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Inequalities: I am apparently bad with them

  1. Sep 3, 2011 #1
    Hello folks :smile:

    I always thought that inequalities could be treated exactly like equations but somehow I seem to be loosing information or something. For example, if I wish to find all values of x for which the following is true:

    1/x + 1/(1-x) > 0

    I would 'solve' it as follows

    1 + x/(1-x) > 0

    x/(1-x) > -1

    x > x-1

    And here is where I fail. Solving any further results in the obvious: 0>-1.

    What is it that I am doing wrong that does not allow me to 'solve' for all x that satisfy the inequality. Am I wrong in thinking that I should be able to.
     
  2. jcsd
  3. Sep 3, 2011 #2
    Inequalities could be treated exactly like equations unless you are multiplying by a negative number, e.g. if we multiply 2>1 by -1, we have to change the "greater than" sign to a "less than" and get -2<-1. When you multiplied your equation by x and (x-1), you have to consider the cases when they are negative.
     
  4. Sep 3, 2011 #3
    Here's how I would attempt this.

    Combine the terms.

    1/x + 1/(1-x) > 0
    1/(x(1-x)) > 0

    Solve for the critical points ( they're 0 and 1).

    Now make a number line and test at each of the intervals (in this case, there are three) to see if they're greater than 0.

    Look at the attachment.:smile: (little barbaric)

    Note: closed interval at 0 and 1 due to it being undefined.
     

    Attached Files:

  5. Sep 4, 2011 #4
    Interesting. I missed that. However I am still having trouble seeing why the x 'falls out' upon solving the inequality in this case, but not in other cases. For example if I want to know all values for which x-1>3 I solve the inequality to get x>4. But in the case above, the x vanishes.

    Interesting. I am however having trouble seeing why there should be a connection between the critical points, which seem to be the points at which the denominator is zero, and the values of x which satisfy the inequality?

    Thanks for the responses so far :smile:
     
  6. Sep 4, 2011 #5
    If the x falls out of the inequality and you reach an inequality that holds, then the original inequality always holds! Trivial example:

    x>=x simplifies to 0>=0, which always holds.
     
  7. Sep 4, 2011 #6

    Char. Limit

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    Critical points are points where the denominator OR the numerator is zero. Basically, for continuous functions, for an interval (a,b) between two adjacent critical points of a function f(x), for any x between a and b, f(x) will have the same sign.

    Thus, if you find all the critical points of a function, and test at some point for every interval between them, you can find where a function is greater than or less than zero.
     
  8. Sep 4, 2011 #7
    Hi dalcde :smile:

    I always thought that too! However, my initial post proves otherwise.

    Hello Char. Limit :smile:

    This makes sense intuitively. However, the problem comes from the 1st chapter of Spivak's Calculus. From what I hear, I am supposed to be able to 'solve' all of these problems using only the 12 properties of numbers and the definitions introduced in this chapter. I am trying to figure out how I can logically arrive at the correct answer to this problem by using only the given information in the chapter.

    Since this is self study, I don't mind taking the extra tine to work through this, step by step. I am just not so sure how Spivak is thinking we should approach it.
     
  9. Sep 4, 2011 #8

    Char. Limit

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    Well, first thing I would do is multiply the first term by (1-x)/(1-x) (multiplying a single term by 1 is acceptable). Now, this wouldn't work if we were working in R, but due to the restrictions on the problem, we're working in R\{0,1}*, so this is acceptable. Multiply the second term by x/x, of course.

    That gets us (1-x)/(x(1-x)) + x/(x(1-x)) > 0. Combining fractions gives us 1/(x(1-x))>0. From here, multiply both sides by x, and here you'll need to split the problem into two parts. The first part, for all x>0, will simply by 1/(1-x)>0. The second part, for all x<0, will be 1/(1-x)<0. Do you see why that is?
     
  10. Sep 4, 2011 #9
    To understand why the result of your solution is not the obvious 0>-1 but the following:

    {[itex]x>0[/itex] AND [itex](1-x)[/itex]}[itex]\Longrightarrow[/itex]{[itex]1+\frac{x}{1-x}>0[/itex] AND [itex]\frac{1-x}{x}+1>0[/itex]}.

    You need to get into the structure of your solution in other words be able to write formal proofs for inequalities .

    Coming now to how one would solve your inequality the usual way is the following:

    (First of all the inequality holds for [itex] x\neq 0 ,x\neq 1[/itex])


    1) Work 1st in the left hand side ,to avoid complications like the one you came upon.

    Hence the equation becomes: [itex]\frac{1}{x(1-x)}>0[/itex].

    2) Now the solution is much simpler and using the theorems in inequalities : 1/Y => Y>0 and AB>0 => (A>0 & B>0) OR ( A<0& B<0) we have:

    [itex]\frac{1}{x(1-x)}>0\Longrightarrow x(1-x)>0\Longrightarrow [(x>0\wedge (1-x)>0)\vee(x<0\wedge (1-x)<0)][/itex][itex]\Longrightarrow (0<x<1)\vee (x<0\wedge1<x)\Longrightarrow 0<x<1[/itex]

    Perhaps you will not undestand the last step
     
  11. Sep 4, 2011 #10
    Dear ,Char

    Can you please finish your way of solving the inequality ,because i tried , and i could not.
     
  12. Sep 4, 2011 #11

    Evo

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    Your not asking him to solve the problem, are you?
     
  13. Sep 4, 2011 #12
    Definetely not.

    I said that i could not finish it the way he started it . I asked him ,if this possible of course , to help me in that respect.
     
  14. Sep 4, 2011 #13
    Hi again Char. Limit :smile:

    The part in bold I don't quite understand. What do you mean that it wouldn't work in R ? Why can't we multiply by (1-x)/(1-x) ? Oh wait, I see what you mean. It does not hold for x = 1, so it would not hold for all R.

    As a side note, what does the notation R\{0,1}* mean exactly? I have interpreted it to mean all of R excluding 1 and 0. Is this formal notation? And where is it typically first encountered (what class, set theory?)? I am trying to fill in a lot of gaps in my math background, so I started with Spivak and will fill in additional holes as I encounter them.

    I will need to work at this a little more to see if I can work through this last part.

    Thanks for your time (and patience :wink: )
     
  15. Sep 4, 2011 #14

    Char. Limit

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    Woops. I meant to have a footnote. (That's why the asterisk is there, it's not part of the notation). However, your interpretation was correct. R\{0,1} means the set of all real numbers, except for 0 and 1.

    And as to the first paragraph, that's right. :)

    Keep pushing through. You just need to remember to consider both positive 1-x (or x) or negative 1-x (or x) in your problem.
     
  16. Sep 10, 2011 #15

    HallsofIvy

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    You are mistaken. If a< b and c is negative then ac> bc. If c is 0, of course, ac= bc. As long as you are multiplying or dividing by constants, that is not a problem. But if you multiply or divide by something involving the "unknown", x, you do not know whether it is positive or negative.

    If f is any continuous function and and takes on values f(a) and f(b) then it must take on all values between f(a) and f(b). Specifically, a function, f(x), can change from "< 0" to "> 0" only at points where f(x)= 0 or where f is NOT continuous. In particular, a rational function is equal to 0 only where the numerator is 0 and is discontinuous only where the denominator is 0.
     
  17. Sep 14, 2011 #16
    Do you approve of the solution that Char.Limit suggested??
     
  18. Sep 14, 2011 #17

    HallsofIvy

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    x/(1- x)> -1 does NOT imply x> x- 1 because if b<0 a/b> c gives a< bc. Multiplying both sides of an inequality by a negative number reverses the direction of an inequality.

    And with problems like this, since you do not know what x is, you do not know whether the quantity you are multiplying by is positive or negative. The best thing to do is look at the cases.

    Since you want to multiply by 1- x, you have to consider:
    1) 1-x> 0 (which is, of course, the same as x< 1)
    Multiplying both sides of the inequality by 1- x is multiplying by a positive number so
    x> -1(1- x)= x- 1. Subtracting x from both sides give 0> -1 which is a true statement no matter what x is. That is, the inequality is satisfied for all x< 1.

    2) 1-x< 0 (which is, of course, the same as x> 1)
    Multiplying both sides of the inequality by 1- x is multplying by a negative number so the inquality reverses. x< -1(1-x)= x- 1 which gives 0< -1 which is a false statement no matter what x is.
     
  19. Sep 14, 2011 #18
    How about the values : -2,-3,-4,-5,-6.............. do they satisfy the inequality??
     
  20. Sep 14, 2011 #19

    symbolipoint

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    I tried the question with simplying and identifying critical points.

    From the given inequality 1/x + 1/(1-x) > 0,
    clearing fractions was done by multiplication by x(1-x). That with just a few simple steps gives
    2x-2x^2 > 0.

    Divide by 2 and factor to get x(1-x)>0

    Critical Points appear to be x at 0 and x at +1. This means we can test points in the intervals of x<0, 0<x<1, and x>0. Which intervals make the Original Inequality to be true?
     
  21. Sep 14, 2011 #20

    Mentallic

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    If you prefer working with polynomials than cases, then you might also prefer to solve it in this manner:

    [tex]\frac{1}{x}+\frac{1}{1-x}>0[/tex]

    Now to rid ourselves of the denominator in the first fraction, if we were to multiply through by x, then we aren't sure of the sign of x and we would have to take cases, but we are sure that x2 is always positive, so there would be no cases involved there.

    So let's multiply through by x2(1-x)2 which is always positive ([itex]x\neq 0,1[/itex]) so we end up with

    [tex]x(1-x)^2+x^2(1-x)>0[/tex]

    And factorizing that further will give you a simple parabola. All you need to do is draw it on paper or in your head to find the solution.
     
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