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Math Help - [SOLVED] Equivalence classes/well definedness, help!

  1. #1
    Member Jason Bourne's Avatar
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    [SOLVED] Equivalence classes/well definedness, help!

    (a) Define a relation ~ on the set \mathbb{R} of real numbers by

    r~ s \Leftrightarrow r-s \in \mathbb{Z}<br />

    Describe the equivalence class of 0.

    (b) Prove that every equivalence class [r] either consists entrely of rational numbers or else entirely of irrational numbers.

    (c) On the set C of all equivalence classes of X with respect to ~ define the operation of addition as follows:

    [r] + [s] = [r+s]

    Prove that this operation is well defined, associative and that [0] is a neutral element.

    (d) Prove that if r is a rational number than there exists a natural number n such that

    \underbrace{[r] + ... + [r]}_n = [0]

    while if r is an irrational number than no such n exists.

    (e) Is the following multiplication if elements of C well defined?:

    [x][y]=[xy]<br />

    -----
    I'm not sure on the above. Thanks for any help. - Jason
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  2. #2
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    Quote Originally Posted by Jason Bourne View Post
    (a) Define a relation ~ on the set \mathbb{R} of real numbers by r~ s \Leftrightarrow r-s \in \mathbb{Z}
    Describe the equivalence class of 0.
    (b) Prove that every equivalence class [r] either consists entrely of rational numbers or else entirely of irrational numbers.
    Jason Bourne; In the past you have gotten rather extensive help. But now you need to at least show some effort on your part. This is simply a list of question leading to an overall result.
    At least show that you understand part (a). How do you think it is worked?

    For part (b), what do you understand about the addition of rational numbers and irrational? What could that answer have to do with part (b)?

    Please tell us what you donít understand.
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  3. #3
    Member Jason Bourne's Avatar
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    okay, sorry.

    for (a) I think that the equivalnece class of 0 is just the set of integers with the set of reals but I wasn't totally sure.

    for (b) I was thinking I could say, if r is rational then [r] consists of all elements , say s, that s-r is an integer, and then prove that s must be rational through supposing that s was irrational. So [r] would be entirely composed of rationals. Similarly for irrational r.

    for (c) I was really not sure how to prove its well defined? Do I go something like this?:

    Let [r]=[r1] and [s]=[s1]
    then r-r_1 \in \mathbb{Z} and s-s1 \in \mathbb{Z}
    so (r+s)-(r_1 + s_1) \in \mathbb{Z}
    so [r+s]=[r_1 + s_1] and so is well defined ?

    Presumable [0] is neutral because [r] + [0] = [r+0] = [r] ?

    in (d) I know I have to use part (b) and (c) but for some reason I can't think how to show this though it seems simple.

    in (e) I think I need to show that given

    [x]=[x_1] and [y]=[y_1] show \ \ \  [xy]=[x_1y_1]

    but I wasn't getting anywhere and couldn't think of a counter example.
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    O.K. Well done so far.
    For part (d), recall that different equivalence classes are pair-wise disjoint.
    Moreover, rational numbers are the ratio of two integers: r \in \mathbb{Q} \Rightarrow \left( {\exists n \in \mathbb{Z}^ +  } \right)\left[ {nr \in \mathbb{Z}} \right]\,\& \,\therefore \quad nr \in \left[ 0 \right]
    Can you see how to finish this one?
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    Member Jason Bourne's Avatar
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    Quote Originally Posted by Plato View Post
    O.K. Well done so far.
    For part (d), recall that different equivalence classes are pair-wise disjoint.
    Moreover, rational numbers are the ratio of two integers: r \in \mathbb{Q} \Rightarrow \left( {\exists n \in \mathbb{Z}^ +  } \right)\left[ {nr \in \mathbb{Z}} \right]\,\& \,\therefore \quad nr \in \left[ 0 \right]
    Can you see how to finish this one?
    sorry I'm still not getting it.

    [0] is the set of all integers in X, but how is it equivalent to [nr]? How is nr \in \mathbb{Z} ?
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    Quote Originally Posted by Jason Bourne View Post
    [0] is the set of all integers in X, but how is it equivalent to [nr]? How is nr \in \mathbb{Z} ?
    Take a rational number, say \frac {p} {q}. Both p\;\&\;q are integers and we can assume q > 0.
    Now {q\left( {\frac{p}{q}} \right) \in \mathbb{Z}} hence q\left( {\frac{p}{q}} \right) \in \left[ 0 \right]\,\& \,\underbrace {\left[ {\frac{p}{q}} \right] + \left[ {\frac{p}{q}} \right] +  \cdots  + \left[ {\frac{p}{q}} \right]}_q = \left[ 0 \right].
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    Member Jason Bourne's Avatar
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    okay thats alot clearer now thanks. What about part (e) , do you think that [x][y]=[xy] would be well defined?
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    Quote Originally Posted by Jason Bourne View Post
    do you think that [x][y]=[xy] would be well defined?
    If it was well-defined then [\sqrt{2}]\cdot [\sqrt{2}] = [\sqrt{2}+1]\cdot [\sqrt{2}].
    But, [\sqrt{2}]\cdot [\sqrt{2}] = [2] = \{ 2 + x |x\in \mathbb{Z} \} = \mathbb{Z}.
    And, [\sqrt{2}+1]\cdot [\sqrt{2}] = [2 + \sqrt{2}] = \{ 2 + x + \sqrt{2} | x\in \mathbb{Z} \}= \sqrt{2} + \mathbb{Z}
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    Member Jason Bourne's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    If it was well-defined then [\sqrt{2}]\cdot [\sqrt{2}] = [\sqrt{2}+1]\cdot [\sqrt{2}].
    But, [\sqrt{2}]\cdot [\sqrt{2}] = [2] = \{ 2 + x |x\in \mathbb{Z} \} = \mathbb{Z}.
    And, [\sqrt{2}+1]\cdot [\sqrt{2}] = [2 + \sqrt{2}] = \{ 2 + x + \sqrt{2} | x\in \mathbb{Z} \}= \sqrt{2} + \mathbb{Z}
    Why does [\sqrt{2}]\cdot [\sqrt{2}] = [\sqrt{2}+1]\cdot [\sqrt{2}]<br />
make it well defined?
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  10. #10
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    Frankly, I did not try to answer because I feared your confusion on this point.
    Do you agree that \left[ 1 \right] = \left[ 0 \right] ?
    If the operation were well defined should \left[ 1 \right]\left[ {\sqrt 6 } \right] = \left[ 0 \right]\left[ {\sqrt 6 } \right]?
    Does it?
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  11. #11
    Member Jason Bourne's Avatar
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    Do you agree that \left[ 1 \right] = \left[ 0 \right] ?
    yes

    If the operation were well defined should \left[ 1 \right]\left[ {\sqrt 6 } \right] = \left[ 0 \right]\left[ {\sqrt 6 } \right]?
    yes

    Does it?
    no
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