- #1
Dustinsfl
- 2,281
- 5
[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?
[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: