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Math Help - Proving two sets equal

  1. #1
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    Unhappy Proving two sets equal

    Here's the problem:

    Let L={(x,y) | 2x+3y=6} and L'={a(3,0) + b(0,2) | a+b=1}
    Prove that L=L'


    What I've done:
    I've picked a few examples, so I believe that this is true. I know that I have to prove that L<L' and L'<L to conclude that L=L'.
    I just can't seem to get how to go about doing it.

    I think i did the second part:
    if (m,n) is in L', then (m,n)=a(3,0)+b(0,2) where a+b=1
    so, m=3a and n=2b
    but a+b=1, so b=1-a
    so (m,n)=(3a,2b)=(3a,2(1-a))
    so then I tried to prove that this point is in L:
    2x+3y=2(3a)+3(2(1-a))=6a+6-6a=6
    I think this proves the second half, L'=L. Does it?

    For the first part, I have:
    (p,q) is in L, so 2p+3q=6
    and I didn't know what to do so I tried to work backwards from the conclusion to get an idea
    (p,q)=(p,0)+(0+q)=(p/3)(3,0)+(q/2)(0,2)
    does p/3 + q/2 =1? but I can't figure out how to get any further.

    Any help or hints or ideas or anything would be greatly appreciated!
    Last edited by sfitz; January 25th 2008 at 12:19 AM. Reason: typo
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  2. #2
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    Well, the first part is to pick a point in L and show that it is in L', and the second part is to pick a point in L' and show that it is in L. That's how I've always learned to prove that one set is equal to another. Does that make sense?
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  3. #3
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    Quote Originally Posted by sfitz View Post
    Well, the first part is to pick a point in L and show that it is in L', and the second part is to pick a point in L' and show that it is in L. That's how I've always learned to prove that one set is equal to another.
    That is the correct method.
    \begin{array}{l}<br />
 2p + 3q = 6 \\ <br />
 \frac{p}{3} + \frac{q}{2} = 1 \\ <br />
 \frac{p}{3}\left( {3,0} \right) + \frac{q}{2}\left( {0,2} \right) = \left( {p,q} \right) \\ <br />
 \end{array}
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  4. #4
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    Quote Originally Posted by Plato View Post
    That is the correct method.
    \begin{array}{l}<br />
 2p + 3q = 6 \\ <br />
 \frac{p}{3} + \frac{q}{2} = 1 \\ <br />
 \frac{p}{3}\left( {3,0} \right) + \frac{q}{2}\left( {0,2} \right) = \left( {p,q} \right) \\ <br />
 \end{array}
    I'm not sure how to go from the first line to the second two. What am I missing?
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  5. #5
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    Quote Originally Posted by sfitz View Post
    I'm not sure how to go from the first line to the second two. What am I missing?
    You are missing division by 6.
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