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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?
May
16
asked Baumslag–Solitar $B(1,2)$ is not hyperbolic
May
9
awarded  Popular Question