My Math Forum Equivalence relation.

 Abstract Algebra Abstract Algebra Math Forum

 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^2-y^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^2-x^2)(x^2*x^2 -1)=0 since expression (x^2-x^2) is always zero, relation is reflexive. 2. xρy <=> (x^2-y^2)(x^2*y^2 -1)=0 <=> -(y^2-x^2)(x^2*y^2 -1)=0 since it's zero, sign doesn't matter, so (y^2-x^2)(x^2*y^2 -1)=0 <=>yρx 3. xρy ^ yρz => xρz xρy ^ yρz <=> (x^2-y^2)(x^2*y^2 -1)=0 ^ (y^2-z^2)(y^2*z^2 -1)=0 since the expression (x^2-y^2)(x^2*y^2 -1) equals zero, it means that either (x^2-y^2) is zero or (x^2*y^2 -1) is zero. If (x^2-y^2) is zero it means that x^2=y^2, in that case in expression (y^2-z^2)(y^2*z^2 -1)=0 I can switch y for x since it's the same, that way I get: (x^2-z^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^2-z^2)(y^2*z^2 -1)=0 I got: (1/x^2-z^2)(1/x^2*z^2 -1)=0 multiplying it by x^2 (1-z^2*x^2)(z^2 -x^2)=0 <=> (x^2*z^2-1)(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: 155 Math Focus: Abstract algebra Prove transitivity as follows: $\displaystyle x\rho y\ \implies\ x=\pm y\ \text{or}\ xy=\pm1$ $\displaystyle y\rho z\ \implies\ y=\pm z\ \text{or}\ yz=\pm1$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:
 Originally Posted by Olinguito Prove transitivity as follows: $\displaystyle x\rho y\ \implies\ x=\pm y\ \text{or}\ xy=\pm1$ $\displaystyle y\rho z\ \implies\ y=\pm z\ \text{or}\ yz=\pm1$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 $\displaystyle \left\{a,\,-a,\,\frac1a,\,-\frac1a\right\}$.
So, basically, what i did was correct, except for the part with equivalence class.

 Tags equivalence, relation

 Thread Tools Display Modes Linear Mode

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

 Contact - Home - Forums - Cryptocurrency Forum - Top