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Math Help - Finding an equivalance class

  1. #1
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    Finding an equivalance class

    Hello, I was given a problem and I listed it below. I have finished everything up to the finding the equivalence class.

    Let f be any function from A A to A. (That is, f is a function of two variables
    defined on the entire plane.) Define a relation Af by the rule: (x,y) Af (z,w) if and only if f(x,y)
    = f(z,w). Show that Af is an equivalence relation. (Hint: first consider a simple specific case,
    such as f(x) = x^2 + y^2.)

    For the special case f(x) = x^2 + y^2 in the previous problem, what are the equivalence classes?

    I have completed proving why the question is reflexive, symmetric, and transitive, but I do not understand how I am supposed to determine what the equivalence classes are since there are no numbers involved. What I have at the moment is [x] = {(x, y)| x belongs to A and y belongs to A}, but I'm not sure if that is correct. If anyone could give me advice or some type of reference to follow it would be greatly appreciated!
    Last edited by Eulavtham; January 13th 2013 at 09:49 PM.
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  2. #2
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    Re: Finding an equivalance class

    Write 5 points equivalent to (1, 0) with respect to Af where f(x, y) = x^2 + y^2.
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    Re: Finding an equivalance class

    your relation is on AxA, it is an equivalence relation between two pairs of elements of A:

    (x,y) ~ (z,w) <=> f(x,y) = f(z,w)

    so [(x,y)] = {(z,w) in AxA : f(z,w) = f(x,y)}.

    i'll get you started: pick an element a of A. suppose f(x,y) = a. then if (z,w) ~ (x,y), don't we have to have f(z,w) = a, as well? what does this means for f(x,y) = x2 + y2?

    show that every equivalence class is of the form f-1({a}) (such equivalence classes are called level-sets). you should notice a certain "similarity" of all the equivalence classes.

    what do the equivalence classes for f(x,y) = xy look like?
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    Re: Finding an equivalance class

    Since AxA then f(x,y) = x^2 + y^2 = z^2 + w^2, correct?

    I believe I understand that part, but I'm not sure what I am looking for when you are looking for the inverse of a. Wouldn't the inverse in this situation be a?
    Last edited by Eulavtham; January 14th 2013 at 12:06 PM.
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    Re: Finding an equivalance class

    Quote Originally Posted by Eulavtham View Post
    Since AxA then f(x,y) = x^2 + y^2 = z^2 + w^2, correct?
    You can't say "Since A x A." This is the same as saying, "since 2 x 2" (and not "since 2 x 2 = 4," which is OK). Also, it is wrong to say f(x,y) = z^2 + w^2 because the right-hand side contains z and w and the left-hand side does not.

    I still suggest finding 5 points equivalent to (1, 0) with respect to Af where f(x, y) = x^2 + y^2.
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    Re: Finding an equivalance class

    Ah, alright. Sorry a little confusing to learn all of this from a less than stellar book (rated 1.5 stars out of 5 on Amazon ), but required for online course. I ended up watching all of to understand more on this matter. So would the equivalent class be [a] = {b | (b, a) in A} assuming a = (x, y) and b = (z, w)
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  7. #7
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    Re: Finding an equivalance class

    Quote Originally Posted by Eulavtham View Post
    So would the equivalent class be [a] = {b | (b, a) in A} assuming a = (x, y) and b = (z, w)
    No, [a] = {b | (b, a) in Af} = {b | f(b) = f(a)}. Here a and b range over A x A. Saying "(b, a) in A" is a type error because when b and a are in A x A, (b, a) is in A x A x A x A, not A. Writing explicitly for f(x, y) = x^2 + y^2 gives (x, y) Af (z, w) iff x^2 + y^2 = z^2 + w^2.
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    Re: Finding an equivalance class

    Ah, alright. I meant Af instead of A, but thanks for the full explanation. All making a lot more sense now I think, but the iff x^2 + y^2 = z^2 + w^2 is something I have not seen before in any of my examples. Would I be writing this as [(x,y)] = {(z,w) | ((z,w), (x, y)) in Af} if and only if x^2 + y^2 = z^2 + w^2 which also means [(z, w)] = {(z, w) | ((x, y), (z, w)) in Af} and this is the "similarity" Deveno hinted in his earlier posting. If that's not nearly correct then I'm afraid I'm more lost than I thought
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    Re: Finding an equivalance class

    Quote Originally Posted by Eulavtham View Post
    but the iff x^2 + y^2 = z^2 + w^2 is something I have not seen before in any of my examples.
    "Iff" is a contraction for "if and only if."

    Quote Originally Posted by Eulavtham View Post
    Would I be writing this as [(x,y)] = {(z,w) | ((z,w), (x, y)) in Af} if and only if x^2 + y^2 = z^2 + w^2
    Yes, except it is not correct to write "S if and only if P" where S is a set and P is a proposition. "If and only if" joins two propositions, which can be true or false, and a set can't be true or false. However, you are right that [(x,y)] = {(z,w) | ((z,w), (x, y)) in Af}, i.e., (z,w) is in [(x,y)] iff x^2 + y^2 = z^2 + w^2.

    Quote Originally Posted by Eulavtham View Post
    which also means [(z, w)] = {(z, w) | ((x, y), (z, w)) in Af} and this is the "similarity" Deveno hinted in his earlier posting.
    This is not really correct because x and y don't occur in the left-hand side or between { and |. In a set-builder construction {u | P(u, v)}, u should be a "fresh" variable name, i.e., a name that does not occur elsewhere, and v is allowed to occur in other places. Then, given v, {u | P(u, v)} denotes the set of all u such that P(u, v) holds. Your statement above: [(x,y)] = {(z,w) | ((z,w), (x, y)) in Af}, is correct since z and w are fresh names.

    In any case, I think Deveno meant not this, but that every equivalence class [(x,y)] corresponds to a number f(x,y); in this case, x^2 + y^2. In fact, [(x,y)] consists of all (z,w) such that z^2 + w^2 = a where a = x^2 + y^2. Or, as Deveno wrote, [(x, y)] = f^{-1}(\{a\}) where a = x^2 + y^2. The similarity is that every equivalence class is produced by some number a in this way.
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    Re: Finding an equivalance class

    Alright after looking over the thread again I believe I understand the majority of the information you've explained to me. I'm going to have to study some of the Discrete Math symbols because our book assumes we already know what they mean (even though this is an intro course o.O). Again, thanks for the help and I'm sure I'll be posting here again with any other issues I have in the next semester :P.
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    Re: Finding an equivalance class

    Responding here for the PM's I sent Emakarov. So you said (0, 5)(5,0)(-5,0)(0,-5) are of the same equivalence class. This would mean points such as (4, -3)(-3, 4) and so on would be of the same equivalence class because x^2 + y^2 = 25, which is the same as 5^2 correct?
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  12. #12
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    Re: Finding an equivalance class

    what i meant was:

    if f(x,y) = x^2 + y^2

    the the equivalence class of (x,y) is a circle of radius \sqrt{x^2 + y^2}. and the equivalence classes are in 1-1 correspondence with the non-negative reals (the range of f). they are "similar" (in the english sense) because they are SIMILAR (in the sense of plane geometry): all circles have the same shape. unless i am badly mistaken, by asking you to plot 5 (not 4) points of [(1,0)], Emakarov was hoping you would NOTICE all 5 points lay on a circle of radius 1.

    suppose that two points (x_1,y_1),(x_2,y_2) both lie on the circle x^2 + y^2 = r^2. if f(x,y) = x^2 + y^2 can you not see that (x_1,y_1) \sim (x_2,y_2)?

    what is a circle? it's nothing but the pre-image under f of the number r^2, all points of the circle all get mapped to one value by f, since f is CONSTANT on the circle.

    you can also look at it THIS way:

    let's say we're in 3 dimensions, with z = f(x,y) = x^2 + y^2. this is a circular paraboloid (like a cone, but with "curved sides"...think of an upside-down hill). the points that are equivalent in the plane below (the xy-plane) correspond to a "level" (height) of z, hence the name "level-set".

    these "level-sets" let us "chop up the plane" into disjoint equivalence classes: each point lies in just ONE "ring of the onion".
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    Re: Finding an equivalance class

    I'm confused because in the second post of the thread it says "Write 5 points equivalent to (1, 0) with respect to Af where f(x, y) = x^2 + y^2." To me this means (-1,1)(0, 1)(1,1)(1,0)(1,-1)(0,-1)(-1,-1) which all belong to the same equivalent class (circle w/ radius of 1). So why wouldn't (4,3)(-3,4) be in the same equivalent class as (5,0)(0,5) which have a radius of 5^2 or 25?
    Last edited by Eulavtham; January 17th 2013 at 07:48 PM.
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  14. #14
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    Re: Finding an equivalance class

    (1,1) is NOT equivalent to (1,0).

    f(1,1) = 1 + 1 = 2.

    f(1,0) = 1 + 0 = 1.

    2 does not equal 1.

    however, (4,3), (-3,4), (3,4), (-3,-4), (-4,3), (5,0) and (0,5) are all in the same equivalence class (the sum of their squares is 25).

    i see nothing in the problem that says x and y have to be integers. another point that is ALSO in the equivalence class of (1,0) is (1/2, √3/2).
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    Re: Finding an equivalance class

    Ah yeah, listing (0,1) and (1,0) was an oversight on my part. I mainly meant the points that equal the radius. I think I can see how the "level-sets" work. So radius 1, 2, 3, 4, 5, etc are all different equivalent classes. A point of the equivalent class would be considered an element of the equivalent class.
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