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Inequality Problem

  1. Jul 3, 2004 #1
    Hi all,

    I'm a math major at Berkeley with a 3.3 GPA average. I've taken calculus, linear and abstract algebra, probability and real analysis. The other day I was applying for a job as a pre-Calculus teaching assistant and my employer gave me this problem with a 2 minute time limit.

    Simplify this expression:

    [tex]\frac{\frac{1}{x}+1}{\frac{1}{x}-1}<2[/tex]


    I couldn't get it.
     
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  3. Jul 3, 2004 #2

    matt grime

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    The LHS is the equal to (1+x)/(1-x) so x is not equal to 1, for x>1, or x<1, clearing denominators correctly will give a simple solution, when such is possbible.
     
  4. Jul 3, 2004 #3

    HallsofIvy

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    First, it's not, strictly speaking, an "expression", it's an inequality. The left hand side,
    [itex]\frac{\frac{1}{x}+1}{\frac{1}{x}-1}[/itex] is an expression. As matt grime said, the obvious thing to do is multiply both numerator and denominator by x, resulting in [itex]\frac{1+x}{1-x}[/itex]. If the problem was really to "simplify this expression", that or [itex]\frac{1+x}{1-x}< 2[/itex] is the best you can do (and should take about 10 seconds).

    If the problem were, instead, to solve the inequality, then:
    a) if 1-x>= 0 (that is, x<= 1), this is the same as 1+x< 2(1-x)= 2- 2x so 3x< 1 or x< 1/3. Since any number < 1/3 IS < 1, one answer is x< 1/3.
    b) if 1-x< 0 (that is, x> 1), this is the same as 1+x> 2(1-x)= 2- 2x so 3x> 1 or x> 1/3.
    But that only works for x> 1 so another answer is x> 1. If x< 1/3 OR x> 1, the inequality is true.

    If you really are "a math major at Berkeley with a 3.3 GPA average" and could not do this problem even after taking a great deal of time, that's should be very embarassing for you (and Berkeley).
     
  5. Jul 4, 2004 #4
    Much thanks for the gracious words.

    Actually, I got the same solution as you did. But we're both wrong.

    You assumed the expressions [tex]\frac{\frac{1}{x}+1}{\frac{1}{x}-1}[/tex] and [tex]\frac{x+1}{x-1}[/tex] are equivalent. But they're not. The former is not defined when x=0 (which is part of your solution) while the later is.

    By the way, we all have our strong and weak points in math. I thought your derision was of poor taste, especially considering you are a "mentor." Please practice a bit more tact in your posts. Thank you.
     
    Last edited by a moderator: Jul 4, 2004
  6. Jul 4, 2004 #5
    Maybe I should have been more direct in my original posts. I might have given the impression that I had no idea what was going on, hence giving HallsOfIvy reason to doubt my claim that I was a Berkeley math student.

    All I want is to confirm that the set of values which satisfies the inequality [tex]\frac{\frac{1}{x}+1}{\frac{1}{x}-1}<2[/tex]

    is

    [tex](-\infty,0) \cup(0,1/3)\cup(1,\infty)[/tex]
     
  7. Jul 4, 2004 #6

    uart

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    Hey I got x<1/3 and x>1 as the regions as well. I would likely have also overlooked the problem at x=0 if I was trying to solve it quickly, so don't feel bad. :)

    Actually I'm hopeless at doing anything quickly. :)
     
  8. Jul 4, 2004 #7
    I may not be a math major from Berkley with a 3.3 gpa but as far as I know [tex]\frac{\frac{1}{x}+1}{\frac{1}{x}-1}[/tex] and [tex]\frac{x+1}{1-x}[/tex]
    are equivalent expressions.
     
    Last edited: Jul 4, 2004
  9. Jul 4, 2004 #8
    hey in Canada 3.3 isn't really good.
    F=0, D=1, C=2, ... , A-=7, A=8, A+=9.
    3.3 would be C+/B-. Maybe that's the system Berkeley uses.

    & given that Berkeley only requires someone to solve problems in a given length of time IS really embarrassing. What does that have to do with being a pre-calc TA?
     
  10. Jul 4, 2004 #9

    AKG

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    JonF, it is clearly explained in post 4 that the expressions are not equivalent, and why. Whenever you do something like divide top and bottom by x, you should always note the exception when x=0.

    For x = 0, you cannot make the second expression from the first, and in that case, the original expression is undefined. For x != 0, you can change first to second.
     
  11. Jul 4, 2004 #10

    AKG

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    The system used in almost all places makes a 3.3 something like 75-80 (correct me if I'm wrong). I think 3.7 is something like 85, 4.0 is 90+, so a 3.3. is not great, but it's not bad. I've never heard a system like yours before, and I'm in Canada (Toronto). I think your system might be unique to your university. Do you know of ther places that use that system? And are you sure you're talking about GPA (grade point average)?
     
  12. Jul 5, 2004 #11
    I wish I was in Berkeley.

    Looks right to me, although your wording of the problem (see first post) is confusing.

    I've been put on the spot several times before by people who have high expectations of me, and every time it happens I draw a blank. It's so wierd. A colleague of mine (who has a maths. Ph.d. from MIT by the way) also had a similar experience. Anyways, it happens.
     
  13. Jul 5, 2004 #12

    But you can always define [tex]\frac{\frac{1}{x}+1}{\frac{1}{x}-1} = 1[/tex] at 0 by means of continuity.
     
    Last edited: Jul 5, 2004
  14. Jul 5, 2004 #13

    AKG

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    First of all, that doesn't make any sense. Second of all, regardless of what you might mean by "by means of continuity," you can't define the expression at x=0, because that would mean that 1/0 would be defined, which it's not.
     
  15. Jul 5, 2004 #14
    Nonsense.
    It's customary to extend functions by means of continuity to points where it isn't defined.
    For instance, [tex] x \sin \frac{1}{x}} [/tex] is defined as equals to 0 at x=0 by continuity. That doesn't mean that I'm defining 1/x in the set of Real numbers, I'm defining the whole function.
     
  16. Jul 5, 2004 #15
    so, what's the final answer??

    i do think that the answer u got is right.
    [tex](-\infty,0) \cup(0,1/3)\cup(1,\infty)[/tex]
     
    Last edited by a moderator: Jul 5, 2004
  17. Jul 5, 2004 #16

    matt grime

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    AKG, there are such things as removable singularities.
     
  18. Jul 5, 2004 #17
    I think U of T is the only place I've heard of that uses the 4-point system. Every other school that I know of uses the 9-point one.
    http://web.uvic.ca/reco/legend/legend.html

    bah Berkeley is a dump. I want to go to a real math school, U of Waterloo. They've got over 5000 math students (1/4 of all at the school), and over 6000 computers. It's the world's largest educational centre for math & comp sci. They've even got a whole faculty for math rather than one dept in the science faculty or whatever.
     
  19. Jul 5, 2004 #18

    AKG

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    Really. I think the States use the 4-point system everywhere.
    I don't understand that concept. sinx/x is not defined at x, but we can "define" it to be 1 and still have continuity and differentiability. But then, isn't it a different function? I mean, aren't:

    f(x) = sinx/x (where the domain is implied to be the R\{0})

    and

    g(x) = sinx/x for x != 0, 1 for x = 0

    two different functions, with two different domains? I mean, by default, isn't the expression that this thread about undefined at x=0 (and 1)? Therefore, how does it make sense to say what the expression is less than or greater than at x=0? We can make sense of it with a similar expression, but we're not asked about a similar expression, we're asked about the given expression, are we not? Or is it that by default removable singularities are always removed (and this is just something they don't tell you till your 3rd or 4th calculus course)? :confused:
     
  20. Jul 5, 2004 #19
    A 3.3 GPA for a Berkeley math student is somewhat impressive (considering the average GPA for math is 2.7). My point was this was a tricky problem, even for a somewhat competent college math student such as myself. In fact, HallsofIvy made the same mistake and I think he's a college professor.
     
  21. Jul 5, 2004 #20
    The time factor is important. When a TA is asked a problem by a student, he can't afford to stand there thinking about it all day. He has to be quick.

    Heh, well the last time I checked the US News rankings, Berkeley's math department is third in the nation (only Stanford and MIT beat it). But of course, we can argue over the legitimacy of these rankings.

    As for the U of Waterloo, it's rather impressive that they manage to accomodate for so many math students but that doesn't necessarily mean it's a good school, does it?
     
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