Proving ~ is an Equivalence Relation of A

[nikie1o2]

Homework Statement

Consider the set A={(u,v,w) in R^3 : u^2+v^2>0} and define a relation ~ on A by (u,v,w)~(u',v',w') IFF there exists a "k" in R, where k doesn't equal 0: (u',v',w')=(ku,kv,kw)

Prove that ~ is an equivalence relation of A

Homework Equations

I honestly don't know where to start, i know i need to satisfy the reflexive, symmetric & transitive requirements but i don't even know what the relation is here. Any help is very much appreciated.

The Attempt at a Solution

 

[vela]

You're given the relation. Two vectors (u,v,w) and (u',v',w') are related if they are non-zero multiples of each other. For example, (1,1,1) ~ (2,2,2) since (2,2,2)=(2x1, 2x1, 2x1), i.e. k=2.

Homework Statement

Consider the set A={(u,v,w) in R^3 : u^2+v^2>0} and define a relation ~ on A by (u,v,w)~(u',v',w') IFF there exists a "k" in R, where k doesn't equal 0: (u',v',w')=(ku,kv,kw)

Prove that ~ is an equivalence relation of A

Homework Equations

I honestly don't know where to start, i know i need to satisfy the reflexive, symmetric & transitive requirements but i don't even know what the relation is here. Any help is very much appreciated.

The Attempt at a Solution

 

[nikie1o2]

You're given the relation. Two vectors (u,v,w) and (u',v',w') are related if they are non-zero multiples of each other. For example, (1,1,1) ~ (2,2,2) since (2,2,2)=(2x1, 2x1, 2x1), i.e. k=2.

Ok, So to show its reflexive i can pick an real numbers? I can say (u,v,w)=(1,1,1) so (1,1,1) R (1,1,1) ?

 

[Mathnerdmo]

No, for reflexive, you'd need to look at a general triple (u,v,w). What does it mean to have
(u,v,w) ~ (u,v,w) ?

Write it out and find a value for k.

You'll need to work with general triples for all of the axioms for an equivalence relation.
(if you really get stuck, you might work with specific points to get a handle on what's happening, but you can't choose specific points for the proof)

 

[vela]

Ok, So to show its reflexive i can pick an real numbers? I can say (u,v,w)=(1,1,1) so (1,1,1) R (1,1,1) ?

No, I was just using specific numbers as an example. As Mathnerdmo said, for a proof, you need to show it's true for the general case.