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 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: 156 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
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 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 Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded 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

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