Proof: 8.3,4,5
Exercise 8.3: For each of the five functions defined on the plane, prove is an onto function or is not onto.
Proof: Let : (,) -> (,) be defined by (x, y) = (x + 2y, y).
Onto: Let (m,n) (,) - [the codomain of ]. In order for to be onto, there must be some (x, y) (,) - [the domain of ] - such that f(x,y) = (m,n). Notice that m = x + 2y and n = y by definition of . By solving for x and y, we see that x = m - 2y and y = n.
Notice, f(m - 2n, n) = (m - 2n + 2n, n) = (m, n). Thus, when x = m - 2y and y = n, f(x, y) = (m, n). Therefore, is onto.
Proof: Let : (,) -> (,) be defined by (x, y) = (x - 2, y + 1).
Onto: Let (m,n) (,) - [the codomain of ]. In order for to be onto, there must be some (x, y) (,) - [the domain of ] - such that f(x,y) = (m,n). Notice that m = x - 2 and n = y + 1 by definition of . By solving for x and y, we see that x = m + 2y and y = n + 1.
Notice, f(m + 2, n + 1) = (m + 2 - 2, n) = (m, n). Thus, when x = m - 2 and y = n + 1, f(x, y) = (m, n). Therefore, is onto.
Exercise 8.4: For each of the five functions defined on the plane, prove is an one-to-one function or is not one-to-one.
Proof: Let : (,) -> (,) be defined by (x, y) = (x + 2y, y).
One-to-one: Let and be points such that . Further, we know that . Comparing individual coordinates, we see that and . However, since , we know that . Now, we see that . Therefore, since and , we know that is one-to-one.
Proof: Let : (,) -> (,) be defined by (x, y) = (x - 2, y + 1).
One-to-one: Let and be points such that . Further, we know that . Comparing individual coordinates, we see that and . Now, we see that . Therefore, since and , we know that is one-to-one.
Exercise 8.5: For each of the five functions defined on the plane, prove is or is not is distance preserving.
Proof: Let : (,) -> (,) be defined by (x, y) = (x + 2y, y). Consider points and . Using the distance formula between two points, we know the distance from to is .
Now, notice that the distance between and is .
If the function is distance-preserving, these distances will be equivalent. Notice the following:
Since , we know that is not distance-preserving.
Proof: Let : (,) -> (,) be defined by (x, y) = (x - 2, y + 1). Consider the points and . Using the distance formula between two points, we know the distance from to is .
Now, notice the distance between and is .
If the function is distance-preserving, these distances will be equivalent. Notice the following:
.
Since , we know that the is distance preserving.