Recent content by scharl4

Ok, so then adding d(z,w) to both sides gives d(x,y) ≤ d(x,z) + d(y,w) + d(z,w), so d(x,z) + d(z,w) + d(y,w) ≥ d(x,w) + d(w,y) by the triangle inequality, and d(x,w) + d(w,y) ≥ d(x,y) by the triangle inequality, so d(x,y) ≤ d(x,z) + d(y,w) + d(z,w). Thanks so much, now I understand where this is...

OK, I used the two cases a) d(x,y) - d(z,w) ≥ 0 and b) d(x,y) - d(z,w) < 0 because of the absolute value in the problem. So for case a) i need to prove d(x,y) - d(z,w) ≤ d(x,z) + d(y,w) and for case b) i need to prove d(z,w) - d(x,y) ≤ d(x,z) + d(y,w)

Homework Statement Using the generalized triangle inequality, prove |d(x,y) - d(z,w)| ≤ d(x,z) + d(y,w) Homework Equations d(x,y) is a metric triangle inequality: d(x,y) ≤ d(x,z) + d(z,y) The Attempt at a Solution I know that this needs to be proved with cases: a) d(x,y) - d(z,w)...

OK thanks. I got a little confused because the other post said I should generate by using (a^n, b^n). I think I have my subgroups correct now.

Ok so I think I found all the subgroups generated by each element of Z6 x Z3. It looks like (0,0), (0,1), (1,0), (1,1), (3,0), (3,1), (4,0), (4,1), (5,0), (5,1) each generate only themselves, while each of the other elements of Z6 x Z3 only generate 2 elements. So, it appears that there are no...

Homework Statement Find all cyclic subgroups of Z6 x Z3. Homework Equations The Attempt at a Solution I understand how to find a cyclic subgroup of a simpler group such as Z4, but having trouble understanding what subgroups look like in a direct product of integer spaces, let alone...