
Abstract Algebra Abstract Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
July 5th, 2015, 03:53 AM  #1 
Newbie Joined: Jul 2015 From: Danube Posts: 2 Thanks: 0  Equivalence relation.
Examine if relation ρ defined as: xρy <=> (x^2y^2)(x^2*y^2 1)=0 is equivalence relation in set or real numbers. If it is, find equivalence classes C0, C1, C2, C3. Now, here is how I solved it: I have to prove that relation is 1. reflexive, 2. symmetric, 3. transitive. 1. xρx <=> (x^2x^2)(x^2*x^2 1)=0 since expression (x^2x^2) is always zero, relation is reflexive. 2. xρy <=> (x^2y^2)(x^2*y^2 1)=0 <=> (y^2x^2)(x^2*y^2 1)=0 since it's zero, sign doesn't matter, so (y^2x^2)(x^2*y^2 1)=0 <=>yρx 3. xρy ^ yρz => xρz xρy ^ yρz <=> (x^2y^2)(x^2*y^2 1)=0 ^ (y^2z^2)(y^2*z^2 1)=0 since the expression (x^2y^2)(x^2*y^2 1) equals zero, it means that either (x^2y^2) is zero or (x^2*y^2 1) is zero. If (x^2y^2) is zero it means that x^2=y^2, in that case in expression (y^2z^2)(y^2*z^2 1)=0 I can switch y for x since it's the same, that way I get: (x^2z^2)(x^2*z^2 1)=0 <=> xρz In case that expression (x^2*y^2 1) equals zero, it means that x^2*y^2=1, which means that y^2=1/x^2, placing it into (y^2z^2)(y^2*z^2 1)=0 I got: (1/x^2z^2)(1/x^2*z^2 1)=0 multiplying it by x^2 (1z^2*x^2)(z^2 x^2)=0 <=> (x^2*z^21)(x^2 z^2)=0 <=> xρz Now, I'm not sure whether this is correct approach, but I can't find any mistakes. For the second part, finding the classes, C0={0}, C1={1,1}, C2={2,2} or C2={1/2,1/2}, C3={3,3} or C3={1/3, 1/3} Now, my question is, is this correct? Last edited by skipjack; July 5th, 2015 at 06:00 AM. 
July 13th, 2015, 02:38 AM  #2 
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra 
Prove transitivity as follows: $\displaystyle x\rho y\ \implies\ x=\pm y\ \text{or}\ xy=\pm1$If $x=\pm y$ then replacing $y$ by $x$ in the second line gives $\pm x=\pm z$ or $(\pm x)z=\pm1$ $\implies$ $x=\pm z$ or $xz=\pm1$ $\implies$ $x\rho z$. Suppose $xy=\pm1$. If $y=\pm z$ then $\pm1=xy=x(\pm z)$ $\implies$ $xz=\pm1$; if $yz=\pm1$ then $\pm z=(xy)z=x(yz)=x(\pm1)$ $\implies$ $x=\pm z$ $\implies$ $x\rho z$. QED. The equivalence class containing $0$ is $\{0\}$; for $a\in\mathbb R,\,a\ne0$, its equivalence class is $\displaystyle \left\{a,\,a,\,\frac1a,\,\frac1a\right\}$. 
July 13th, 2015, 08:24 AM  #3  
Newbie Joined: Jul 2015 From: Danube Posts: 2 Thanks: 0  Quote:
 

Tags 
equivalence, relation 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Equivalence relation  tom33  Algebra  3  January 17th, 2014 04:30 PM 
Equivalence Relation  Taladhis  Abstract Algebra  2  February 11th, 2013 08:20 AM 
Equivalence Relation  jrklx250s  Real Analysis  3  December 7th, 2011 10:42 AM 
Equivalence relation  Dontlookback  Abstract Algebra  1  April 20th, 2010 11:52 AM 
equivalence relation  tinynerdi  Abstract Algebra  1  January 11th, 2010 09:24 AM 