# Proof that $(x,y)\sim (x',y') \iff x=x' , y-y'=n2\pi$ where $n\in\mathbb Z$ is an equivalence relation?

Have I made any mistakes in the following proof?

THEOREM:

If $P=\{(x,y)\in\mathbb R^2|x>0 \}$ and if there exists a relation $$\sim|(x,y)\sim(x',y') \iff x=x' , y-y'=n2\pi$$ where $n\in\mathbb Z$ then $\sim$ is an equivalence relation on $P$.

PROOF:

If $(x,y)\sim(x,y)$ then $x=x$ and $y-y=0$ .

Since for all $x\in \mathbb R, x=x$ and $0=n$ in $n2\pi$, $\sim$ is reflexive.

Now if $(x,y)\sim(x',y')$ then $x=x', y-y'=n2\pi$ so if $(x',y')\sim(x,y)$ then $x'=x , y'-y=-n2\pi$ since $-n\in \mathbb Z$, $\sim$ is symmetric.

If $(x,y)\sim(x',y')$ and $(x',y')\sim(x'',y'')$ then

$x=x'$ and $x'= x''$ and $y-y'=n2\pi$ and $y'-y''=n'2\pi$. So $x=x''$ by the transitivity of $=$ and $y-y''= (n-n')2\pi$ and $(n-n')\in\mathbb Z$. So, $\sim$ is transitive.

Therefore $\sim$ is an equivalence relation on $P$.

QED

How would I show that that each equivalence class contains a unique pair $(x,y)$ with $0\leq y<2\pi$?

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What is your question ? –  Belgi Sep 13 '12 at 16:21
You've got the right idea, it's just a little turned around. To show reflexivity, first show that $x=x$ and $y-y=n2\pi$; this implies $(x,y) \sim (x,y)$. Similarly, show that $x=x$' implies $x'=x$ and $y-y'=n2\pi$ implies $y'-y=m2\pi$ for some $m$; then state that this means $(x,y) \sim (x',y')$ Transitivity is shown similarly. –  Tarnation Sep 13 '12 at 16:25
@tarnation Reflexivity and transitivity look OK, but yeah there is a hiccup in symmetry. –  rschwieb Sep 13 '12 at 16:51
why ? (the hiccup ) –  Joshua Rocky Lizardi Sep 13 '12 at 17:39
How would I show that that each equivalence class contains a unique pair (x,y) with 0≤y<2π?... –  Joshua Rocky Lizardi Sep 13 '12 at 17:53

The proofs of reflexivity and transitivity are basically okay, though they could be stated a little better, but there is indeed a hiccup in the proof of symmetry. Here’s what you have:

Now if $(x,y)\sim(x',y')$ then $x=x', y-y'=n2\pi$ so if $(x',y')\sim(x,y)$ then $x'=x$, $y'-y=-n2\pi$ since $-n\in \mathbb Z$, $\sim$ is symmetric.

Look at the logical structure: if $(x,y)\sim(x',y')$, then (something), so if $(x',y')\sim(x,y)$, then (something else). That isn’t the structure of the statement of symmetry: you should be proving that if $(x,y)\sim(x',y')$, then $(x',y')\sim(x,y)$. You have all of the pieces needed to do this; you just haven’t put them together correctly. Here’s one way to do that:

If $(x,y)\sim(x',y')$, then by definition $x=x'$ and $y-y'=2n\pi$ for some integer $n$. But then $y'-y=2(-n)\pi$, and $-n$ is also an integer, so $x'=x$ and $y'-y=2(-n)\pi$ with $-n$ an integer, and therefore by definition $(x',y')\sim(x,y)$. Therefore $\sim$ is symmetric.

To show that each equivalence class contains a unique pair $(x,y)$ with $0\le y<2\pi$, you must show two things: that each equivalence class contains at least one such pair, and that each equivalence class contains at most one such pair. It’s often easier in such arguments to prove uniqueness (at most one), so let’s start there.

Suppose that $0\le y,y'< 2\pi$ and that $(x,y)$ and $(x',y')$ are in the same equivalence class. Then $(x,y)\sim(x',y')$ so $x=x'$ and $y-y'=2n\pi$ for some $n\in\Bbb Z$. Now $y<2\pi$ and $y'\ge 0$, so $y-y'<2\pi-0=2\pi$. Similarly, $y\ge 0$ and $y'<2\pi$, so $y-y'>0-2\pi=-2\pi$. That is, $-2\pi<y-y'<2\pi$, and $y-y'=2n\pi$ for some $n\in\Bbb Z$, so $-2\pi<2n\pi<2\pi$. Dividing through by $2\pi$, we see that $-1<n<1$, and $n\in\Bbb Z$, so $n$ must be $0$, and $y=y'$. We already knew that $x=x'$, so $(x,y)=(x',y')$. This shows that an equivalence class cannot contain two different pairs $(x,y)$ and $(x',y')$ with $0\le y,y'<2\pi$.

To show that each equivalence class does contain such a pair, start with an arbitrary equivalence class $C$, and let $(x,y)$ be any member of $C$. We want to find a pair equivalent to $(x,y)$ whose second element is in the interval $[0,2\pi)$. Any pair equivalent to $(x,y)$ will have $x$ as its first element, so we just have to find its second element. We want a real number $y'\in[0,2\pi)$ such that $y-y'=2n\pi$ for some $n\in\Bbb Z$. Intuitively it should be clear that such a $y'$ exists: if $y\ge 2\pi$, we just keep subtracting $2\pi$ until we get into the interval $[0,2\pi)$, and if $y<0$, we keep adding $2\pi$ until we do so. The problem is to make this rigorous; the trick is to write $y$ as an integer multiple of $2\pi$ plus a remainder that’s less than $2\pi$.

Let $z=\frac{y}{2\pi}$, and let $n=\lfloor z\rfloor$, the greatest integer less than or equal to $z$. Then $n\le z<n+1$, so $2\pi n\le 2\pi z<2\pi(n+1)$, i.e., $2n\pi\le y<2(n+1)\pi$. Let $y'=y-2n\pi$; clearly $y=2n\pi+y'$, and $0\le y'<2\pi$. You should have no trouble verifying that $(x,y')\sim(x,y)$ and hence that $(x,y')$ is a member of $C$ whose second element is in the interval $[0,2\pi)$. That will show that each equivalence class has at least one such member, and together with the previous result will show that each equivalence class has exactly one such member.

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