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Homework Help: Basic real number proof

  1. Jul 23, 2012 #1
    1. The problem statement, all variables and given/known data

    This is an exercise from the book 'Mathematical Proofs: A transition to advanced mathematics'.

    Prove that if r is in ℝ and 0<r<1, then 1/r(1-r)≥4

    3. The attempt at a solution

    My guess is that it would be best to use a proof by contrapositive. So I start by assuming that 1/r(1-r)<4 then rearrange the equation to find that r≤0 and r≥1. If that approach is right then my poor algebra is the reason I can't complete the proof. Is this the right approach?
  2. jcsd
  3. Jul 23, 2012 #2
    If you mean [itex]1/(r(1-r)) \geq 4[/itex] (which is not what you wrote), then you can rearrange into a quadratic and then differentiate to find the minimum...

    I'm sure there must be easier ways, but this way is simple enough. :smile:

    EDIT: Just remembered this is PRE-calculus. :redface:
    Last edited: Jul 23, 2012
  4. Jul 23, 2012 #3
    Hi privyet
    the "right" approach, if there is any, will be certainly hinted at by the context if you are following a book, that is, you almost certainly have to follow the same line of reasoning that you have seen used during the course (the ones used in the current/previous chapter(s))..

    As for the proof (or better said, 'a' proof), you could see that -r²+r is an inverted parabola with a maximum at 1/2 being 1/4, but this is not the right approach if your book is trying to make you follow certain lines of reasoning.
    in the end, what matter is the proof itself and if it is valid, well, whatever works works, but when following a book, you are typically being taught a technique which you are very much welcome to apply so as to make yourself comfortable with a tool that will certainly come in handy later.

  5. Jul 23, 2012 #4
    As the others have pointed out, the approach you should use depends entirely on what skill this problem is supposed to develop. A purely algebraic (no calculus, no geometry) solution could be approached as follows:

    Since [itex]0<r<1[/itex], at no point is [itex]r(1-r)[/itex] equal to zero. This is important because it means that expression behaves properly if we move it around the inequality. Furthermore, you can easily check that [itex]r(1-r)[/itex] is strictly positive for [itex]0<r<1[/itex]. Thus, if you multiply both sides of the inequality by [itex]r(1-r)[/itex], it's still valid for r in this range (remember that multiplying both sides of an inequality by a negative number flips the direction of the inequality).

    Thus, the expression can be rewritten:

    If you use what you know about factoring expressions, you will be able to show this inequality (and hence the original inequality) holds. Based on how tidily it works out with this method (and the fact that this a pre-calculus forum), I strongly suspect this is the way the book intends you to do it.
  6. Jul 23, 2012 #5
    Thank you all for the advice.

    First of all, I did mean 1/(r(1−r))≥4. The exercise comes from the 4th chapter of the book, which focuses on applying direct proof and contrapositive proof methods to basic proofs of integers, real numbers and sets.

    So I guess the approach suggested by LastOneStanding would be considered a direct proof ie. Assume 0<r<1 and prove that 1/(r(1−r))≥4.

    I can see how to arrive at 4(r−r2)−1≤0 and if I factorise it I find that r=0.5, which if put into the original inequality gives a the result 1/(0.5(1-0.5))=4. This looks like the kind of result I need but I'm note sure how to put it in a proper proof form.
  7. Jul 23, 2012 #6
    You can't solve for r- it's an inequality, not an equation. You're almost there, though. You clearly factorized it correctly, based on the result you stated. Now, question: if x is a real number, when is the equation [itex]x^2≥0[/itex] valid? Now compare this to the factored inequality and draw the same conclusion. With the additional restriction that r must be between 0 and 1 (since you used this fact earlier), you should be able to finish it up.
  8. Jul 24, 2012 #7

    Ray Vickson

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    You need to show that r(1-r) ≤ 1/4 for 0 ≤ r ≤ 1. This would be easy if you were allowed to use Calculus, but if not, you need a direct argument. Try to complete the square, so write [itex] r(1-r) = a - (r-b)^2[/itex] for some a and b that you need to determine. Now finding the maximum value of [itex] a - (r-b)^2[/itex] on [itex] r \in [0,1][/itex] is easy.

  9. Jul 24, 2012 #8
    Thanks LastOneStanding, another rookie mistake on my part. I have to admit that I cheated with the factorisation (thanks WolframAlpha). I don't know if there's a specific method I should follow to factorize something like [itex]4(r−r2)−1[/itex], but I still find it very difficult to just guess at what the factorisation might be and go from there, except for the most obvious ones. Hopefully practice will help with that.

    Anyway, rearranging the equation to [itex]-(2r-1)^2≤0[/itex], I find that the inequality is valid for all [itex]r \in R[/itex]. Is this sufficient to say that I've proved that if [itex]0<r<1[/itex], then [itex]1/(r(1-r))≥4[/itex].

    My latest attempt (with the help of Ray Vickson's comment) goes like this:

    Assume [itex]r \in R[/itex] and [itex]0<r<1[/itex], then [itex]r(1-r)>0[/itex] and we can multiply the inequality by [itex]r(1-r)[/itex]. Therefore [itex]1/4≥r(1-r)[/itex]. Since [itex]r(1-r)=r-r^2[/itex] is a quadratic equation the maximum value can be found at [itex]r=-b/2a[/itex]. Putting the values of [itex]a[/itex] and [itex]b[/itex] into the equation gives [itex]r=1/2[/itex]. Putting this value into the original inequality [itex]1/(r(1-r))≥4[/itex] gives [itex]4≥4[/itex], which is true, and since [itex]1/(r(1-r))≥4[/itex] is true for all [itex]r<1/2[/itex] the statement is proved.

    I'm confident that the above attempt is very different from what the author hoped when he wrote the article. The proofs in this early stage of the book are generally of a few lines, occasionally more for proofs involving cases. Anyway I've learnt/remembered a lot from working on the question and the advice given here. Thanks.
  10. Jul 25, 2012 #9
    Correct, in that form the inequality is valid for all r. However, recall that to get to that stage, you had to use the restriction 0<r<1. Thus, you can state that the final result is valid for r in that range as required (i.e. yes, you can say you've proved what the question is asking). For r outside the range, the conclusion will not hold; even though the factorized inequality holds for all r, some of the steps needed to get there would be invalid. Hope that makes sense.

    As for how to do the factorization: you should review "factoring by decomposition".
  11. Jul 25, 2012 #10

    Ray Vickson

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    You need to find a and b so that [itex] r(1-r) = a - (r-b)^2.[/itex] In other words, you need
    [itex] r - r^2 = a - b^2 + 2br - r^2[/itex] for all r, so (equating the coefficients of [itex] r^0, r^1 \text{ and } r^2[/itex]) we have [itex] a - b^2 = 0[/itex] and [itex]2b = 1.[/itex] Thus, [itex] r(1-r) = 1/4 - (r-1/2)^2.[/itex] Now (r - 1/2)2 ≥ 0 for all r, and = 0 when r = 1/2 (which is inside the region 0 ≤ r ≤ 1). So, indeed, we have r(1-r)≤ 1/4 for all real r.

  12. Jul 26, 2012 #11
    Thanks again for the help. Factoring by decomposition will be a big help.
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