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Is x<=ky equivalent to x<y?

  1. Jan 14, 2013 #1
    I have a question on something that has come up many times for me in other contexts, such as Lipschitz transforms, and recently contraction mappings and the fixed point theorem.

    Is the following statement always false, and why?

    Let x and y be two positive real numbers.
    [itex]\exists k<1 \,\, s.t. \,\, x \leq k\cdot y \, \Leftrightarrow \, x<y[/itex]

    So far, I've always seen statements like this to be false, but I'm not sure why.

  2. jcsd
  3. Jan 14, 2013 #2


    Staff: Mentor

    is k also a positive real?
  4. Jan 14, 2013 #3
    Sorry, yes. 0<k<1.
  5. Jan 14, 2013 #4


    Staff: Mentor

    I cant say if its true or not but try to visualize it on a xy plane where the area under the line x=y is x<y versus the line x=ky.

    Perhaps that will help you decide.
  6. Jan 14, 2013 #5
    I guess the thing that I don't understand is that if k is strictly less than 1, can't we still make k as close to 1 as we want? Then wouldn't saying [itex]x \leq k \cdot y[/itex] where k is arbitrarily close to 1 (but not equal) be the same as saying [itex]x<1 \cdot y[/itex]?
    It seems to me that changing the [itex]\leq[/itex] to [itex]<[/itex] would account for k not being equal to 1, only close.

    In the example with the lines, the same idea would hold. We can make k arbitrarily close to 1, and thus make the line x=ky arbitrarily close to the line x=y.
  7. Jan 14, 2013 #6


    User Avatar
    Science Advisor

    If k< 1 then, mutiplying both sides by the positive number y, ky< y. So if x< ky, it is certainly true that x< y.

    Conversely, if x< y, then x/y< 1 and so there exist k such that x/y< k< 1. That is, we have k< 1 and, multiplying both sides of x/y< k, x< ky.
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