Equivalence Relations

**dwsmith** · April 29th 2010, 02:44 PM

$x,y,z\in\mathbb{R}$
$x\sim y$ iff. $x-y\in\mathbb{Q}$

Prove this is an equivalence relation.

Reflexive:
$a\sim a$
$a-a=0$ ; however, does $0\in\mathbb{Q}$ ? I was under the impression $0\notin\mathbb{Q}$

**Plato** · April 29th 2010, 02:48 PM

Originally Posted by dwsmith

$x,y,z\in\mathbb{R}$
$x\sim y$ iff. $x-y\in\mathbb{Q}$

Prove this is an equivalence relation.

Reflexive:
$a\sim a$
$a-a=0$ ; however, does $0\in\mathbb{Q}$ ? I was under the impression $0\notin\mathbb{Q}$

$0=\frac{0}{1}$ the ratio of two integers.

**dwsmith** · April 29th 2010, 02:50 PM

Originally Posted by Plato

$0=\frac{0}{1}$ the ratio of two integers.

I understand that but I thought $\mathbb{Q}$ were all $\frac{a}{b}$ such that $a\neq 0$

**dwsmith** · April 29th 2010, 02:59 PM

Symmetric:
$a\sim b$ , then $b\sim a$
Since $a,b\sim\mathbb{Q}$ , then a and b can expressed as $a=\frac{c}{d}$ and $b=\frac{e}{f}$

$\frac{c}{d}-\frac{e}{f}\rightarrow \frac{cf-de}{df}$

How can I get than in the form of $\frac{e}{f}-\frac{c}{d}$ ?

Would it be allowable to multiple through by a -1 and then swap cd and ef to obtain:
$\frac{ef-cd}{df}\rightarrow\frac{e}{f}-\frac{c}{d}$ ?

**Plato** · April 29th 2010, 03:00 PM

Originally Posted by dwsmith

I understand that but I thought $\mathbb{Q}$ were all $\frac{a}{b}$ such that $a\neq 0$

It is $\color{red}b\neq 0$ .

**dwsmith** · April 29th 2010, 03:31 PM

Originally Posted by dwsmith

Symmetric:
$a\sim b$ , then $b\sim a$
Since $a,b\sim\mathbb{Q}$ , then a and b can expressed as $a=\frac{c}{d}$ and $b=\frac{e}{f}$

$\frac{c}{d}-\frac{e}{f}\rightarrow \frac{cf-de}{df}$

How can I get than in the form of $\frac{e}{f}-\frac{c}{d}$ ?

Would it be allowable to multiple through by a -1 and then swap cd and ef to obtain:
$\frac{ef-cd}{df}\rightarrow\frac{e}{f}-\frac{c}{d}$ ?

Transitive:
$a\sim b, b\sim c$ , then $a\sim c$
$c=\frac{g}{h}$
$\frac{c}{d}-\frac{e}{f}$
$\frac{e}{f}-\frac{g}{h}$

add together
$\frac{c}{d}-\frac{g}{h}\rightarrow\frac{ch-gd}{dh}\in\mathbb{Q}$ $a\sim c$

Equivalence class of $\sqrt{2}$ and a
$[\sqrt{2}]=(x\in\mathbb{R}|x\sim\sqrt{2})$
$x=\frac{a}{b}$ and $a,b\in\mathbb{Z}$
$[\sqrt{2}]=(x\in\mathbb{R}|\frac{a}{b}+\sqrt{2}\sim\sqrt{2}$
Correct?

**Plato** · April 29th 2010, 03:45 PM

The sum of two rationals is rational.
So it is transitive.
Way to go!

**dwsmith** · April 29th 2010, 03:45 PM

What about what I did for symmetric? Is that legal?

**Plato** · April 29th 2010, 03:49 PM

Originally Posted by dwsmith

What about what I did for symmetric? Is that legal?

Of good grief, come on THINK.
If $r\in \mathbb{Q}$ then $-r\in \mathbb{Q}$

**Defunkt** · April 29th 2010, 04:21 PM

Originally Posted by dwsmith

Symmetric:
$a\sim b$ , then $b\sim a$
Since $a,b\sim\mathbb{Q}$ , then a and b can expressed as $a=\frac{c}{d}$ and $b=\frac{e}{f}$

$\frac{c}{d}-\frac{e}{f}\rightarrow \frac{cf-de}{df}$

How can I get than in the form of $\frac{e}{f}-\frac{c}{d}$ ?

Would it be allowable to multiple through by a -1 and then swap cd and ef to obtain:
$\frac{ef-cd}{df}\rightarrow\frac{e}{f}-\frac{c}{d}$ ?

If $a,b \in \mathbb{Z}$ and $a \sim b$ it means, by definition, that $\exists \ m,n \in \mathbb{Z}, \ n \neq 0,$ such that $a - b = \frac{m}{n} \in \mathbb{Q}$ . What can you say about $b-a$ then?

Your proofs for symmetry and transitivity are both incorrect -- $a \sim b$ means that $a-b \in \mathbb{Q}$ , but that doesn't necessarily mean that $a, b \in \mathbb{Q}$ -- take, for example, $a = \pi - 1, b = \pi$ ...

**dwsmith** · April 29th 2010, 04:26 PM

Originally Posted by Defunkt

If $a,b \in \mathbb{Z}$ and $a \sim b$ it means, by definition, that $\exists \ m,n \in \mathbb{Z}, \ n \neq 0,$ such that $a - b = \frac{m}{n} \in \mathbb{Q}$ . What can you say about $b-a$ then?

Your proofs for symmetry and transitivity are both incorrect -- $a \sim b$ means that $a-b \in \mathbb{Q}$ , but that doesn't necessarily mean that $a, b \in \mathbb{Q}$ -- take, for example, $a = \pi - 1, b = \pi$ ...

I understand what you mean; however, I am not sure on how to show it if what I have done is incorrect.

**Defunkt** · April 29th 2010, 04:30 PM

For symmetry you only have to notice what Plato said - if $r \in \mathbb{Q}$ then $-r \in \mathbb{Q}$ .

For transitivity, again, as Plato said -- the sum of two rationals is rational. How can you express $a-c$ in terms of two numbers you know are rational?

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