Functions (Home Work)

Question

1. Define the following functions on the integers by f(k) = k + 1, g(k) = 2k, and h(k) = ⌈k/2⌉
(a) Which of these functions are one-to-one?
(b) Which of these functions are onto?
(c) Express in simplest terms the compositions f °g, g ° f , g ° h, h ° g, and h2.

2. Inverse images. If f is any function from A into B, we can describe the inverse image of from B into P(A), which is also commonly denoted
f -1. If b ∈ B, f -1(b) = {a ∈ A | f(a) = b). If f does have an inverse, the inverse image of b is {f-1(b)}.
(a) Let g : R → R be defined by g(x) = x2. What are g-1(4), g -1(0) and g-1(-1)?
(b) If r : R → Z, where r(x) = Γr⌉, what is r-1(1)?

3. If f , u, d : N → N, where f(a) = 2a, u(a) = a + 1, and d(a) = max(0, a - 1), calculate
(a) (f °u)(a).
(b) f2(a).
(c) (d °( f°u))(a).
(d) (d °u)(a) = a; therefore, what is d °u ?
(e) Explain why d is not the inverse of u.

Answer 1

1. Let Z be the set of integers.

    Each of these functions map from Z→Z

 

    (a) Only f(k) and g(k) are one-to-one.

         h(k) is not one-to_one because the integer h(2) = ⌈2/2⌉ = ⌈1⌉ = 1

         h(1) = ⌈1/2⌉ =  ⌈0.5⌉ = 1.  Since there are two values in the domain

         that have the same value, the function is not one-to-one.

 

    (b) Only f(k) and h(k) are onto

         g(k) is not onto, since there is no integer such that g(k) = 1

 

    (c) (f °g)(k) = f(g(k)) = g(k) + 1 = 2k + 1

         (g ° f)(k) = g(f(k)) = 2f(k)= 2(k + 1) = 2k + 2

         (g ° h)(k) = g(h(k)) = 2 *h(k) = 2⌈k/2⌉

        (h ° g)(k) = h(g(k) = ⌈g(k)/2⌉= ⌈2k/2⌉ = k, since the ceiling function always results in an integer

        h²(k) = (h ° h)(k) = ⌈h(k)/2⌉ = ⌈⌈(k/2⌉/2⌉

 

2. Using the definition of inverse defined above.

 

     (a) g R → R , g(x) = x²

         Then g^-1(b) = {a ∈ R | a² = b}, b ∈ R

 

         g^-1(4) = {a ∈ R | a² = 4} = {-2, 2}

         g^-1(0) = {a ∈ R | a²= 0} = {0}

         g^-1(-1) = {a ∈ R | a²= -1} = ∅, since the square of a real number is always non-negative..

 

    (b) r:R → Z,  r(x) = Γx⌉,

         Then r^-1(b) = {a ∈ R | Γa⌉ = b}, b ∈ Z

         r^-1(1) = {a ∈ R | Γa⌉ = 1} = [1, 2)

         This is the same 1 ≤ a < 2

 

3.  Assuming N is the set of positive integers

 

     (a) (f °u)(a) = f(u(a)) = 2u(a) = 2(a + 1) = 2a + 2

 

     (b) f²(a) = (f °f)(a) = f(f(a)) = 2f(a) = 2(2a) = 4a

 

     (c) (d °( f°u))(a) = d(f(u(a))

                              = d(2a + 2) from (a)

                              = max(0, 2a+2 -1) = max(0, 2a+1)

                              = 2a +1, since 2a + 1 > 0

 

    (d) (d °u)(a) = d(u(a)) = max(0, u(a) -1)

                                      = max(0, a + 1 - 1)

                                      = max(0, a) = a

          This is true since a > 0.

          u is the inverse of d

 

    (e) Since a ∈ N, let a > 1.

         Then d(a) = max(0, a -1) = a -1 since this

         would be a positive integer.

 

         (u °d)(a) = u(d(a))

                       = d(a) + 1

                       = a - 1 + 1 = a

 

        Since 1 ∈ N, d(1) = max(0, 0) = 0

        But this is not possible because the image/co-domain

        of d is N and 0 ∉ N.  Therefore d(1) is not defined and

        has no value when a = 1.

 

       (u °d)(1) = u(d(1)) = d(1) + 1

      Since d(1) is undefined, this composite function is also

      not defined when a = 1.

 

      Because there is one value that does map from d to u

      d is not the inverse of u.  For all other values the inverse

      mapping does  exists.