x−yx-y is rational only if both xx and yy are rational. Isn’t xx then equivalent to every rational provided the result falls in [0,1)[0,1)? What then are the equivalence classes?

Thanks!

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Your first sentence is wrong. Think about α\alpha and α+1/100\alpha + 1/100 when α\alpha is irrational.

– Ethan Bolker

2 days ago

Well that clears that up. Silly mistake.

– Chris

2 days ago

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2 Answers

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The rationals in the interval are one class. π\pi plus rationals which lie in the interval are another (π−3\pi-3, π−22/7\pi-22/7, etc). √2−1\sqrt{2}-1 represents a class, so does e−2e-2. But I can’t construct a list of representatives of all equivalence classes without appealing to the axiom of choice.

The easiest way to get the list using the axiom of choice is to view the reals as a vector space over the rationals, and remember that the axiom of choice is equivalent to the assertion that every vector space has a basis. A basis for the reals over the rationals gives representatives for the equivalence classes under your equivalence relation. You’ll want to check that you can arrange for them to be in the prescribed interval of course.

You say x−yx-y is rational only if both xx and yy are rational. This is not true. Consider π\pi and π+1\pi+1. Neither are rational but their difference is rational.

For all irrational I∈[0,1)I\in [0,1) you have by definition

˜I= class of I={I+r; r∈Q}\widetilde I=\text { class of } I=\{I+r;\space r\in \Bbb Q\}