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July 5th, 2015, 03:53 AM   #1
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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.
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July 13th, 2015, 02:38 AM   #2
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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 $\displaystyle \left\{a,\,-a,\,\frac1a,\,-\frac1a\right\}$.
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July 13th, 2015, 08:24 AM   #3
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Quote:
Originally Posted by Olinguito View Post
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.
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