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Proving A is a proper subset of B

  1. Jul 17, 2012 #1
    1. The problem statement,

    A = {(x, y) belongsto R^2 : (x−2)^2 + (y + 1)^2 < 9},
    B = {(x, y)belongs to R^2 : y − 2x < 2}.
    Show that A is a proper subset of B.


    3. The attempt at a solution
    Now I know that (x−2)^2 + (y + 1)^2 is an equation of a circle with centre translated to (2to the right on x axis and 1 down on the y axis. since its <9 its the interior area of the circle excluding actual circumference of the circle.

    and graph of y-2x<2 has intercept y=2 and x=-1 and slopes from first qudrant to the forth and all the area above is in the set ( I am not sure)- excluding the actual line itself. but I have no idea how to prove this any help would be appriciated.
     
    Last edited: Jul 17, 2012
  2. jcsd
  3. Jul 17, 2012 #2

    Curious3141

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    Should that be area above, or area below the line? Think about it: y is LESS than something.

    It's always good to start with definitions. For example, what's a proper subset?
     
  4. Jul 17, 2012 #3
    So it's the area below. I thought something wasn't adding up. But as far as writing is concern. Do you have any formal way of writing this!! Because I don't see how else would you write it out.
     
  5. Jul 17, 2012 #4

    ehild

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    Make a figure to see the problem more clearly.

    Find if he centre of the circle belongs to A. If yes, find if the distance of any point of the line from the centre is greater then the radius of the circle. If this is true the circle is a subset of A.

    ehild

    I am too late ...Curious beat me.:tongue2:
     
  6. Jul 17, 2012 #5
    Ehild, you lost me completely. This topic is fairly new to me so I will be happy to have bit more explaination. If you could expand or give an example that be great.
     
  7. Jul 17, 2012 #6

    ehild

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    First answer to Curious question first: What is a subset? What is a proper subset?
    Is B a subset of A in the drawing? Is it a proper subset? Is C a subset of A?

    ehild
     

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  8. Jul 17, 2012 #7
    Ehild, you lost me completely. This topic is fairly new to me so I will be happy to have bit more explaination. If you could expand or give an example that be great.
     
  9. Jul 17, 2012 #8
    C is not a proper subset of a as part of C lies outside A however B is a proper subset as all of b lies within A
     
  10. Jul 17, 2012 #9

    ehild

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    How your problem looks like? What part of C lies outside A? How can you specify it?
    How can you formulate that all points b lie within A?

    ehild
     
  11. Jul 17, 2012 #10
    Centre of the circle is (2,-1)in this case so if you take the point x=2 & y=-1 plug it in the y-2x<2 ten it is true.

    Any good?
     
  12. Jul 17, 2012 #11

    ehild

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    Yes, the centre of the circle belongs to A. B means those points which are closer to point (2,-1) than r=3. To be not element of A the distance of a point from (2,1) should be greater than the shortest distance of the border line of A. What is the shortest distance of the line from the centre of the circle?

    ehild
     
  13. Jul 17, 2012 #12

    Curious3141

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    I would just make an accurate sketch of the areas corresponding to each inequality. It is important to prove that the line lies completely above the circle and there's no intersection. This is easy to do by solving the equations of the line and circle simultaneously and proving that the resulting quadratic has a negative discriminant.

    At this point, it is obvious that:

    a) the circle lies completely below the line (hence every element of A is an element of B).

    b) there are points (an infinite number of them, technically an uncountable number of them) that lie outside the circle but below the line (hence there are elements of B that are not elements of A).

    Observation a) establishes that A is a subset of B.

    Observation b) establishes that A is a proper subset of B.

    You can express this in formal set theoretic notation to complete the proof.
     
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