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 May 23 revised Showing that $\mathbb{R}\times [0,1]$ is quasi-isometric to $\mathbb{R}$ added 398 characters in body May 23 asked Showing that $\mathbb{R}\times [0,1]$ is quasi-isometric to $\mathbb{R}$ May 22 comment Do we sometimes have to go “each way” separately for iff proofs? I find this question interesting, although it may have a simple answer, I could imagine there being some crazy counterexample- I'm thinking of some sort of independence proof? May 22 comment Showing that triangles in $\mathbb{Z}$ are thin @DerekHolt Ah yes of course, thanks for the help! :) If you wish to post as an answer I would gladly accept May 22 comment Showing that triangles in $\mathbb{Z}$ are thin @DerekHolt thanks. Could you explain your first line a bit please? May 22 asked Showing that triangles in $\mathbb{Z}$ are thin May 20 awarded Electorate May 19 comment Showing geodesics in $\mathbb{H}^2$ have unique common perpendicular or common endpoint OK, but how do I go about doing this? May 19 comment Showing geodesics in $\mathbb{H}^2$ have unique common perpendicular or common endpoint Is there no way I can do this with intermediate value theorem like you suggested? Something like taking a $\gamma_3$ that intersects $\gamma_2$ perpendicularly and then vary the endpoints of $\gamma_3$ somehow to get the intersection with $\gamma_1$ to be perpendicular? May 19 comment Showing geodesics in $\mathbb{H}^2$ have unique common perpendicular or common endpoint I'm not sure sorry, my basic algebra is pretty bad! How do I calculate that? May 19 comment Showing geodesics in $\mathbb{H}^2$ have unique common perpendicular or common endpoint Sorry I is quite hard to explain what I mean without pictures! May 19 comment Showing geodesics in $\mathbb{H}^2$ have unique common perpendicular or common endpoint My geodesics are circles that intersect the boundary perpendicularly so I was being stupid taking straight line not through the center as it does not intersect perpendicularly sorry. So do I take $\gamma_3$ which intersects $\gamma_2$ perpendicularly. Then vary the other endpoint and by IVT there will be a point which intersects $\gamma_1$ perpendicularly. May 19 comment Showing geodesics in $\mathbb{H}^2$ have unique common perpendicular or common endpoint Ok so I say that $\gamma_1$ is the horizontal line through the origin and take $\gamma_2$ such that they don't share endpoints on $S_\infty$. Now I need to find geodesic intersecting them both perpendicularly. So any vertical line will intersect $\gamma_1$ perpendicularly can I then show that it intersects $\gamma_2$ perpendicularly at some point? May 19 revised Showing geodesics in $\mathbb{H}^2$ have unique common perpendicular or common endpoint added 73 characters in body May 19 comment Showing geodesics in $\mathbb{H}^2$ have unique common perpendicular or common endpoint No I am using the interior of the unit disc in $\mathbb{R}^2$ sorry I should have mentioned that I will edit my question May 19 asked Showing geodesics in $\mathbb{H}^2$ have unique common perpendicular or common endpoint May 18 accepted $\mathbb{C}\otimes_\mathbb{C} \mathbb{C} \cong \mathbb{R}\otimes _\mathbb{R} \mathbb{C}$ May 18 accepted Question on Frobenius Reciprocity May 17 accepted Baumslag–Solitar $B(1,2)$ is not hyperbolic May 16 comment Baumslag–Solitar $B(1,2)$ is not hyperbolic @N.Owad No, could you give me reference/ explain to me?