Proving Equivalence Relations for Real Numbers x, y, z in R

[Dustinsfl]

$$x,y,z\in\mathbb{R}$$

$$x\sim y$$ 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

$$0\notin\mathbb{Q}$$

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?

 

[LCKurtz]

$$x,y,z\in\mathbb{R}$$

$$x\sim y$$ 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

$$0\notin\mathbb{Q}$$

Remember the rationals are the reals that can be written m/n where m and n are integers.

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 $$b=\frac{e}{f}$$

No. You aren't given that a and b are in Q; they are two real numbers. You are given that a ~ b. What does that mean? And you are supposed to show that b ~ a. Write down what that means. Then see if you can show it.

Transitive:

Same suggestion. Write down what you are given and what you need to prove.