- #1
Bleys
- 74
- 0
The problem is in an Algebra book but the jist of it is about calculus/analysis. If G is the interval (-c,c) and the operation is addition of velocities in Special Relativity, (that is:
[tex]x \circ y = \frac{x+y}{1+\frac{xy}{c^2}}[/tex]) then decide whether G is a group. At first I thought it wasn't; figured either associativity or closure would fail. But when I tried some strange values it actually worked so I went about proving it. Associativity was fine. The identity is zero, inverse of x is -x. Closure is what I'm having trouble with. Everytime I try to use the fact |x|<c and |y|<c I lose information and I get upper bounds that are way too high to be of any use. I noticed things looked 'somewhat' (which is an overstatement) similar to what i was trying to prove if I used [tex] (x+y)^2 = x^2 + 2xy + y^2 [/tex], played around with it, but again once I started using inequalities i get too far. Is there some piece of information other than -c<x,y<c that I'm missing which would be key to solving this?
[tex]x \circ y = \frac{x+y}{1+\frac{xy}{c^2}}[/tex]) then decide whether G is a group. At first I thought it wasn't; figured either associativity or closure would fail. But when I tried some strange values it actually worked so I went about proving it. Associativity was fine. The identity is zero, inverse of x is -x. Closure is what I'm having trouble with. Everytime I try to use the fact |x|<c and |y|<c I lose information and I get upper bounds that are way too high to be of any use. I noticed things looked 'somewhat' (which is an overstatement) similar to what i was trying to prove if I used [tex] (x+y)^2 = x^2 + 2xy + y^2 [/tex], played around with it, but again once I started using inequalities i get too far. Is there some piece of information other than -c<x,y<c that I'm missing which would be key to solving this?