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Given the open unit ball in $\mathbb{R}^2$ (or for any other $\mathbb{R}^n$ as well), if I use the function $f:B^2 \rightarrow \mathbb{R}^2$ that does the following: $f(x,y)= (\frac{x}{2}, \frac{y}{2} )$, will the result on the entire ball, i.e. $f(B^2)$, be the ball with a radius of $\frac{1}{2}$ (with same center)? Or perhaps with a radius of $\frac{1}{4}$?

Edit: And another question is how can I move the ball to another ball with a different center point? should I add to just one ( $x$ or $y$ ) coordinate or to both? What I mean is, if I have a ball with radius $r$ and center point $x_0$ and I want to transfer this ball with to a ball with same radius $r$, just with a center $x_1$, what type of function should I need?

(the same questions applies in general to $\mathbb{R}^n$, just thought $\mathbb{R}^2$ would be easier to see)

Zhan I.s.
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  • Think at the radius as the segment from $(0,0)$ to $(1,0)$: what happen to that segment? – Dario Nov 16 '14 at 08:51
  • In $\mathbb{R}$ I can see this obviously (will be $(0,\frac{1}{2})$ ), but I wasnt sure if it holds in other $\mathbb{R}^n$ as well? – Zhan I.s. Nov 16 '14 at 08:53
  • A cleaner way: your map send $(x,y)$ to $\left(x'=\frac{x}{2},y'=\frac{y}{2}\right)$. Hence $$x^2+y^2=1\quad\rightarrow\quad (2x')^2+(2y')^2=1\ .$$ Can you obtain the radius of the new circle from this equation? – Dario Nov 16 '14 at 08:59
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    Yes, thanks! As for my other question - if I define $f(x,y)=(x+k,y)$, so $f(B^2)$ will be a ball with radius $1$ with center $(k,0)$? and so if I take a function $g(x,y)=(x+k_1,y+k_2)$ then the result $f(B^2)$ will be a ball with radius $1$ and center point $(k_1,k_2)$? – Zhan I.s. Nov 16 '14 at 09:07
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    Yes, you can verify it in the same way. And in the same way you can verify it in $\mathbb{R}^n$. – Dario Nov 16 '14 at 09:12

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