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Homework Help: Equivalence Relations

  1. Apr 29, 2010 #1
    [tex]x,y,z\in\mathbb{R}[/tex]

    [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?
     
    Last edited: Apr 29, 2010
  2. jcsd
  3. Apr 29, 2010 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

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

    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.

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