Question

Prove that the relation is an equivalence relation.
The Relation Ton R X R given by (x,y)T(a,b) iff \( x^2 +y^2=a^2+b^2 \).
Sketch the equivalence classes of (1,2); of (4,0)

Answer 1

Copied from wikipedia:

A given binary relation ~ on a set A is said to be an equivalence relation if and only if it is reflexive, symmetric and transitive. Equivalently, for all a, b and c in A:

a ~ a. (Reflexivity)
if a ~ b then b ~ a. (Symmetry)
if a ~ b and b ~ c then a ~ c. (Transitivity)

Answer 2

So, on the set RxR, we need to show reflexivity, symmetry, and transitivity.

Answer 3

so its reflexive and syymetric

Answer 4

ohhhh and transitivity

Answer 5

Yeah, I mean, that's obvious. You could write a proof for it pretty easily, I'm guessing.

Answer 6

ya i am having problems with teh second part i am not sure what they want

Answer 7

The second part? Sketch the equivalence classes of...?

Answer 8

ya like wth does it want me exactly to do?

Answer 9

An equivalence relations takes a set and breaks it up into partitions, labeling big chunks of the set as "equivalent." So when they ask about the equivalence class, they want to know what things in the set are being grouped together.
In this case, the set is RxR, so ordered pairs. And the way we call things equivalent is if a^2+b^2 is the same. So RxR is getting broken up into equivalence classes based on what the result is when you do a^2+b^2.

Answer 10

So let's look at the ordered pair (1,2)

Answer 11

I get the impression the equivalence class of (1,2) is the circle on the x,y plane centered at 0,0 with radius sqrt(5)

Answer 12

That's correct... answer guy.

Answer 13

its ok if phi gives me an answer lol
like I am studying online so I am teaching all this to myself so sometimes i need to see an example to clarify things

Answer 14

I'm just fundamentally opposed to giving answers, particularly when someone else is working with someone and then you just hop in and drop an answer.

Answer 15

ya thats true

Answer 16

wait but i still dont get y that is the answer let me work that thru

Answer 17

ohhhh ok i get it

Answer 18

But anyways, here's the explanation
What are we doing when we do a^2+b^2?
Well, if we square rooted that, we'd have the distance from the origin. So we're finding the square of the distance.

Answer 19

So the equivalence relation that has all a^2+b^2 = 5 means that all of the points have to have the same distance from the origin.

Answer 20

ok so the equivalence class of (4,0) is a circle with the origin being (0,0) and a radius of 4 lol ok this one is simple

Answer 21

Yes. It's pretty simple. Ask another one. =)

Answer 22

okkk well no questions that r this style but i found smth similar

Answer 23

Let R be the relation on Q defined by \( \frac{p} {q} R \frac{s} {t}\) iff pt=qs. Show that R is an equivalence relation. describe all ordered pairs in the equivalence class \( \frac{2} {3} \)

Answer 24

idk this one is a lil bit harder

Answer 25

Well I would say that the proof part is harder than naming the equivalence class.

Answer 26

ok so the ordered pairs wld be multiples of (2/3)

Answer 27

Actually maybe it's not so bad...

\(\Large \frac{p}{t} R \frac{p}{t} \)?
Yes, because pt = pt.
\(\Large \frac{p}{t} R \frac{r}{s} \rightarrow \frac{r}{s} R \frac{p}{t}\)?
Yes, because ps = rt \(\rightarrow\) rt = ps

Answer 28

okkkkkk

Answer 29

And transitivity:
\(\Large \frac{p}{t} R \frac{r}{s}, and \frac{r}{s} R \frac{q}{v}\)
ps = rt, and rv = qs
so
p/t = r/s and r/s = q/v, therefore p/t = q/v and pv = qt

Answer 30

yaaaaaaaaaaa

Answer 31

i solved tran

Answer 32

transitivity in a backwards manner but it worked too

Answer 33

And yeah, I think it should just be multiples of 2/3.
\(\Large \frac{2}{3} R \frac{a}{b} \rightarrow 2a = 3b\)
So 2/3 = a/b, which just meants that a/b is some multiple of 2/3.

Answer 34

So that gives you an idea of what this equivalence relation is actually doing. It's calling two fractions equivalent if one is a multiple of the other.

Answer 35

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Answer 36

That was really clear

Answer 37

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Answer 38

you're*

Answer 39

Good =D