since G={(1,1),(2,3),(3,4),(4,5),(5,2)}

The its inverse is G-1={(1,1),(3,2),(4,3),(5,4)(2,5)

Since H={(1,2),(2,4),(3,1),(4,3),(5,5)}

The inverse of H is H-1={(2,1),(4,2),(1,3),(3,4),(5,5)}

The mapping F-1 is not a function since for the element 2 , we have two images like F-1(2)=2 and F-1(2)=3

The mapping G-1 is a function from X to X since it has for each element in X a unique image in X.

The mapping H-1 is a function from X to X since it has for each element in X a unique image in X.