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Math Help - equality condition that need reasoning

  1. #1
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    equality condition that need reasoning

    I have been following Krizalid's lesson here. http://www.mathhelpforum.com/math-he...es-132202.html

    I do not understand this part.

    equality condition that need reasoning-untitled.jpg

    Why is x-2\leq0
    or why can't x-2>0 too?

    By substituting x=3 and 4, I know it doesn't fit the inequality.

    But can anyone explain using reasoning?
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  2. #2
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    Quote Originally Posted by stupidguy View Post
    I have been following Krizalid's lesson here. http://www.mathhelpforum.com/math-he...es-132202.html

    I do not understand this part.

    Why is x-2\leq0
    or why can't x-2>0 too?

    By substituting x=3 and 4, I know it doesn't fit the inequality.

    But can anyone explain using reasoning?
    To anyone else wanting to comment on this thread, it helps to view Krizalid's .pdf here, and see the context on page 4.

    LHS = left hand side, RHS = right hand side, used below in reference to the inequality you're trying to solve.

    So Krizalid isn't placing a restriction x-2\le0 but rather is putting this forward as a candidate solution set. Since the radicand is nonnegative for this entire interval, and because furthermore the RHS is nonpositive for this interval, thus the LHS is greater than or equal to the RHS and the inequality holds. Thus (-\infty, 2] is part of the solution set.

    You said x=3 doesn't satisfy the inequality but actually it does.. see the steps that occur after both sides are squared.

    Does this answer your question?
    Last edited by undefined; September 12th 2010 at 04:49 AM. Reason: fixed wording
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    Quote Originally Posted by undefined View Post
    To anyone else wanting to comment on this thread, it helps to view Krizalid's .pdf here, and see the context on page 4.

    LHS = left hand side, RHS = right hand side, used below in reference to the inequality you're trying to solve.

    So Krizalid isn't placing a restriction x-2\le0 but rather is putting this forward as a candidate solution set. Since the radicand is defined for this entire interval, and because the LHS is therefore nonnegative and the RHS nonpositive for this interval, the LHS is greater than or equal to the RHS and the inequality holds. Thus (-\infty, 2] is part of the solution set.

    You said x=3 doesn't satisfy the inequality but actually it does.. see the steps that occur after both sides are squared.

    Does this answer your question?
    yes, 3 can. I forgot that LHS can be equal to RHS. My bad.

    Why cant x-2>0 be another candidate solution set?
    Last edited by stupidguy; September 12th 2010 at 04:18 AM.
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    Also, I do not understand what is ' [ ', ' ] ', ' ( ', ' ) '.

    Why cant he just adopt structures like '3>x>2',  x-2\leq0?
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    Quote Originally Posted by stupidguy View Post
    yes, 3 can. I forgot that LHS can be equal to RHS. My bad.

    Why cant x-2>0 be another candidate solution set?
     x-2\ge0 was used to get the 2 in [2,13/4]. We know the inequality must fail when x > 13/4 because then the radicand will be negative.
    Quote Originally Posted by stupidguy View Post
    Also, I do not understand what is ' [ ', ' ] ', ' ( ', ' ) '.

    Why cant he just adopt structures like '3>x>2',  x-2\leq0?
    http://zonalandeducation.com/mmts/mi...lNotation.html

    Interval notation is useful for writing concisely.
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    Quote Originally Posted by undefined View Post
     x-2\ge0 was used to get the 2 in [2,13/4]. We know the inequality must fail when x > 13/4 because then the radicand ceases to be defined.



    Interval Notation

    Interval notation is useful for writing concisely.
    Why did he "]-infinity,-3]" ?
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    Quote Originally Posted by stupidguy View Post
    Why did he "]-infinity,-3]" ?
    The ISO notation

    Interval (mathematics) - Wikipedia, the free encyclopedia

    A bit confusing to have so much notation, I know.
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    how did he find [-3,3]\cap[2,\frac{13}{4}]=[2,3]?

    I am using number line....but he din show....what other special methods are available?
    Last edited by stupidguy; September 12th 2010 at 06:31 AM.
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    Quote Originally Posted by undefined View Post
    The ISO notation

    Interval (mathematics) - Wikipedia, the free encyclopedia

    A bit confusing to have so much notation, I know.
    What is the difference between '(2,3)' and ']2,3[' ?
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    ]-\infty, 2] and (-\infty, 2] can be used to indicate that " -\infty is not in the set.

    how did he find [3, 3]\cap[2,\frac{13}{4}]= [2, 3]?
    He didn't- you've miswritten the first part. He found that [-\infty, 3]\cap [2, \frac{13}{4}]= [2, 3]. That is because 13/4= 3\frac{1}{4}, slightly larger than three. Because " \cap", the intersection of two sets or those point in both, the intersection does not contain any numbers less than 2, since they are not in the interval [2, \frac{13}{4}] and does not contain any numbers larger than 3 since they are not in ]-\infty, 3].
    Last edited by HallsofIvy; September 12th 2010 at 12:38 PM.
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  11. #11
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    Quote Originally Posted by stupidguy View Post
    how did he find [3,3]\frown[2,\frac{13}{4}]=[2,3]?

    I am using number line....but he din show....what other special methods are available?
    You made a typo and missed the negative sign in [-3, 3] ... basically if you think about it enough you'll realize [a,b] \cap [c,d] = [max(a,c), min(b,d)] and if max(a,c) > min(b,d) then you get the empty set.

    A LaTeX tip: use \cap instead of \frown

    Quote Originally Posted by stupidguy View Post
    What is the difference between '(2,3)' and ']2,3[' ?
    These are the same thing, just different notation.
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    ]-\infty, 2] and (-\infty, 2] can be used to indicate that " -\infty is not in the set.

    how did he find [3, 3]\cap[2,\frac{13}{4}]= [2, 3]?
    He didn't- you've miswritten the first part. He found that [-\infty, 3]\cap [2, \frac{13}{4}]= [2, 3]. That is because 13/4= 3\frac{1}{4}, slightly larger than three. Because " \cap", the intersection of two sets or those point in both, the intersection does not contain any numbers less than 2, since they are not in the interval [2, \frac{13}{4}] and does not contain any numbers larger than 3 since they are not in ]-\infty, 3].
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    my bad for not adding a minus to 3. nvrtheless, i understand, he did the intersection in his head! I usually draw a number line.

    now that i understand the basics.

    can case 1 be x-2<0 instead?
    then, the solution be (-\infty,2)

    As for the condition for case 2, [2,\frac{13}{4}], I do not know how he get that, though i know it is related to eg 6.

    Someone pls elaborate.
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    basically if you think about it enough you'll realize [a,b] \cap [c,d] = [max(a,c), min(b,d)] and if max(a,c) > min(b,d) then you get the empty set.
    I dun understand wat u mean. If this is important, pls elaborate. thx.
    Last edited by stupidguy; September 12th 2010 at 09:32 AM.
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    Quote Originally Posted by stupidguy View Post
    my bad for not adding a minus to 3. nvrtheless, i understand, he did the intersection in his head! I usually draw a number line.

    now that i understand the basics.

    can case 1 be x-2<0 instead?
    then, the solution be (-\infty,2)
    Yes it's possible to divide into cases x-2<0 and x-2\ge0. Notice that this completely covers the real number line; using x-2\le0 and x-2\ge0 counts \,x=2 twice (which isn't a problem since we're taking a union anyway).

    Quote Originally Posted by stupidguy View Post
    As for the condition for case 2, [2,\frac{13}{4}], I do not know how he get that, though i know it is related to eg 6.

    Someone pls elaborate.
    What happens when you solve 13x-4\ge0?

    Quote Originally Posted by stupidguy View Post
    I dun understand wat u mean. If this is important, pls elaborate. thx.
    x\in[a,b]\cap[c,d] means

    x\ge a\ \text{AND}\ x\le b\ \text{AND}\ x\ge c\ \text{AND}\ x\le d.

    Consider just the part x\ge a\ \text{AND}\ x\ge c. Can you see why this is equivalent to x\ge\max(a,c)?
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