equivalence class of function, picking proper x

Question

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 $y=\alpha$?

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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 $\sqrt5$?

Answer 1

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 $[\sqrt 2] = A$.

Answer 2

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.