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Problems on Inequalities and Analysis

  1. Oct 15, 2009 #1
    1. The problem statement, all variables and given/known data

    I encountered a few problems while attending a problem solving seminar. Abstract mathematics and real analysis is not my forte and haven't really taken any courses in that regard. Thought maybe someone here could offer some help in better understanding about the following topics. Hope I am posting in the right section too.

    1) Let r > 0. Prove that: |x| < r <==> -r < x < r

    2) Solve & write the result in interval notation:
    |3 / (2x-1)| < 1

    3) Let S (not a null set) be a bonded(below & above) subset of R. Denote a = inf S, b = sup S. Is it true that "for every 'n' (element of N) there exists an 'x' (element of S) such that:
    a) a > x - (1/n)
    b) b < x + (1/n)


    2. Relevant equations

    I have no idea whether any equations could be used to solve this kind of a proof based problem.

    3. The attempt at a solution

    For the first question, I am quite not sure, whether it is to prove P implies Q or Q implies P or both. This is the first time, with that "<=>" operator.

    Second question, I don't know how the problem solving approach changes when doing in the perspective of real analysis.
    -1 < 3 / (2x-1) < 1
    -2x+1 < 3 < 2x-1
    x> -1/4 and in the interval notation: (-1/4, infinity)
    Hope that's the way. Please enlighten me if it is to be attempted in some other way.

    Third question, I dont have the slightest idea about it.
     
  2. jcsd
  3. Oct 15, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Think of "if |x|< r then -r< x< r" as solving an inequality rather than a "proof".
    The best way to solve inequalities is to first solve the associated equality.
    What are the values of x that satisfy |x|= r? There are two roots that divide the line into three intervals. |x|< r will be uniformly true or false in each interval. Check one value of x in each interval to see whether the inequality is true or false in that interval.

    Now, the other way. Break -r< x< r into two cases.
    a) Suppose [itex]0\le x< r. Then x is non-negative so |x|= x. That makes it pretty easy to show that |x|< r, doesn't it?

    b) Suppose -r< x< 0. Then x is negative so |x|= -1. What do you get by multiplying each part of -r< x< 0 by -1?

    What are the roots of |3/(2x-1)|= 1? You should consider the cases 2x-1> 0 and 2x-1< 0. After you have found the roots of that equation, do the same as in (1).

    (a) a> x- (1/n) is the same as a+(1/n)> x. Suppose there were NO x satisfying that. What would that tell you about a+(1/n)? Remember that a is the greatest lower bound.

    (b) Pretty much the same thing. b< x+ (1/n) is the same as x> b- (1/n). If there were NO x satisying that, what would be true of b- (1/n)?


     
  4. Oct 15, 2009 #3
    Thanks a lot for that...

    Forgive me for asking, but I didn't quite understand this part:
    Thank you once again...
     
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