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Help solving problem involving inequalities

  1. May 31, 2008 #1
    Hi all,

    I'm trying to teach myself analysis using the book "A Companion to Analysis: A Second First and First Second Course in Analysis" by T. W. Körner.

    There's an inequality problem in there that's used to prove a statement about the continuity of a function, that I've got stuck in (problem 1.16, part (iii), in case you happen to have the book):

    Working in [itex]\mathbb{Q}[/itex] (the space of rational numbers), if [itex]x^2<2[/itex] and [itex]\delta=\frac{(2-x^2)}{6}[/itex], show that [itex]y^2<2[/itex] whenever [itex]|x-y| < \delta[/itex]

    Any help would be greatly appreciated!
  2. jcsd
  3. May 31, 2008 #2


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    How about |x-y|<(2-x^2)/6<1/3, thus y<1/3-x<1/3+sqrt(2)<2 since sqrt(2)=1.41.. and 1/3=0.33..
  4. Jun 1, 2008 #3
    That doesn't seem to be quite correct...
    It needs to be proven that [itex]y^2 < 2[/itex], not [itex]y < 2[/itex] - the latter doesn't imply the former.
    Also, I have the feeling the author expects a solution that doesn't explicitly using the value of [itex]\sqrt{2}[/itex], since no such rational number [itex]x[/itex] exists such that [itex]x^2=2[/itex].
  5. Jun 2, 2008 #4


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    The best way to approach it is to find the region in the x,y plane that satisfies the inequalies.

    With the |x-y| term you're best off to consider the two cases (x>y and x<y) seperately.

    For example.

    Case1. Assume y>x

    y < 1/3 + x - x^2 /6.

    That is, we are looking for the region of the x,y plane where both y>x and y < 1/3 + x - x^2 /6 are satisfied. It's pretty straight forward to find, just the region between a parabola and a straight line.

    Repeat for the other case (y<x) and you'll soon know everything that you could wist to kinow about the solution region and the inequality y^2<2 will be immediately apparent.
    Last edited: Jun 2, 2008
  6. Jun 2, 2008 #5
    Thanks for the replies... I posted the same problem on Usenet sci.math, and was offered a simpler solution (although I managed to solve the problem myself in pretty much the same way immediately after posting the question on sci.math - funny that).
    |x-y| < d (d is delta)
    => |y| < |x| + d
    => y^2 < x^2 + d^2 + 2|x|d < x^2 + d^2 + 4d
    and then showing d^2 + 4d < (2-x^2)
    from which the result follows.
  7. Jun 4, 2008 #6
    Will this way work?:

    |x-y| < (2-x^s)/2

    Substitute sqrt (2) for x, and you get

    sqrt (2) - y < 0

    - y < - sqrt (2)

    y > sqrt (2)

    y^2 > 2


    sqrt (2) - y > 0

    - y > - sqrt (2)

    y < sqrt (2)

    y^2 < 2

    So therefore, y^2 does not equal 2, and you get your answer of y^2 < 2
    Last edited: Jun 4, 2008
  8. Jun 4, 2008 #7
    Ah no, you have to show it for all values of x and y that satisfy the inequalities, not just a specific set.
  9. Jun 5, 2008 #8
    Nah, that's not right, for at least two reasons: (1) It is clearly specified that x^2 < 2, i.e. -sqrt(2) < x < sqrt(2), so a substitution of x = sqrt(2) is not justified, and in fact leads to |sqrt(2)-y|<0 , which can never be true (because the absolute value of a number is either greater than or equal to zero). (2) A second reason is that we're dealing in rational numbers (as specified in the original problem): sqrt(2), for us, "doesn't exist", so such a substitution would be "cheating" (even if problem (1) didn't exist).

    IMO, the easiest-to-follow solution is the one outlined in my previous post. I'll "latexify" it and fill in the missing steps in my next post, in case someone's interested.
    Last edited: Jun 5, 2008
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