0
$\begingroup$

A relation is defined on the set $A=\{a + b\sqrt{2} \; : \; a, b \in \mathbb{Q} \text{ and } a + b\sqrt{2} \neq 0\}$ by $xRy$ if $x/y$ is in $\mathbb{Q}$. Show that $R$ is an equivalence relation and determine the distinct equivalence classes.

I showed that $R$ is reflexive, symmetric, and transitive already, and so I know that $R$ is an equivalence relation. My issue is finding the equivalence classes. Since $x,y$ are coming from set $A$ but set $A$ uses $a,b$ which come from $\mathbb{Q}$, how do I define the equivalence classes?

I tried this: $[1] = \{x \in A \; : \; xR1\} = \{x \in A \; : \; x \in \mathbb{Q}\} = \{a + b\sqrt{2} \; : \; a \in \mathbb{Q}, b=0\}$.

Is that the correct way to define it?

$\endgroup$
4
  • $\begingroup$ By "rad(2)" do you mean $\sqrt{2}$? You write that as $\backslash\text{sqrt}\{2\}$. $\endgroup$ Commented Apr 13, 2015 at 16:10
  • $\begingroup$ Yes sorry I'm not too good with the symbols. $\endgroup$ Commented Apr 13, 2015 at 16:11
  • $\begingroup$ Put math stuff between dollar signs like $\$\backslash\text{sqrt}\{2\}\$$. $\endgroup$ Commented Apr 13, 2015 at 16:13
  • $\begingroup$ You might want to see this to gain more understanding about equivalence classes math.stackexchange.com/questions/1232413/… $\endgroup$ Commented Apr 13, 2015 at 16:52

1 Answer 1

Reset to default
1
$\begingroup$

I believe there is one equivalence class for each rational $r\in\mathbb Q$. That's because $a+b\sqrt{2}$ is equivalent to $r+\sqrt{2}$ where $r=a/b$. Now just show $r+\sqrt{2}$ is equivalent to $s+\sqrt{2}$ iff $r=s$.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.