[b][u]Exercise 8.3:[/u][/b] For each of the five functions [math]f_1,f_2,f_3,f_4,f_5[/math] defined on the plane, prove [math]f_n[/math] is an onto function or [math]f_n[/math] is not onto.[br][br][b][u]Proof:[/u][/b] Let [math]f_3[/math] : ([math]\mathbb{R}[/math],[math]\mathbb{R}[/math]) -> ([math]\mathbb{R}[/math],[math]\mathbb{R}[/math]) be defined by [math]f_3[/math](x, y) = (x + 2y, y).[br][br][u]Onto:[/u] Let (m,n) [math]\in[/math] ([math]\mathbb{R}[/math],[math]\mathbb{R}[/math]) - [the codomain of [math]f_3[/math]]. In order for [math]f_3[/math] to be onto, there must be some (x, y) [math]\in[/math] ([math]\mathbb{R}[/math],[math]\mathbb{R}[/math]) - [the domain of [math]f_3[/math]] - such that f(x,y) = (m,n). Notice that m = x + 2y and n = y by definition of [math]f_3[/math]. By solving for x and y, we see that x = m - 2y and y = n. [br][br]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, [math]f_3[/math] is onto.[br][br][br][b][u]Proof:[/u][/b] Let [math]f_4[/math] : ([math]\mathbb{R}[/math],[math]\mathbb{R}[/math]) -> ([math]\mathbb{R}[/math],[math]\mathbb{R}[/math]) be defined by [math]f_4[/math](x, y) = (x - 2, y + 1).[br][br][u]Onto:[/u] Let (m,n) [math]\in[/math] ([math]\mathbb{R}[/math],[math]\mathbb{R}[/math]) - [the codomain of [math]f_4[/math]]. In order for [math]f_4[/math] to be onto, there must be some (x, y) [math]\in[/math] ([math]\mathbb{R}[/math],[math]\mathbb{R}[/math]) - [the domain of [math]f_4[/math]] - such that f(x,y) = (m,n). Notice that m = x - 2 and n = y + 1 by definition of [math]f_4[/math]. By solving for x and y, we see that x = m + 2y and y = n + 1. [br][br]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, [math]f_4[/math] is onto.[br][br][br][b][u]Exercise 8.4:[/u][/b] For each of the five functions [math]f_1,f_2,f_3,f_4,f_5[/math] defined on the plane, prove [math]f_n[/math] is an one-to-one function or [math]f_n[/math] is not one-to-one.[br][br][b][u]Proof:[/u][/b] Let [math]f_3[/math] : ([math]\mathbb{R}[/math],[math]\mathbb{R}[/math]) -> ([math]\mathbb{R}[/math],[math]\mathbb{R}[/math]) be defined by [math]f_3[/math](x, y) = (x + 2y, y).[br][br]One-to-one: Let [math](x_1,y_1)[/math] and [math](x_2,y_2)[/math] be points such that [math]f(x_1,y_1)=f(x_2,y_2)[/math]. Further, we know that [math](x_1+2y_1,y_1)=(x_2+2y_2,y_2)[/math]. Comparing individual coordinates, we see that [math]y_1=y_2[/math] and [math]x_1+2y_1=x_2+2y_2[/math]. However, since [math]y_1=y_2[/math], we know that [math]x_1+2y_1=x_2+2y_1[/math]. Now, we see that [math]x_1=x_2[/math]. Therefore, since [math]x_1=x_2[/math] and [math]y_1=y_2[/math], we know that [math]f_3[/math] is one-to-one.[br][br][br][b][u]Proof:[/u][/b] Let [math]f_4[/math] : ([math]\mathbb{R}[/math],[math]\mathbb{R}[/math]) -> ([math]\mathbb{R}[/math],[math]\mathbb{R}[/math]) be defined by [math]f_4[/math](x, y) = (x - 2, y + 1).[br][br]One-to-one: Let [math](x_1,y_1)[/math] and [math](x_2,y_2)[/math] be points such that [math]f(x_1,y_1)=f(x_2,y_2)[/math]. Further, we know that [math](x_1-2,y_1+1)=(x_2-2,y_2+1)[/math]. Comparing individual coordinates, we see that [math]y_1=y_2[/math] and [math]x_1-2=x_2-2[/math]. Now, we see that [math]x_1=x_2[/math]. Therefore, since [math]x_1=x_2[/math] and [math]y_1=y_2[/math], we know that [math]f_4[/math] is one-to-one.[br][br][br][b][u]Exercise 8.5:[/u][/b] For each of the five functions [math]f_1,f_2,f_3,f_4,f_5[/math] defined on the plane, prove [math]f_n[/math] is or is not [math]f_n[/math] is distance preserving.[br][br][b][u]Proof:[/u][/b] Let [math]f_3[/math] : ([math]\mathbb{R}[/math],[math]\mathbb{R}[/math]) -> ([math]\mathbb{R}[/math],[math]\mathbb{R}[/math]) be defined by [math]f_3[/math](x, y) = (x + 2y, y). Consider points [math]\left(x_1,y_1\right)[/math] and [math]\left(x_2,y_2\right)[/math]. Using the distance formula between two points, we know the distance from [math]\left(x_1,y_1\right)[/math] to [math]\left(x_2,y_2\right)[/math] is [math]\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}[/math]. [br][br]Now, notice that the distance between [math]f_3\left(x_1,y_1\right)=\left(x_1+2y_1,y_1\right)[/math] and [math]f_3\left(x_2,y_2\right)=\left(x_2+2y_2,y_2\right)[/math] is [math]\sqrt{\left(x_2+2y_2-x_1-2y_1\right)^2+\left(y_2-y_1\right)^2}[/math]. [br][br]If the function is distance-preserving, these distances will be equivalent. Notice the following:[br][math]\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt{\left(x_2-x_1+2y_2-2y_1\right)^2+\left(y_2-y_1\right)^2}[/math][br][math]\Longrightarrow\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2=\left(x_2-x_1+2y_2-2y_1\right)^2+\left(y_2-y_1\right)^2[/math][br][math]\Longrightarrow\left(x_2-x_1\right)^2=\left(x_2-x_1+2y_2-2y_1\right)^2[/math][br][math]\Longrightarrow x_2-x_1=x_2-x_1+2y_2-2y_1[/math][br][br]Since [math]x_2-x_1\ne x_2-x_1+2y_2-2y_1[/math], we know that [math]f_3[/math] is not distance-preserving. [br][br][u][b]Proof:[/b][/u] Let [math]f_4[/math] : ([math]\mathbb{R}[/math],[math]\mathbb{R}[/math]) -> ([math]\mathbb{R}[/math],[math]\mathbb{R}[/math]) be defined by [math]f_4[/math](x, y) = (x - 2, y + 1). Consider the points [math]\left(x_1,y_1\right)[/math] and [math]\left(x_2,y_2\right)[/math]. Using the distance formula between two points, we know the distance from [math]\left(x_1,y_1\right)[/math] to  [math]\left(x_2,y_2\right)[/math] is [math]\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}[/math]. [br][br]Now, notice the distance between [math]f_4\left(x_1,y_1\right)=\left(x_1-2,y_1+1\right)[/math] and [math]f_4\left(x_2,y_2\right)=\left(x_2-2,y_2+1\right)[/math] is [math]\sqrt{\left(x_2-2-x_1+2\right)^2+\left(y_2+1-y_1-1\right)^2}[/math].[br][br]If the function is distance-preserving, these distances will be equivalent. Notice the following:[math]\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt{\left(x_2-2-x_1+2\right)^2+\left(y_2+1-y_1-1\right)^2}[/math][br][math]\Longrightarrow\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}[/math].[br][br]Since [math]\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}[/math], we know that the [math]f_{\text{4}}[/math] is distance preserving.