[tex]x,y,z\in\mathbb{R}[/tex](adsbygoogle = window.adsbygoogle || []).push({});

[tex]x\sim y[/tex] iff. [tex]x-y\in\mathbb{Q}[/tex]

Prove this is an equivalence relation.

Reflexive:

[tex]a\sim a[/tex]

[tex]a-a=0[/tex]; however, does [tex]0\in\mathbb{Q}[/tex]? I was under the impression

[tex]0\notin\mathbb{Q}[/tex]

Symmetric:

[tex]a\sim b[/tex], then [tex]b\sim a[/tex]

Since [tex]a,b\sim\mathbb{Q}[/tex], then a and b can expressed as [tex]a=\frac{c}{d}[/tex] and [tex]b=\frac{e}{f}[/tex]

[tex]\frac{c}{d}-\frac{e}{f}\rightarrow \frac{cf-de}{df}[/tex]

How can I get than in the form of [tex]\frac{e}{f}-\frac{c}{d}[/tex]?

Would it be allowable to multiple through by a -1 and then swap cd and ef to obtain:

[tex]\frac{ef-cd}{df}\rightarrow\frac{e}{f}-\frac{c}{d}[/tex]?

Transitive:

[tex]a\sim b, b\sim c[/tex], then [tex]a\sim c[/tex]

[tex]c=\frac{g}{h}[/tex]

[tex]\frac{c}{d}-\frac{e}{f}[/tex]

[tex]\frac{e}{f}-\frac{g}{h}[/tex]

add together

[tex]\frac{c}{d}-\frac{g}{h}\rightarrow\frac{ch-gd}{dh}\in\mathbb{Q}[/tex] [tex]a\sim c[/tex]

Equivalence class of [tex]\sqrt{2}[/tex] and a

[tex][\sqrt{2}]=(x\in\mathbb{R}|x\sim\sqrt{2})[/tex]

[tex]x=\frac{a}{b}[/tex] and [tex]a,b\in\mathbb{Z}[/tex]

[tex][\sqrt{2}]=(x\in\mathbb{R}|\frac{a}{b}+\sqrt{2}\sim\sqrt{2})[/tex]

Correct?

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# Homework Help: Equivalence Relations

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