Question

For each of the functions below, indicate whether the function is onto, one-to-one, neither or both. If the function is not onto or not one- to-one, give an example showing why.

(a) f: R → R. f(x) = x²
(b) g: R → R. g(x) = x³
(c) h: Z → Z. h(x) = x³
(d) f: Z → Z, f(x) = [x/5] - 4
(e) f: Z → Z, f(x) = 5x - 4
(f) f: Z → Z, f(x) = x-4
(g) f: Z x Z → Z x Z, f(x, y) = (x+1, 2y)
(h) f: Z x Z → Z x Z, f(x, y) = ([x/5],y-2)
(i) f: Z x Z → Z x Z, f(x, y) = ([x/5],5y-2)
(j) f: Z x Z → Z, f(x, y) = (1-y, 1-x)
(k) f: Z⁺ x Z⁺ → Z⁺, f(x, y) = 2ˣ + y.
I have to solve (b),(d),(f),(g),(i)

Answer 1

For function g(x) = x³, it is both onto and one-to-one. Function f(x) = [x/5] - 4 is neither onto nor one-to-one. Function f(x) = x-4 is onto but not one-to-one. Function f(x, y) = (x+1, 2y) is onto but not one-to-one. Function f(x, y) = ([x/5],5y-2) is neither onto nor one-to-one.

Step-by-step explanation:

(b) g: R → R. g(x) = x³:
The function g(x) = x³ is both onto and one-to-one. It is onto because for every real number y, there exists a real number x such that x³ = y. It is one-to-one because every distinct input x corresponds to a distinct output x³. Therefore, g(x) = x³ is both onto and one-to-one.

(d) f: Z → Z, f(x) = [x/5] - 4:
The function f(x) = [x/5] - 4 is neither onto nor one-to-one. It is not onto because there are certain integers that can never be reached as output, such as multiples of 5. It is not one-to-one because different inputs can result in the same output, for example f(4) = f(9) = -3.

(f) f: Z → Z, f(x) = x-4:
The function f(x) = x-4 is onto but not one-to-one. It is onto because every integer y can be obtained by adding 4 to an input x (e.g., f(x) = y+4). However, it is not one-to-one because different inputs can yield the same output, for example f(3) = f(7) = -1.

(g) f: Z x Z → Z x Z, f(x, y) = (x+1, 2y):
The function f(x, y) = (x+1, 2y) is onto but not one-to-one. It is onto because any pair of integers (x, y) can be obtained by incrementing x by 1 and doubling y. However, it is not one-to-one because different pairs of inputs can yield the same pair of outputs, for example f(1, 2) = f(2, 1) = (2, 4).

(i) f: Z x Z → Z x Z, f(x, y) = ([x/5],5y-2):
The function f(x, y) = ([x/5],5y-2) is neither onto nor one-to-one. It is not onto because there are certain pairs of integers that cannot be obtained as output, such as (x, y) where x is not divisible by 5. It is not one-to-one because different pairs of inputs can result in the same pair of outputs, for example f(3, 1) = f(8, 0) = (0, -2).