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Math Help - Prove that the set of points on any two line segments are equivalent.

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    Prove that the set of points on any two line segments are equivalent.



    I'm guessing this is what the question is asking?

    I think its pretty show that there is an equivalence relation here. And then the implication would be that since you can take a line segment inside another line segment and come up with an equivalence relation, the set of points on any given line segment is infinite.

    Right?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by davismj View Post


    I'm guessing this is what the question is asking?

    I think its pretty show that there is an equivalence relation here. And then the implication would be that since you can take a line segment inside another line segment and come up with an equivalence relation, the set of points on any given line segment is infinite.

    Right?
    Let \ell_1,\ell_2 be the two lines. Translate \ell_1 over however many units so that it is not above \ell_2. Consider the projections \pi:\ell_1,\ell_2\mapsto(a,b),(c,d). This is clearly a bijection. Thus, we must merely show that given two segments (a,b),(c,d) that (a,b)\simeq (c,d). Conjecture there exists a linear bijection between them. Then f(x)=mx+k. So then we need that {f(a)=ma+k=d}\brace{f(b)=mb+k=d}. Solving this system of equations will give you a bijective function between the two which is actually a homeomorphism (that last part isn't necessary)

    And, I don't see how defining an equivalence relation helps.


    Wait....I read the question. But now I am having second thoughts about it's meaning. I thought you meant to show that any two lines has the same number of points("equivalent" in the category of sets). Is there somethign else?
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    Quote Originally Posted by Drexel28 View Post
    Let \ell_1,\ell_2 be the two lines. Translate \ell_1 over however many units so that it is not above \ell_2. Consider the projections \pi:\ell_1,\ell_2\mapsto\mathbb{R}. This is clearly a bijection. Thus, we must merely show that given two segments (a,b),(c,d) that (a,b)\simeq (c,d). Conjecture there exists a linear bijection between them. Then f(x)=mx+k. So then we need that {f(a)=ma+k=d}\brace{f(b)=mb+k=d}. Solving this system of equations will give you a bijective function between the two which is actually a homeomorphism (that last part isn't necessary)

    And, I don't see how defining an equivalence relation helps.


    Wait....I read the question. But now I am having second thoughts about it's meaning. I thought you meant to show that any two lines has the same number of lines ("equivalent" in the category of sets). Is there somethign else?
    No, that's it for this question. To my understanding, the problem is to show that for any two arbitrary lines p and q in a plane, there is a bijection between P = {x : x is on p} and Q = {y : y is on q}.

    That's why I parametrized each line. Each line is a bijection, from [0,1] to P and [0,1] to Q. Therefore, we can show that P ~ [0,1] ~ Q so P ~ Q. What do you think?

    "First, we recall that set A is equivalent to set B if and only if there is a one-to-one mapping of A onto B."
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by davismj View Post
    No, that's it for this question. To my understanding, the problem is to show that for any two arbitrary lines p and q in a plane, there is a bijection between P = {x : x is on p} and Q = {y : y is on q}.

    That's why I parametrized each line. Each line is a bijection, from [0,1] to P and [0,1] to Q. Therefore, we can show that P ~ [0,1] ~ Q so P ~ Q. What do you think?

    "First, we recall that set A is equivalent to set B if and only if there is a one-to-one mapping of A onto B."
    Oh maybe...write out the bijection and I'll check it.

    (what I said works then btw)
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    Quote Originally Posted by Drexel28 View Post
    Oh maybe...write out the bijection and I'll check it.

    (what I said works then btw)
    F: f(p) = q_0 + \frac{q_0 - q_1}{p_0 - p_1}(p - p_0) = q

    and

    G: g(q) = p_0 + \frac{p_0 - p_1}{q_0 - q_1}(q - q_0) = p
    Last edited by davismj; March 4th 2010 at 05:20 PM.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by davismj View Post
    F: f(p) = q_0 + \frac{q_0 - q_1}{p_0 - p_1}(p - p_0) = q

    and

    G: f(q) = p_0 + \frac{p_0 - p_1}{q_0 - q_1}(q - q_0) = p
    That's a mapping between...
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    Quote Originally Posted by Drexel28 View Post
    That's a mapping between...
    f(g(q)) = q

    g(f(p)) = p

    These statements imply that these functions are bijective, since both f(p) and g(q) have inverses.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by davismj View Post
    f(g(q)) = q

    g(f(p)) = p

    These statements imply that these functions are bijective, since both f(p) and g(q) have inverses.
    Yes. But to have a mapping you need to have a rule of assignment, a domain, and a codmain. You've only given the rules of assignment. But I get what you're saying
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