# Math Help - Equivalence Relations

1. ## Equivalence Relations

$x,y,z\in\mathbb{R}$
$x\sim y$ 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 $0\notin\mathbb{Q}$

2. 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.

3. 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 $a\neq 0$

4. 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:
$\frac{ef-cd}{df}\rightarrow\frac{e}{f}-\frac{c}{d}$?

5. Originally Posted by dwsmith
I understand that but I thought $\mathbb{Q}$ were all $\frac{a}{b}$ such that $a\neq 0$
It is $\color{red}b\neq 0$.

6. 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 $a\sim c$
$c=\frac{g}{h}$
$\frac{c}{d}-\frac{e}{f}$
$\frac{e}{f}-\frac{g}{h}$

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

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?

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

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

9. 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 $-r\in \mathbb{Q}$

10. 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, $a = \pi - 1, b = \pi$...

11. 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.

12. 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?