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For a pair of points $p_1 = (x_1, y_1)$ and $p_2 = (x_2, y_2)$ define $$d(p_1, p_2) = \sqrt{ (x_1 - x_2)^{2} + (y_1 - y_2)^{2} }$$ the distance between p1 and p2. Then the function f satisfies $$d(f(p_1), f(p_2)) = d(p_1, p_2)$$ $\forall p_1, p_2 \in \mathbb{R}^{2}$

Prove that $f$ is a bijection, thus it has an inverse.


So I know a bijection is an injecction that is also surjective. I.e., a mapping which is both one to one and onto.

So to prove it's injective I believe all we need to do is basically set $d(p_1, p_2) = 0$ then show that, that implies $p_1 = p_2$?

To prove its surjective I'm having a little more trouble conceptualizing. If anyone could help that would be exceptional.

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  • $\begingroup$ You are misstating the problem. Please check. Surely, the intended problem is to show that $f$ is bijective. $\endgroup$ – quasi Apr 27 '17 at 0:09
  • $\begingroup$ You mean to show that $f$ (not $d$) is a bijection. $\endgroup$ – Ted Shifrin Apr 27 '17 at 0:10
  • $\begingroup$ Oh my apologies, I shall make the correction. Thanks guys! $\endgroup$ – Kyle O'Connell Apr 27 '17 at 0:11
  • $\begingroup$ Rather than just edit it, why not take a cut at the corrected problem? $\endgroup$ – quasi Apr 27 '17 at 0:12
  • $\begingroup$ I wasn't asking for help because I thought it was d I had to prove. It was simply a typo, my question is more about proving the surjection with this type of function. $\endgroup$ – Kyle O'Connell Apr 27 '17 at 0:18

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