Proving equivalence classes bijective to the set of points on the unit circle

**ams2990** · November 1st 2009, 12:11 PM

Define a relation on R as follows. Two real numbers $x, y$ are
equivalent if $x - y$ $\epsilon Z$ . Show that the set of equivalence classes of this relation is bijective to the set of points on the unit circle.

A part of the problem that I've omitted asked us to prove that the relation is an equivalence one -- I've done that. I've also defined the set of points on the unit circle, which is $\{ a,b \epsilon R | \sqrt{x^{2}+y^{2}} \}$ I don't know where to go from here, though.

**Plato** · November 1st 2009, 02:35 PM

Originally Posted by ams2990

Define a relation on R as follows. Two real numbers $x, y$ are
equivalent if $x - y$ $\epsilon Z$ . Show that the set of equivalence classes of this relation is bijective to the set of points on the unit circle.

A part of the problem that I've omitted asked us to prove that the relation is an equivalence one -- I've done that. I've also defined the set of points on the unit circle, which is $\{ a,b \epsilon R | \sqrt{x^{2}+y^{2}} \}$

Here are some observations.
$\Phi (t) = \left( {\cos (2\pi t),\sin (2\pi t)} \right)$ is a bijection of $[0,1)$ to the unit circle.

Using the floor function, $\left( {\forall r \in R} \right)\left[ {r - \left\lfloor r \right\rfloor \in [0,1)} \right]$

But you should see that $\left( {\forall r \in R} \right)\left[ {r - \left\lfloor r \right\rfloor } \right]$ is an equivalence class for this equivalence relation.

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