Equivalence relations problem #2 (alg)

Pearce_09
Messages
71
Reaction score
0
R = the real numbers

A = R x R; (x,y) \equiv (x_1,y_1) means that
x^2 + y^2 = x_1^2 + y_1^2; B= {x is in R | x>= 0 }

Find a well defined bijection sigma : A_\equiv -> B

like the last problem, I just can't seem to find the right way to solve this??
 
Physics news on Phys.org
hint 1 x^2+y^2 is a positive real number.

hint 2 What is an equivalence class under this relation, think geometrically.

hint 3 radius.
 
Pearce_09 said:
R = the real numbers

A = R x R; (x,y) \equiv (x_1,y_1) means that
x^2 + y^2 = x_1^2 + y_1^2; B= {x is in R | x>= 0 }

Find a well defined bijection sigma : A_\equiv -> B

like the last problem, I just can't seem to find the right way to solve this??

did you get it?
 
ya i did thanks fourier
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top