MHB Binary Relations and Equivalence Classes | Proving R is an Equivalence Relation

[JProgrammer]

So the question I am trying to solve is this:

Define a binary relation R on R as follows: R={(x,y)∈ R×R:cos⁡(x)=cos⁡(y)}
Prove that R is an equivalence relation, and determine its equivalence classes.

I've figured out the first two requirements for being a binary relation:

1. cos(x) = cos(x)
2. cos(x) = cos(x + 2kpi)

I don't know how to go about solving the third requirement for being a binary relation because there is no z to work with.
If someone could show me how to find the third requirement, that would be great.

Thank you

 

[I like Serena]

So the question I am trying to solve is this:

Define a binary relation R on R as follows: R={(x,y)∈ R×R:cos⁡(x)=cos⁡(y)}
Prove that R is an equivalence relation, and determine its equivalence classes.

I've figured out the first two requirements for being a binary relation:

1. cos(x) = cos(x)
2. cos(x) = cos(x + 2kpi)

I don't know how to go about solving the third requirement for being a binary relation because there is no z to work with.
If someone could show me how to find the third requirement, that would be great.

Thank you

Hi JProgrammer! Welcome to MHB! ;)

From wiki, an equivalence relation requires:

For all a, b and c in X:
1. a ~ a. (Reflexivity)
2. a ~ b if and only if b ~ a. (Symmetry)
3. if a ~ b and b ~ c then a ~ c. (Transitivity)

The equivalence class of a under ~, denoted [a], is defined as $[a]=\{b\in X\mid a\sim b\}$.​

You have shown the first, since indeed for any $x$ we have $\cos x = \cos x$.
Can you show 2 and 3, and can you find the equivalence classes?

 

[JProgrammer]

I don't understand what you mean. For the symmetric requirement, I have shown it as: cos(x) = cos(x + 2kx). As for the transitive requirement, that is what I need help with. I don't know what that would be since there is no z to work with.

Also, I don't know how to find the equivalence classes.

 

[I like Serena]

I don't understand what you mean. For the symmetric requirement, I have shown it as: cos(x) = cos(x + 2kx). As for the transitive requirement, that is what I need help with. I don't know what that would be since there is no z to work with.

Also, I don't know how to find the equivalence classes.

Ah, but that is not symmetry.
For symmetry we need that for all $x,y$ we have that $xRy$ if and only if $yRx$.
That means that for all $x,y$ we need that $\cos x = \cos y$ if and only if $\cos y = \cos x$.
This is trivially true. It follows from how the equal sign ($=$) is defined, which is also an equivalence relation.

For transitivity we need that for all $x,y,z$ we have that if $\cos x = \cos y$ and $\cos y=\cos z$, that we also have that $\cos x = \cos z$.
Again, this is trivially true.

That leaves the equivalence classes.
You have found that $x$ is equivalent to $x+2k\pi$, since $\cos(x) = \cos(x+2k\pi)$.
So that might suggest that $[x]=\{x+2k\pi \mid k\in \mathbb Z\}$.
But how about $-x$? Don't we have that $\cos(x)=\cos(-x)$?
That would imply that $-x$ should also be in the equivalence class $[x]$. (Worried)