[proofplan] We prove that the affine rescaling $F(y)=x_0+ry$ sends the unit ball into the ball of radius $r$ centered at $x_0$, and that the proposed inverse $G(x)=(x-x_0)/r$ sends that ball back into the unit ball. The positivity of $r$ is used both to divide by $r$ and to preserve strict inequalities after scaling. Finally, we compute the two compositions $G \circ F$ and $F \circ G$ and conclude that the maps are mutual inverses. [/proofplan] [step:Show that the rescaling map sends the unit ball into $B(x_0,r)$] Let $y \in B(0,1)$. By the definition of $B(0,1)$, we have $|y| < 1$. Since $r > 0$, the homogeneity of the Euclidean norm gives \begin{align*} |F(y)-x_0| = |x_0 + ry - x_0| = |ry| = r|y| < r. \end{align*} Therefore $F(y) \in B(x_0,r)$. Hence $F: B(0,1) \to B(x_0,r)$ is well-defined. [/step] [step:Show that the inverse candidate sends $B(x_0,r)$ into the unit ball] Let $x \in B(x_0,r)$. By the definition of $B(x_0,r)$, we have $|x-x_0| < r$. Since $r > 0$, division by $r$ is valid, and homogeneity of the Euclidean norm gives \begin{align*} |G(x)| = \left|\frac{x-x_0}{r}\right| = \frac{|x-x_0|}{r} < 1. \end{align*} Thus $G(x) \in B(0,1)$, so $G: B(x_0,r) \to B(0,1)$ is well-defined. [guided] We must first check that the formula for $G$ actually lands in the claimed codomain. Take an arbitrary point $x \in B(x_0,r)$. By the definition of the open Euclidean ball centered at $x_0$ with radius $r$, this means \begin{align*} |x-x_0| < r. \end{align*} The hypothesis $r>0$ is essential here: it permits division by $r$, and dividing a strict inequality by $r$ preserves the direction of the inequality. Using the homogeneity property of the Euclidean norm, we compute \begin{align*} |G(x)| = \left|\frac{x-x_0}{r}\right| = \frac{|x-x_0|}{r}. \end{align*} Since $|x-x_0|<r$, we obtain \begin{align*} \frac{|x-x_0|}{r} < 1. \end{align*} Therefore $|G(x)|<1$, which is exactly the condition $G(x)\in B(0,1)$. Hence the proposed inverse map $G: B(x_0,r)\to B(0,1)$ is well-defined. [/guided] [/step] [step:Compute both compositions and identify the inverse] Let $y \in B(0,1)$. Using the definitions of $F$ and $G$, we have \begin{align*} G(F(y)) = \frac{F(y)-x_0}{r} = \frac{x_0+ry-x_0}{r} = y. \end{align*} Thus $G \circ F = \operatorname{id}_{B(0,1)}$. Let $x \in B(x_0,r)$. Again using the definitions of $F$ and $G$, we have \begin{align*} F(G(x)) = x_0 + rG(x) = x_0 + r\frac{x-x_0}{r} = x. \end{align*} Thus $F \circ G = \operatorname{id}_{B(x_0,r)}$. [/step] [step:Conclude that $F$ is a bijection with inverse $G$] The identities $G \circ F = \operatorname{id}_{B(0,1)}$ and $F \circ G = \operatorname{id}_{B(x_0,r)}$ show that $F$ and $G$ are mutual inverses. Therefore $F$ is bijective from $B(0,1)$ onto $B(x_0,r)$, and $F^{-1}=G$. [/step]