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Proofs, [itex]\exists[/itex] x [itex]\in[/itex] (1, [itex]\infty[/itex]) such that

  1. Feb 7, 2012 #1
    Hey! I tried to make the title as descriptive as possible, but ran out of characters. Im trying to prove that..

    1. The problem statement, all variables and given/known data

    "There exists x [itex]\in[/itex] (1, [itex]\infty[/itex]) such that for all y [itex]\in[/itex] (0,1), xy[itex]\geq[/itex]1.

    [itex]\exists[/itex] x [itex]\in[/itex] (1, [itex]\infty[/itex]) s.t. [itex]\forall[/itex] y [itex]\in[/itex] (0,1), xy[itex]\geq[/itex]1.


    2. Relevant equations

    none.

    3. The attempt at a solution

    I say 'false' because when the entire statement is negated, the working negation is true.

    [itex]\neg[/itex]{[itex]\exists[/itex] x [itex]\in[/itex] (1, [itex]\infty[/itex]) s.t. [itex]\forall[/itex] y [itex]\in[/itex] (0,1), xy[itex]\geq[/itex]1.} (Negating line)

    [itex]\forall[/itex] x [itex]\in[/itex] (1, [itex]\infty[/itex]), [itex]\exists[/itex] y [itex]\in[/itex] (0,1), s.t. xy < 1. (This is the working negation of original statement)

    Now looking at this statement, since x can be infinitely large, and I can pick an infinitely smaller y, the negation would be true, making the original statement false.

    But if I look at the original statement, can't I do the same thing? Would this be a paradox?




    Also, there is a similar problem, except the original statement is "For all y's in the element (0,1) there exists an x in the element (1, infinity) such that xy < 1 ." I get the same result, except in this one, since x and y can get infinitely close to 1, albeit on either side, they will cancel eachother out, making the working negation true, and the statement false.
     
  2. jcsd
  3. Feb 7, 2012 #2

    Mark44

    Staff: Mentor

    Re: Proofs, [itex]\exists[/itex] x [itex]\in[/itex] (1, [itex]\infty[/itex]) such tha

    It is helpful to look at a graph of y = 1/x. If x > 0, y = 1/x is equivalent to xy = 1. Also, for x > 0, the inequality xy >= 1 is all the points on or above the graph of y = 1/x.

    For any given x > 1, there is one y value (call it y*) in (0, 1) for which xy* = 1. y* divides the interval (0, 1) into two pieces. What can you say about each piece relative to the inequality xy >= 1?
     
  4. Feb 7, 2012 #3
    Re: Proofs, [itex]\exists[/itex] x [itex]\in[/itex] (1, [itex]\infty[/itex]) such tha

    What about this intention of a contrapositive:


    If there are no x belonging to (1, oo), meaning that x would belong

    to (-oo, 1], then xy would have to be less than 1,

    with y belonging to (0, 1).
     
  5. Feb 7, 2012 #4
    Re: Proofs, [itex]\exists[/itex] x [itex]\in[/itex] (1, [itex]\infty[/itex]) such tha

    Hmm.. Im trying hard to understand what you're trying to tell me, but it's slipping me..

    I understand that there is one y value (y*) for every x, to make xy*=1 .. The y* value divides the interval into 2 pieces. y* gets increasingly smaller as x gets bigger.. Is that what you were asking?

    Im not sure about the relation of the inequality and the two pieces.. I'm sorry, I feel like I should understand this, but it's just not coming yet. I appreciate your help Mark44.


    A contrapositive is logically equivalent to the original statement. We went over contrapositives briefly in class, and I'm not sure how to get a contrapositive of my original statement. I could take yours at face value, but no offense here, how do I know if you're right? I don't know if it's right by my knowledge and what we've learned so far, so I can't comprehend your statement yet.

    I know that [If x, then y] is logically equivalent to x [itex]\Rightarrow[/itex]y . and then the working negation would be x[itex]\wedge\neg[/itex]y . So I guess the contrapositive would be [itex]\neg[/itex]x[itex]\vee[/itex]y

    Now I'm really lost.. I'm not sure I should use a contrapositive to prove this one.. Thanks though, for your input, checkitagain. I'd like to try the contrapositive to prove this once I fully understand the concept.
     
  6. Feb 7, 2012 #5

    Mark44

    Staff: Mentor

    Re: Proofs, [itex]\exists[/itex] x [itex]\in[/itex] (1, [itex]\infty[/itex]) such tha

    Yeah, pretty much. The y* value divides the interval (0. 1) into two pieces.
    I hope you have sketched a graph of y = 1/x, since that would make things a lot clearer.

    Note: I have changed my notation so that I am not using * any more.
    For a given x value greater than 1, say x0, there is a number y0 such that x0y0 = 1.

    As you already recognized, y0 divides the interval (0, 1) on the y-axis. What can you say about all of the numbers in the interval (y0, 1), relative to the value of y*x0?
    What can you say about all of the numbers in the interval (0, y0), relative to the value of y*x0?

    If you have the graph to look at, this is fairly simple, but if you're merely wrestling with logical symbolism, and aren't thinking about the basic underlying geometry, it's much harder.
    Although you can do this using the contrapositive, I don't see any need for it. I would work directly with the statement as given and show a counterexample.
     
  7. Feb 8, 2012 #6
    Re: Proofs, [itex]\exists[/itex] x [itex]\in[/itex] (1, [itex]\infty[/itex]) such tha

    Ok, I'm with you on this so far, but what value would I use for 'y' when you say "...relative to the value of y*x0?"

    In my head, y0 and y are the same thing..

    you know what I mean? I'm just super lost.. I'm looking at the graph, and I can see that the interval (0,y0) would be decreasing and the interval (y0,1) would be increasing as x0 gets larger and larger.. I guess I'm completely missing the point you're going for.
     
  8. Feb 8, 2012 #7

    Mark44

    Staff: Mentor

    Re: Proofs, [itex]\exists[/itex] x [itex]\in[/itex] (1, [itex]\infty[/itex]) such tha

    It doesn't matter, as long as you pick a y value in one of the two intervals (0, y0) or (y0, 1). Every y value in the first interval has the same behavior relative to the expression x0*y. And every y value in the second interval has the same behavior relative to the expression x0*y, but that behavior is different from that of the first interval.
    y0 is supposed to represent the specific y value that is associated with x0.
    This makes no sense. An interval is not increasing or decreasing. The graph of y = 1/x is decreasing everywhere on the interval (0, ∞).
    Maybe it will help to look at a concrete example. Let x0 = 2. Then y0 = 1/2. From this, we see that x0 * y0 = 1.

    So now we have two intervals along the y-axis: (0, .5) and (.5, 1).

    For the same value of x, namely x0 = 2, if you pick a y-value in (0, .5), will x0y be larger than 1 or less than 1? (It can't possibly be equal to 1, since that occurs only for y0 = 1/2.)

    OTOH, if you pick a y-value in the other interval, (.5, 1), will x0y be larger than 1 or less than 1?
     
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