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Math Help - Use the triangle inequality to show...

  1. #1
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    Use the triangle inequality to show...

    Use the triangle inequality to show that if a and be are in R, with a not equal to b, then there exists an open interval U centered at a and V centered at b, both of radius epsilon= 1/2 abs[a-b] , with U/\V= 0 (intersection)

    I really don't know what I wanna do here. I'm imagining a line centered at A and another at b with radius epsilon something like this =>

    ----- (-e)-------a-------(e)------

    -----(-e)--------b-------(e)-----
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by mtlchris View Post
    Use the triangle inequality to show that if a and be are in R, with a not equal to b, then there exists an open interval U centered at a and V centered at b, both of radius epsilon= 1/2 abs[a-b] , with U/\V= 0 (intersection)

    I really don't know what I wanna do here. I'm imagining a line centered at A and another at b with radius epsilon something like this =>

    ----- (-e)-------a-------(e)------

    -----(-e)--------b-------(e)-----
    We may assume wlog that a<b. Thus the open interval around a is U=\left(a-\frac{1}{2}(b-a),a+\frac{1}{2}(b-a)\right)=\left(\frac{1}{2}(3a-b),\frac{1}{2}(a+b)\right)

    The open interval around b is V=\left(b-\frac{1}{2}(b-a),b+\frac{1}{2}(b-a)\right)=\left(\frac{1}{2}(a+b),\frac{1}{2}(-a+3b)\right)

    From this it's clear that U and V are disjoint, so U\cap V=\emptyset.
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  3. #3
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    what about showing the interval for both U and V to be of radius epsilon= 1/2*absolute value[a-b]?
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by mtlchris View Post
    what about showing the interval for both U and V to be of radius epsilon= 1/2*absolute value[a-b]?
    It doesn't have to be; take any c\leq\frac{1}{2}, and taking the radius to be c|a-b|, will guarantee U\cap V=\emptyset.
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  5. #5
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    I don't really follow...how come we can assume that a < b ?
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  6. #6
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by mtlchris View Post
    I don't really follow...how come we can assume that a < b ?
    One of them has to be greater than the other. If it's not a, then it's b. The reason we can assume wlog is that if we assumed b<a, we would still get the same result.
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  7. #7
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    and we make this assumption so we can relate U and V?
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  8. #8
    Super Member redsoxfan325's Avatar
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    I made the assumption so I didn't have to deal with the absolute value sign.
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  9. #9
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    ok I follow this, but if I need to use the triangle property how do i go about using it?
    A friend told me:
    let u= (a-e, a+e) = absolute value(c-a) < e for c in R
    let v= (b-e, b+e) = absolute value(d-a) < e for d in R

    then absloutevalue(x-a) < e = 1/2absloute(a-b)
    absloute value(x-b) < e = 1/2absloute(a-b)

    is this correct? and if so....how does he go from abs(c-a) < e to
    abs(x-a) < e
    and then to = 1/2 abs(a-b)
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  10. #10
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by mtlchris View Post
    ok I follow this, but if I need to use the triangle property how do i go about using it?
    A friend told me:
    let u= (a-e, a+e) = absolute value(c-a) < e for c in R
    let v= (b-e, b+e) = absolute value(d-a) < e for d in R
    U is a set; |c-a| is a number. How can those be equal?
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  11. #11
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    ok i should have written
    { c in R : abs(c-x) < e }
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  12. #12
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by mtlchris View Post
    ok I follow this, but if I need to use the triangle property how do i go about using it?
    A friend told me:
    let u= (a-e, a+e) = absolute value(c-a) < e for c in R
    let v= (b-e, b+e) = absolute value(d-a) < e for d in R

    then absloutevalue(x-a) < e = 1/2absloute(a-b)
    absloute value(x-b) < e = 1/2absloute(a-b)

    is this correct? and if so....how does he go from abs(c-a) < e to
    abs(x-a) < e
    and then to = 1/2 abs(a-b)
    Your friend introduced more variables than are necessary (namely c and d). If U=\{x\in\mathbb{R}:|a-x|<\epsilon\} and V=\{x\in\mathbb{R}:|x-b|<\epsilon\}, then assume for a contradiction that U\cap V\neq\emptyset; that is, \exists x such that x\in U and x\in V. Then |a-x|<\epsilon and |x-b|<\epsilon.

    Then by the triangle inequality |a-b|\leq|a-x|+|x-b|<\epsilon+\epsilon=2\epsilon=|a-b|.

    So we have |a-b|<|a-b|, which is a contradiction. So U\cap V=\emptyset.
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