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Equivalence relation problem

  1. Feb 19, 2012 #1
    1. The problem statement, all variables and given/known data

    The question is to show that Cos^2(x) + Sin^2(y) = 1 is an equivalence relation.

    3. The attempt at a solution

    I know that there are three conditions which the equation must satisfy. (reflexivity, symetry, transitivity)

    For reflexivity I tried: Cos^2(x) - Cos^2(x) = Sin^2(y) - Sin^2(y) = 0.

    For symmetry: ( I don't fully understand this condition) I think that maybe Sin^2(y) + Cos^2(x) = 1 will do. Why? I'm not sure. Any explanation will be very apreciated.

    Transitivity: Cos^2(x) + Sin^2(y) = 1, Sin^2(y) + Cos^2(z) = 1, Cos^2(z) + Sin^2(h) = 1.

    If I add all o them I'll get Cos^2(x) + Sin^2(y) + Sin^2(y) + Cos^2(z) + Cos^2(z) + Sin^2(h) = 1 + 1 + 1

    I was tempted to say that Sin^2(y) = Sin^2(z) = Sin^2(h), and that cos^2(x) = cos^2(z) (my thought was: "since they are supposed to be equivalent...") and rewrite it as:

    3Cos^2(x) + 3Sin^2(h) = 3 (and then I divide it by 3)

    Cos^2(x) + Sin^2(h) = 1.

    I think my methods are wrong.

    Any help will be very appreciated.
     
  2. jcsd
  3. Feb 19, 2012 #2
    It will help to write the problem more precisely, then say exactly what you're trying to show in each step.

    The problem should be stated like this:

    Define x ~ y if cos^2(x) + sin^2(y) = 1. Show that ~ is an equivalence relation.

    Now, write down exactly what is meant by reflexivity, transitivity, and transitivity in terms of ~. This will get you on the right track regarding what you need to prove. Start with reflexivity.
     
  4. Feb 19, 2012 #3
    Hello there!

    well, I have read many times the definitions of each, but still don't get it.

    a) x ~ x, (reflexivity)

    Cos^2(x) = Cos^2(x) (?)

    b) if x ~ y and y ~ z then x ~ z. (transitivity)

    Cos^2x + Sin^2y, (?)

    c) if x ~ y then y ~ x (symmetry)

    Cos^2x = Cos^2y, Sin^2y = Cos^2x (?)

    Why you mentioned asscociativity?
     
    Last edited: Feb 19, 2012
  5. Feb 19, 2012 #4
    a) x ~ x, (reflexivity)

    Cos^2(x) + Sin^2(x)= 1, Cos^2x = 1 - (1 - Cos^2x), Cos^2x = Cos^2x. (?)
     
  6. Feb 19, 2012 #5
    Yes, one of the above is correct but why?

    What is the exact definition of ~? What would the exact definition of x ~ x be? Is it true?
     
  7. Feb 19, 2012 #6
    a) x ~ x, (reflexivity)

    Cos^2x + Sin^2x = 1

    b) if x ~ y and y ~ z then x ~ z. (transitivity)

    Cos^2x + Sin^2y = 1, Cos^2y + Sin^2z = 1,

    Cos^2x + Sin^2y + Cos^2y + Sin^2z = 2,

    (Cos^2x + Sin^2z) + (Cos^2y + Sin^2y) = 2

    c) if x ~ y then y ~ x (symmetry)

    Cos^2x + Sin^2y = 1, Cos^2y + Sin^2x = 1,

    Right?
     
  8. Feb 19, 2012 #7
    eh, I guess because there is an x, for which the definition of ~ (in this case) holds true, namely Cos^2x + Sin^2x =1, right?
     
  9. Feb 19, 2012 #8
    Do you understand that reflexivity says that x ~ x must be true for ALL x in order for ~ to be an equivalence relation?

    And in this specific case, is x ~ x always true for any x?
     
  10. Feb 19, 2012 #9
    I would say yes...
     
  11. Feb 19, 2012 #10
    So, I guess now my answer is correct, or? I'll try another example, and see if I understand this.

    x ~ y if e^(x^2)= e^(y^2). Show that ~ is an equivalence relation.


    a) x ~ x, (reflexivity)

    e^(x^2)= e^(x^2), for all x.

    b) if x ~ y and y ~ z then x ~ z. (transitivity)

    e^(x^2)= e^(y^2),
    e^(y^2)= e^(z^2),

    e^(x^2)+ e^(y^2)= e^(y^2) + e^(z^2), (I cancel the e^(y^2))

    e^(x^2)= e^(z^2),

    c) if x ~ y then y ~ x (symmetry)

    e^(x^2)= e^(y^2).

    Is this correct?
     
  12. Feb 19, 2012 #11
    Did your teacher explain to you that you have to say WHY something is true?

    Now you have to do transitivity and symmetry, right? Before you go on to the next one.

    But you haven't completed reflexivity yet.

    We have to show that for all x, x ~ x. That means we have to show that for all x, cos^2(x) + sin^2(x) = 1. And we know this is true because ...
     
  13. Feb 19, 2012 #12
    I know that it comes from the Pythagoras formula x^2 + y^2 = r^2. I substitute, Cos(t) = x/r, Sin(t)=y/r. And,

    (x/r)^2+(y/r)^2 = (x^2+y^2)/r^2 = r^2/r^2 = 1.

    Cos^2t + Sin^2t = 1

    So, for reflexivity it will work for every t.
     
  14. Feb 19, 2012 #13
    Yes I think this one's good.
     
  15. Feb 19, 2012 #14
    Yes, in fact that's a little overkill. It's sufficient to note that cos^2(x) + sin^1(x) = 1 for every x. I think you've got the idea now.
     
  16. Feb 19, 2012 #15
    c) if x ~ y then y ~ x (symmetry)

    e^(x^2)= e^(y^2).

    e^(y^2)= e^(x^2).
     
  17. Feb 19, 2012 #16
    but does transitivity and symmetry are correct on the first example?
     
  18. Feb 19, 2012 #17
    You wrote:

    "I think that maybe Sin^2(y) + Cos^2(x) = 1 will do. Why? I'm not sure. Any explanation will be very apreciated."

    That's not right. You should write down EXACTLY what it is you're trying to show, starting with defining what it means for ~ to be symmetric.
     
  19. Feb 19, 2012 #18
    Hello there! I did indeed rewrote it. It's post #6. I will rewrite it here:

    a) x ~ x, (reflexivity)

    Cos^2x + Sin^2x = 1, for all x.

    b) if x ~ y and y ~ z then x ~ z. (transitivity)

    Cos^2x + Sin^2y = 1, Cos^2y + Sin^2z = 1,

    Cos^2x + Sin^2y + Cos^2y + Sin^2z = 2,

    (Cos^2x + Sin^2z) + (Cos^2y + Sin^2y) = 2

    c) if x ~ y then y ~ x (symmetry)

    Cos^2x + Sin^2y = 1, Cos^2y + Sin^2x = 1,

    Is this fine?
     
  20. Feb 19, 2012 #19
    Well you wrote down what you have to prove, but you didn't prove it. I didn't look at the transitivity proof.

    What kind of class is this for?
     
  21. Feb 20, 2012 #20
    Hello there!

    Let's see:

    c) if x ~ y then y ~ x (symmetry)

    Cos^2x + Sin^2y = 1, Cos^2y + Sin^2x = 1,

    (1 - Sin^2x) + (1 - Cos^2y) = 1

    2 - Sin^2x - Cos^2y = 1, - Sin^2x - Cos^2y = -1 (I multiply by -1)

    Sin^2x + Cos^2y = 1,

    Is this fine?

    What kind of class? What do you mean? Course? I'm learning this because I want to.
     
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