Proof: Define [math]f_3\left(x,y\right)=\left(x+2y,y\right)[/math] where [math]f_3:\left(\mathbb{R},\mathbb{R}\right)\longrightarrow\left(\mathbb{R},\mathbb{R}\right)[/math]. Pick a point [math](q,r)[/math]. In order to be considered onto, there should be a point [math]\left(x,y\right)[/math] for which [math]f_3\left(x,y\right)=\left(q,r\right)[/math]. [math][/math] This means [math]x+2y=q[/math] and [math]y=r[/math]. By solving for [math]x[/math] and [math]y[/math], we know that [math]x=q-2y[/math] and [math]y=r[/math]. Notice, [math]f_3\left(q-2y,r\right)=\left(q-2y+2y,r\right)=\left(q,r\right)[/math]. Therefore, we have found the preimage of [math]f_3[/math]. Since there is a point in the preimage of [math]f_3[/math] that maps to an arbitrary point, we can conclude that the function [math]f_3[/math] is onto.[br][math]\diamondsuit[/math]

Proof: Define [math]f_4\left(x,y\right)=\left(x-2,y+1\right)[/math] where [math]f_4:\left(\mathbb{R},\mathbb{R}\right)\longrightarrow\left(\mathbb{R},\mathbb{R}\right)[/math]. Pick a point [math](q,r)[/math]. In order to be considered onto, there should be a point [math]\left(x,y\right)[/math] for which [math]f_4\left(x,y\right)=\left(q,r\right)[/math]. This means [math]x-2=q[/math] and [math]y+1=r[/math]. By solving for [math]x[/math] and [math]y[/math], we know that [math]x=q+2[/math] and [math]y=r-1[/math]. Notice, [math]f_4\left(q+2,r-1\right)=\left(q+2-2,r-1+1\right)=\left(q,r\right)[/math]. Therefore, we have found the preimage of [math]f_4[/math]. Since there is a point in the preimage of [math]f_4[/math] that maps to an arbitrary point, we can conclude that the function [math]f_4[/math] is onto. [math]\diamondsuit[/math][br]

Proof: Define [math]f_3\left(x,y\right)=\left(x+2y,y\right)[/math]. Pick two points [math]\left(x_1,y_1\right)[/math] and [math]\left(x_2,y_2\right)[/math]. If [math]f_3\left(x_1,y_1\right)=f_3\left(x_2,y_2\right)[/math] then it is implied that [math]\left(x_1+2y_1,y_1\right)=\left(x_2+2y_2,y_2\right)[/math]. If we consider the x and y coordinates separately, we notice that [math]y_1=y_2[/math] and [math]x_1+2y_1=x_2+2y_2[/math] which implies [math]x_1+2y_1=x_2+2y_1[/math] from the previous statement. From this we see, [math]x_1=x_2[/math]. Since [math]x_1=x_2[/math] and [math]y_1=y_2[/math], we know [math]f_3[/math] is one-to-one.[math]\diamondsuit[/math]

Proof: Define [math]f_4\left(x,y\right)=\left(x-2,y+1\right)[/math]. Pick two points [math]\left(x_1,y_1\right)[/math] and [math]\left(x_2,y_2\right)[/math]. Notice if[math]f_4\left(x_1,y_1\right)=f_4\left(x_2,y_2\right)[/math], then it is implied that [math]\left(x_1-2,y_1+1\right)=\left(x_2-2,y_2+1\right)[/math]. If we consider the x and y coordinates separately, we notice that [math]x_1-2=x_2-2[/math] which implies [math]x_1=x_2[/math] and [math]y_1+1=y_2+1[/math] which implies [math]y_1=y_2[/math]. 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.[math]\diamondsuit[/math]

Proof: Define [math]f_3\left(x,y\right)=\left(x+2y,y\right)[/math]. Consider points [math]\left(x_1,y_1\right)[/math] and [math]\left(x_2,y_2\right)[/math]. Determine the distance between the points using the distance formula. Note, 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]. 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]. If the function is in fact distance-preserving, these distances would be equivalent. Notice, [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]\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]\left(x_2-x_1\right)^2=\left(x_2-x_1+2y_2-2y_1\right)^2[/math][br][math]x_2-x_1=x_2-x_1+2y_2-2y_1[/math][br]Since the sides are not equal to one another, we know that the function is not distance-preserving. [math]\diamondsuit[/math]

Proof: Define [math]f_4\left(x,y\right)=\left(x-2,y+1\right)[/math]. Consider the points [math]\left(x_1,y_1\right)[/math] and [math]\left(x_2,y_2\right)[/math]. Determine the distance between the points using the distance formula. Note, 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]. 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]. If the function is distance-preserving, then the distances would be the same. Notice, [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]\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]Since the distances are equivalent, we know that the function is distance preserving. [math]\diamondsuit[/math]