Answer :
Final answer:
It involves the determination of bijection for back-specified functions, the formulation of a linear and a non-linear bijection between given intervals, and the development of a bijection between Natural and Integer number sets.
Explanation:
A bijection, or bijective function, is a function that is both injective (or one-to-one) and surjective (or onto). To solve each part of the question, we'll have to determine whether the definitions of bijections apply in each case.
(a) The function f(x) = x is a bijection of [0,1] to [0,1].
(b) Let's look at the function f(x) = x² on the interval [-1,1]. For this function to be bijective, every element from the domain [-1,1] must map to a unique element in the range [0,1] and every element in the range must be mapped to. However, the function is not injective or one-to-one because both x=1 and x=-1 map to the same element y=1 in the range. Therefore, f(x) = x² is not a bijection of [-1,1] to [0,1].
(c) For a linear bijection of [0,1] onto the interval [3,6] we can find a linear function in the form f(x) = mx+b that satisfies necessary conditions. We can use the endpoints of our provided domain [0,1] as our x-values, and [3,6] as the corresponding y-values. Solving these equations, we find that m=3 and b=3. Thus, the function f(x)=3x + 3 is a linear bijection of [0,1] onto the interval [3,6].
(d) To find a non-linear bijection between [0,1] and [3,6], we can create a function that respects the end points, but has a non-linear form. A simple example is the function f(x) = 3sin²(πx/2) + 3. You can verify it as one-to-one and onto.
(e) For a bijection of N onto Z, we might use a piecewise function that splits the Natural numbers into two sets (even and odd) with each one mapping to negative and non-negative integers respectively. A possible function can be f(n) = n/2 for n even and f(n)=-(n+1)/2 for n odd.
To know more about Bijection:
https://brainly.com/question/34638830
#SPJ11
