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Proving that a real number exists in between a real number,

  1. Feb 25, 2008 #1
    Ok I'm supposed to prove that if for any real x&y s.t x<y there exist a real z s.t

    x<z<y

    I'm supposed to use the following;

    Axiom: every nonempty subset S of real numbers which is bounded above has a supremum: that is there is a real number B s.t B=sup(S)

    1) every nonempty subset S that is bounded below has a greatest lower bound; that is there is a real number L s.t L=inf(S)

    2) The set P of positive integers (i.e 1,2,3,4..n) is unbounded from above.

    3) For every real number x there exists a positive integer n s.t n>x

    4) If x>0 and y is an arbitrary real number, there exists a positive integer n such that nx>y

    5) If three real numbers a,x,y satisfy; a=< x=<a + y/n for any n>=0, then x=a

    6) If x has a supremum, then for some x in S we have x>sup(S)-h

    7) If x has an infimum, then for some x in S we have x<inf(S)+h

    8) Given 2 nonempty subsets S and T of R such that s=<t Then for every s in S and t in T, S has supremum and T has an infimum, and they satisfy sup(S)=<inf(T)


    really I have no clue how to start this problem. Alot of the inequalities are useless because they only involve integer n. I'm looking for a real z.
     
  2. jcsd
  3. Feb 25, 2008 #2
    Why not just show that (X1+X2)/2 is a real because the reals are closed under multiplication. Then show:

    That if X1 is less then X2 then:

    X1<(X1+X2)/2<x2
     
  4. Feb 25, 2008 #3
    Ooo. I just thought of a solution. Its a bit different from yours. Here goes.

    For x there is a z such that x<nz 4) arcimedian property

    For z there is a y such that z<ny 4) archmedian property

    set n=1 and we have x<z and z<y. Therefore, x<z<y

    Well, how did I do? is it rigorous enough?
     
  5. Feb 25, 2008 #4
    You can't arbitrarily pick n.
     
    Last edited: Feb 25, 2008
  6. Feb 25, 2008 #5
    No, not only is it not rigorous but it doesn't prove what you are trying to prove.


    What you have to start with is fixing to reals x,y and then for these (arbitrary, but fixed) x,y you have to find a real z such that x < z < y.

    You started with x, then you constructed z which is fine, und then you re-adjusted your y to be larger than z. It has to work the other way round. Given x and y you are to find a z inbetween the two.
     
  7. Feb 26, 2008 #6
    I'm getting nowhere with the Archimedean property. I'm using the last theorem. 8)

    Say x belongs in X and y belongs in Y. With both X & Y subsets of R(reals). Then x<y => sup(X)<inf(Y).

    since x<sup(X) and inf(Y)<y => x<sup(X)<inf(Y)<y.

    just chose z to be either sup(X) of inf(Y).

    z=sub(X) works.

    Does this work?
     
  8. Feb 26, 2008 #7
    Why not use dedekindi cuts to prove this, it is much easier i guess, and it is valid for any real nr, including irrationals!
     
  9. Feb 26, 2008 #8
    What are dedekindi cuts? I'm reading out of Apostol( which is supposed to be totally rigorous btw) and there was no mention of dedekini cuts. He just uses the least upper bound axiom to demonstrate that irrationals exist.
     
  10. Feb 26, 2008 #9
    No, unless you specify what X and Y are. For X={x}, Y={y} your argument clearly doesn't work. Proving that there ARE sets X containing x and Y containing y, such that x < sup(X) < inf(Y) < y, would prove your problem indeed:smile: Note, that you need strict inequalities here, which do not follow from the definition of sup, inf and the fact that x(y) belongs to X(Y).
     
  11. Feb 26, 2008 #10

    HallsofIvy

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    Is there a problem with using John Creighto's suggestion in the very first response to your question?

    If x and y are real numbers, with x< y, then so is (x+ y)/2 and it is easy to prove that x< (x+y)/2< y.

    You don't need any deep properties of the real numbers because this theorem is also true of the rational numbers that don't have those properties.
     
  12. Feb 26, 2008 #11
    Probbably what he is looking to prove is that between every two rational numbers there is an irrational one. So obviously in this case John Creighto's method does not work!
     
  13. Feb 26, 2008 #12
    Well dedekindi cuts are called those cuts on a set of numbers that have the following proberties:
    1.We cut that set into two classes, the lower class A, and the upper class A'
    2. Every number of that set should fall either in the lower calss A, or in the upper class A'
    3. Every number r of the lower class has the property that it is smaller than any number r' of the upper class, also vice versa.

    This dedekindi cut is first defined for rationals. that is
    1. We cut the set of rationals into two classes, the lower class A, and the upper class A'
    2. every rational number should fall either in the lower class A, or the upper class A'
    3. Every rational number r that falls in the lower class is smaller than every number r that falls in the upper class, and vice versa!

    After that we define a rational number to be that number that is as a result of such a cut, if R is that rational nr, than it must fall either in the lower class where it is the maximum, or in the upper class where it is its minimum.
    Next case, if a nr, say b, is a result of such a cut, but on the other hand it falls neither in the lower class nor in the upper class, then we define that number to be an irrational one. so this way we extend these cuts to all real numbers.
    After that using these properties we can show that between any two real numbers x,y, including the case when they are irrational, there exists a number c such that
    if x<y, then x<c<y
     
  14. Feb 26, 2008 #13

    morphism

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    Dedekind, after the mathematician, and not "dedekindi". And they're not necessary for this problem at all. John Creighto's suggestion is more than adequate.
     
  15. Feb 26, 2008 #14
    Yeah i know, but i just said that if one needed to show that between every two rationals there exists an irrational, than Jon Creighto's suggestioni would not work!!
     
  16. Feb 26, 2008 #15
    let x<y add an x to both sides we get

    2x<x+y, deviding by 2 we get

    x<(x+y)/2, also x<y adding a y on both sides we get x+y<2y => (x+y)/2<y, so from both of these we get
    x<(x+y)/2 =c<y, c=(x+y)/2 is also a rational

    However i think that this method is not that consistent if either x or y where inrrationals!
     
  17. Feb 26, 2008 #16

    morphism

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    Who said anything about rationals and irrationals?

    You don't even need Dedekind cuts to prove that there's an irrational between any two rationals. For instance if x<y are rationals, then [itex]\frac{1}{\sqrt{2}}x + \left(1 - \frac{1}{\sqrt{2}}\right)y[/itex] is an irrational that sits between them.
     
  18. Feb 26, 2008 #17
    I guess it's just me! However i was just trying to introduce another way of proving it!
    Wow, how could i prove that, i mean how would i show that [itex]\frac{1}{\sqrt{2}}x + \left(1 - \frac{1}{\sqrt{2}}\right)y[/itex] is an irrational that sits between x and y?
     
  19. Feb 26, 2008 #18

    Well it's obviously irrational if x,y are rational..

    Assume x<y

    Then
    [tex]
    \frac{1}{\sqrt{2}}x + \left(1 - \frac{1}{\sqrt{2}}\right)y-x
    =\left(\frac{1}{\sqrt{2}}-1\right)x + \left(1 - \frac{1}{\sqrt{2}}\right)y
    = \left(1 - \frac{1}{\sqrt{2}}\right)(y-x)
    >0[/tex]

    Therefore the number in question is greater than x; analogously you can show that it is less than y:

    [tex]
    \frac{1}{\sqrt{2}}x + \left(1 - \frac{1}{\sqrt{2}}\right)y-y
    =\frac{1}{\sqrt{2}}x - \frac{1}{\sqrt{2}}y
    = \frac{1}{\sqrt{2}}(x-y)<0[/tex]
     
    Last edited: Feb 26, 2008
  20. Feb 26, 2008 #19
    NIce! However, if one had never seen this before, how is he supposed to think to come up with such a number like [itex]\frac{1}{\sqrt{2}}x + \left(1 - \frac{1}{\sqrt{2}}\right)y[/itex], because it did not look obvious to me before!
     
  21. Feb 26, 2008 #20

    morphism

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    Well, here's how I came up with it:
    2x = x + x < x + y < y + y = 2y

    This gives us a real number between x and y. I wanted to generalize this. So let u be an irrational number (to be determined), and note that
    x = x + ux - ux = ux + (1-u)x < ux + (1-u)y < uy + (1-u)y = y

    Of course for the first inequality to hold, we'd better have u between 0 and 1, hence my choice of 1/sqrt(2).
     
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