# Equivalence Relations - I am stuck, help!

• Apr 11th 2009, 07:37 PM
spearfish
Equivalence Relations - I am stuck, help!
Hey all,

I am getting stuck proving the following relation:

x, y are elements of Q^+. x~y iff x = y(2^n) for some integer n.

Here's what I have so far:

Reflexive:
Let x be an element in Q^+.
x = x(2^n).
x/x = 2^n
1 = 2^n
1 = 2^0
1 = 1, Ok so does this prove xRx or do I need further work?

Symmetric:
Let x,y be elements in Q+.
Suppose xRy.
x = y(2^n)
Now suppose yRx.
y = x(2^m)
Plugging in for y and solving for x, I get
x = (x(2^n))(2^m)
x = x(2^n+m)
x/x = 2^(n+m), This is what I don't know what to do. What exactly does this result tell me? Do I need further work, or is this completely the wrong approach? Please somebody.

Transitive:
Let x,y,z be elements in Q+
Suppose xRy.
x = y(2^n)
Suppose yRz.
y = z(2^m).
Pluggin in and solving.
x = (z(2^m))(2^n)
x = z(2^m+n), Once again, now what?

Please explain my faults and show how to correct. Explanations would be very helpful. Thanks
• Apr 11th 2009, 07:52 PM
K A D C Dilshan
Quote:

Originally Posted by spearfish
Hey all,

I am getting stuck proving the following relation:

x, y are elements of Q^+. x~y iff x = y(2^n) for some integer n.

Here's what I have so far:

Reflexive:
Let x be an element in Q^+.
x = x(2^n).
x/x = 2^n
1 = 2^n
1 = 2^0
1 = 1, Ok so does this prove xRx or do I need further work?

Symmetric:
Let x,y be elements in Q+.
Suppose xRy.
x = y(2^n)
Now suppose yRx.
y = x(2^m)
Plugging in for y and solving for x, I get
x = (x(2^n))(2^m)
x = x(2^n+m)
x/x = 2^(n+m),
This is what I don't know what to do. What exactly does this result tell me? Do I need further work, or is this completely the wrong approach? Please somebody.

Transitive:
Let x,y,z be elements in Q+
Suppose xRy.
x = y(2^n)
Suppose yRz.
y = z(2^m).
Pluggin in and solving.
x = (z(2^m))(2^n)
x = z(2^m+n), Once again, now what?

Please explain my faults and show how to correct. Explanations would be very helpful. Thanks

Actually what you have done to prove reflexive and transient property of R is correct ,
But you have made a mistake on the proof of symmetric property of R .
You have taken that if (x,y) belongs to R x = y (2^n) , y = x (2^m)
Mistake is there
if (x,y) belongs to R x = y (2^n) ,
y = x (2^-n) we know that -n is also an Integer
therefore R is symmetric .
• Apr 11th 2009, 07:58 PM
spearfish
Thanks a bunch for the help.