Let $\mathbb{R}^2 = \{Q = (a,b) | a,b\in \mathbb{R}\}$. Prove that if $q_1 = (a_1,b_1)$ and $q_2=(a_2,b_2)$ are equivalent, meaning $a_1^2+b_1^2 = a_2^2 +b_2^2$, then this gives an equivalence relation on $\mathbb{R}^2$. What is $[(1,0)], [(0,1)],[(2,2)],[(0,0)]?$ What does an equivalence class look like?

I am not sure how to do the second part of the question?

First part proof: Let $q_1$ and $q_2$ be equivalent then:

Reflexive: Let $a\sim b$ then $a \sim a$. So $a_1^2+a_1^2 = a_2^2 +a_2^2$ implies $2a_1^2 = 2a_2^2$ implies $a_1^2 = a_2^2$.

Symmetry: We must show $a\sim b$ and $b\sim a$. Thus let $a\sim b$ then we have $a_1^2+b_1^2 = a_2^2 +b_2^2$ are equivalent thus $b\sim a$ implies $b_1^2+a_1^2 = b_2^2 +a_2^2$ which are equivalent.

Transitive: $a\sim b$ and $b\sim c$ implies $a\sim c$. So $a_1^2+b_1^2 = a_2^2 +b_2^2$ and $b_1^2+c_1^2 = b_2^2 +c_2^2$ thus if we add them we have $a_1^2+2b_1^2+c_1^2 = a_2^2 +2b_2^2+c_2^2$ which implies $a_1^2+c_1^2 = a_2^2 +c_2^2$.

  • $\begingroup$ The statement of Reflexivity is not correct. For a relation to be reflexive, $a \sim a$ has to be true for all $a$. (Unlike the other two properties, it's not an implication.) You need to verify that $a = (a_1,a_2)$ is related to $a=(a_1,a_2)$ for EVERY point in $\mathbb{R}^2$. You shouldn't get an equation like $a_1^2 = a_2^2$, since that forces $a$ to be a very specific type of point. $\endgroup$ – Braindead Jan 30 '14 at 22:36
  • $\begingroup$ For symmetry, you are trying to show that $a\sim b$ IMPLIES $b\sim a$, which is different from "$a\sim b$ AND $b\sim a$". That is, under the assumption that $a\sim b$ is true, you are proving $b\sim a$ also holds. You have the right formulas, but the argument is a bit mixed up and it needs to be cleaned up. $\endgroup$ – Braindead Jan 30 '14 at 22:38
  • $\begingroup$ Something went very wrong with transitivity. It looks like you are mixing up your letters and your indices. Clearly write out what $a$,$b$, and $c$ are. $\endgroup$ – Braindead Jan 30 '14 at 22:43
  • $\begingroup$ @Braindead For reflexivity I am not sure I understand. If they are equivalent then there relation is $a_1^2+b_1^2 = a_2^2 +b_2^2$ so if a $a \sim a$ instead of $a\sim b$ then $a_1^2+a_1^2 = a_2^2 +a_2^2$ which implies $a_1^2 = a_2^2$. Is that not correct? If it isn't, then what will be the correct way of doing reflexivity? $\endgroup$ – Alex Jan 30 '14 at 23:59
  • $\begingroup$ If $a_1^2 = a_2^2$, then you can't have points like $(1,2)$, since $1^2\ne 2^2$. This is too long to explain so I'll write it in an answer. $\endgroup$ – Braindead Jan 31 '14 at 1:35

To show that something is an equivalence relation, you need to show that it satisfies:

  1. Reflexivity
  2. Symmetry
  3. Transitivity

I will edit this post as needed. For now, let's focus on:

Reflexive Property

A relation $\sim$ is called reflexive if $a\sim a$ for all $a$ in your set.

In your problem, the set is $\mathbb{R}^2.$ In order for your $\sim$ to be an equivalence relation, no matter which point $a$ I pick from $\mathbb{R}^2$, $a\sim a$ has to be true.

The biggest problem you are having is that you are mixing up the indices and different letters.

Pick a point in $\mathbb{R}^2$. To avoid confusion with indices, let's just call this point $(x,y)$.

Let's verify and see if $(x,y)\sim (x,y)$.

According to the definition you have, $(a_1,b_1)\sim (a_2,b_2)$ means

$$a_1^2 + b_1^2 = a_2^2 + b_2^2$$

See how everything on the left had side of the equality comes from only the first point $(a_1,b_1)$? (And everything on the right hand side of the equality comes from only the second point $(a_2,b_2)$?)

With that in mind, can you rewrite $(x,y)\sim (x,y)$? Think about what should be on the left side of the "=" sign and what should be on the right side of the "=" sign.

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  • $\begingroup$ If I rewrite $(x,y)\sim (x,y)$ then wouldn't it be $x^2+y^2 = x^2+y^2$? $\endgroup$ – Alex Feb 2 '14 at 3:31
  • $\begingroup$ Exactly! $(x,y) ~ (x,y)$ is true no matter what $x$ and $y$ are, because $x^2 + y^2 = x^2 + y^2$ is true no matter what $x$ and $y$ are! $\endgroup$ – Braindead Feb 2 '14 at 6:11
  • $\begingroup$ Thank you for the clarification! $\endgroup$ – Alex Feb 2 '14 at 7:57

A geometric interpretation of each equivalence class is being asked in the second part. What can you say about the sets of points on the plane which are equivalent?

Hint: Look at the distance (or really distance squared) of those points from the origin

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  • $\begingroup$ So $[(0,0)]$ is just the origin and for $[(1,0)]$ and $[(0,1)]$ are equivalent but what geometric interpretation will that produce? Just the lines at the points $[(1,0)]$ and $[(0,1)]$? $\endgroup$ – Alex Jan 17 '14 at 2:34
  • $\begingroup$ think sir :). You have to look at the distances- that's the key. $\endgroup$ – voldemort Jan 17 '14 at 2:35
  • $\begingroup$ Did I do the first part correctly? $\endgroup$ – Alex Jan 17 '14 at 2:49
  • $\begingroup$ Yes you did. Also, Think of circles, and not lines. $\endgroup$ – voldemort Jan 17 '14 at 3:43

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