1. ## Equivalence Relations

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

Prove this is an equivalence relation.

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

2. Originally Posted by dwsmith
$\displaystyle x,y,z\in\mathbb{R}$
$\displaystyle x\sim y$ iff. $\displaystyle x-y\in\mathbb{Q}$

Prove this is an equivalence relation.

Reflexive:
$\displaystyle a\sim a$
$\displaystyle a-a=0$; however, does $\displaystyle 0\in\mathbb{Q}$? I was under the impression $\displaystyle 0\notin\mathbb{Q}$
$\displaystyle 0=\frac{0}{1}$ the ratio of two integers.

3. Originally Posted by Plato
$\displaystyle 0=\frac{0}{1}$ the ratio of two integers.
I understand that but I thought $\displaystyle \mathbb{Q}$ were all $\displaystyle \frac{a}{b}$ such that $\displaystyle a\neq 0$

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

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

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

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

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

6. Originally Posted by dwsmith
Symmetric:
$\displaystyle a\sim b$, then $\displaystyle b\sim a$
Since $\displaystyle a,b\sim\mathbb{Q}$, then a and b can expressed as $\displaystyle a=\frac{c}{d}$ and $\displaystyle b=\frac{e}{f}$

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

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

Would it be allowable to multiple through by a -1 and then swap cd and ef to obtain:
$\displaystyle \frac{ef-cd}{df}\rightarrow\frac{e}{f}-\frac{c}{d}$?
Transitive:
$\displaystyle a\sim b, b\sim c$, then $\displaystyle a\sim c$
$\displaystyle c=\frac{g}{h}$
$\displaystyle \frac{c}{d}-\frac{e}{f}$
$\displaystyle \frac{e}{f}-\frac{g}{h}$

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

Equivalence class of $\displaystyle \sqrt{2}$ and a
$\displaystyle [\sqrt{2}]=(x\in\mathbb{R}|x\sim\sqrt{2})$
$\displaystyle x=\frac{a}{b}$ and $\displaystyle a,b\in\mathbb{Z}$
$\displaystyle [\sqrt{2}]=(x\in\mathbb{R}|\frac{a}{b}+\sqrt{2}\sim\sqrt{2}$
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 $\displaystyle r\in \mathbb{Q}$ then $\displaystyle -r\in \mathbb{Q}$

10. Originally Posted by dwsmith
Symmetric:
$\displaystyle a\sim b$, then $\displaystyle b\sim a$
Since $\displaystyle a,b\sim\mathbb{Q}$, then a and b can expressed as $\displaystyle a=\frac{c}{d}$ and $\displaystyle b=\frac{e}{f}$

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

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

Would it be allowable to multiple through by a -1 and then swap cd and ef to obtain:
$\displaystyle \frac{ef-cd}{df}\rightarrow\frac{e}{f}-\frac{c}{d}$?
If $\displaystyle a,b \in \mathbb{Z}$ and $\displaystyle a \sim b$ it means, by definition, that $\displaystyle \exists \ m,n \in \mathbb{Z}, \ n \neq 0,$ such that $\displaystyle a - b = \frac{m}{n} \in \mathbb{Q}$. What can you say about $\displaystyle b-a$ then?

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

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

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

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