MHB Wizard1's question at Yahoo Answers (Isometry)

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The discussion confirms that the translation map T: H→H, defined as T(x,y) = (x, y+1), is an isometry of the set H = {(x,y) ∈ ℝ^2 : y > 0}. It establishes that T is well-defined since for any point (x,y) in H, the output (x, y+1) remains in H. The proof shows that the distance between points after applying T remains equal to the original distance, confirming T's isometric property. The mathematical verification involves calculating the Euclidean distance before and after the transformation, which yields the same result. Thus, T is indeed an isometry of H.
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Here is the question:

Let H = {(x,y) ∈ ℝ^2 : y > 0}, with the usual Euclidean distance. Let T: H→H be the translation map: T(x,y) = (x, y+1) for all (x,y) ∈ H. Verify that T is an isometry of H.
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Here is a link to the question:

Verify that T is an isometry of H...? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Hello wizard1,

If $(x,y)\in H$ then, $y>0$ which implies $y+1>0$. As a consequence, the map $T:H\to H$ is well defined. Now, for all $(x,y)$ and $(x',y')$ points of $H$: $$\begin{aligned}d[T(x,y),T(x',y')]&=d[(x,y+1),(x',y'+1)]\\&=\sqrt{(x'-x)^2+(y'+1-x'-1)^2}\\&=\sqrt{(x'-x)^2+(y'-x')^2}\\&=d[(x,y),(x',y')]\end{aligned}$$ That is, $T$ is an isometry.
 
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