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Defining R to be the relationship on real numbers given by xRy iff x-y is rational, I've been asked to find the equivalence class of $\sqrt2$. My instincts say that the equivalence class of $\sqrt2$ would just be the empty set. But after a riveting conversation on a similar subject subtraction of two irrational numbers to get a rational

I'm curious to if I am able to pick and choose x to suite my needs? say could I define x to be $\alpha+\sqrt2$ so the equivalence class would be $y=\alpha$?

EDIT: This is just to see if I have the principle of equivalence classes down..

given the Relation R xRy s.t. x-y is rational, the following equivalence classes are..

[0] = {y: y is rational}

[1] = {y: y is rational}

[$\sqrt2$] = {y: y = $\alpha+\sqrt2$ s.t. $\alpha$ is rational}

To summarize to find the above equivalence classes on xRy iff x-y is rational

I would look at all of the ordered pair (0,y),(1,y),($\sqrt2$,y) fulfills the above stipulation?

Or as to ensure further confidence that I understand (I don't want to be spoon, fed but confirmation that I'm on the right track would be greatly appreciated.

Given the relation T given by (x,y)T(a,b) iff $x^2+y^2=a^2+b^2$

would the equivlanece class of (1,2) be the circle of radius $\sqrt5$?

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user3256725
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    Your instincts need some training because the equivalence class of something is never the empty set -- at the very least the equivalence class of $\sqrt 2$ will need to contain $\sqrt 2$ itself, if you have an equivalence relation at all! – hmakholm left over Monica Oct 22 '15 at 08:13
  • would you be able to take a look at my edit and tell me if I'm on the right path to understanding this concept? – user3256725 Oct 22 '15 at 08:31
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    x @user, yes all of the edit seems to be right. Instead of ${y:y=\alpha+\sqrt2 \text{ where }\alpha\text{ is rational}}$ I would just write ${\alpha+\sqrt2\mid \alpha\in\mathbb Q}$, but that is just cosmetics. – hmakholm left over Monica Oct 22 '15 at 08:33
  • Thank you, It is always humbling to see what a strong community we have here dedicated to the advancement of human intellect. I can't thank you enough for your help – user3256725 Oct 22 '15 at 08:35

2 Answers2

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Remember that, by definition, the equivalence class of $\sqrt 2$ is the set $$[\sqrt 2] = \{y\in\mathbb R~|~\sqrt 2 - y \in \mathbb Q\}$$

Pose $A = \{\alpha + \sqrt 2~|~\alpha\in\mathbb Q\}$.

  • Pick an element $x\in A$ : there is $\alpha \in\mathbb Q$ such that $x = \alpha + \sqrt 2$. Then $\sqrt 2 - x = -\alpha \in \mathbb Q$, thus $x\in[\sqrt 2]$.
  • Pick an element $x\in[\sqrt 2]$. Then, by definition of $[\sqrt 2]$, there is $\alpha \in \mathbb Q$ such that $\sqrt 2 - x = \alpha$, ie $x = -\alpha = \sqrt 2$. Thus $x\in A$.

We just showed that $[\sqrt 2] = A$.

Kevin Quirin
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  • Ahh, I think I had an intrinsic misunderstanding of equivalence classes. I was close, but you just clarified. once able to, I will mark your answer. @DRF, that being said, I do greatly appreciate your input too. The one downfall of this site is only being able to select one answer.. – user3256725 Oct 22 '15 at 08:17
  • while I have you, would you be able to look at my edit and make sure I'm on the right track? – user3256725 Oct 22 '15 at 08:22
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Hint first thing you need to do is figure out what an equivalence class is. Once you have an understanding of what it is that result should be in general terms (real number, rational number, set of real numbers, set of rational numbers etc.) you can continue.

DRF
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