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.