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Tricky proof?

  1. Oct 30, 2012 #1
    The problem is "For every real number x, there exists integers a and b such that a≤x≤b
    and b-a=1"
    Im stuck on the first part of the proof. So in my proof I let a=x and b=x+1. Then x+1-x=1=b-a. But what I don't get why is that is safe for me assume here that a and b are not always going to be integers since x is a real number? I was thinking that even when the hypothesis is false the implication always true vacuously.
     
  2. jcsd
  3. Oct 30, 2012 #2

    haruspex

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    I'm unclear what you're saying in the last couple of sentences, but your proof does not work because, as you say, if x is not an integer then your a and b won't be.
    You just need to find a better choice for a and b.
     
  4. Oct 30, 2012 #3
    I'm stuck. Like take for example x=2.5. Since a≤x then a can equal x which is 2.5 which is not an integer. If a=2.5 then b has to equal 3.5 in order for b-a=1 but a and b are not integers. I don't know how to get around it?
     
  5. Oct 30, 2012 #4

    Mark44

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    It isn't safe for you to assume that. Part of the hypothesis is "there exist[STRIKE]s[/STRIKE] integers a and b such that ..."
    An implication is true by definition when the hypothesis is false, no matter what the concluson is. For example, "if 1 = 2, then my dog has five legs" is a true statement, but it's neither useful or interesting.

    What the theorem you're trying to prove is that any real number x is straddled by two successive integers. If x happens to be an integer, then you can let a = x and b = x + 1.
     
  6. Oct 30, 2012 #5

    SammyS

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    No. a cannot be 2.5 .

    Reread the statement of your problem (bold for emphasis).
    "For every real number x, there exists integers a and b such that a≤x≤b and b-a=1"​

    Is 2.5 an integer?
     
  7. Oct 30, 2012 #6

    haruspex

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    You seem to be misreading the question as "for all a and b s.t. a≤x≤b...". You only have to show that for any given x there exists some pair a and b such that etc. So e.g., given x = 2.5, what pair of integers surrounds it?
     
  8. Oct 30, 2012 #7
    Well 2 and 3 surround it. So in my proof I should say "When x is an integer then let a=x and b=x+1. x+1-x=1=b-a. Otherwise when x is not an integer then there exists integers a and b where b=a+1. Thus a<x<a+1 and a+1-a=1=b-a. Hence for every real number x there exists integers a and b where a≤x≤b and b-a=1."
     
  9. Oct 30, 2012 #8

    Mark44

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    You can tighten up the part where x is not an integer by using the greatest integer function ## \lfloor x \rfloor##. This function evaluates to the largest integer that is less than or equal to x.

    In essence, it strips off any fractional part of a number, so that, for example,
    ##\lfloor 2.35\rfloor = 2##.
     
  10. Oct 30, 2012 #9

    haruspex

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    To complete the proof, there are a couple more details.
    Shouldn't really assume the floor function is an accepted and well-defined function. What is to be proved here is very low level so the proof should only make use of the simplest tools. You can define the floor function as above, provided you demonstrate that there is such an integer and it is unique.
    Secondly, it remains to provide a formula for b, and to show that it satisfies requirements.
     
  11. Oct 30, 2012 #10

    Mark44

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    Why not? It seems to be accepted enough that there is notation for it. As I recall, the first I heard of it was in a college-level math class.
    The OP has done that. Once you find a, by whatever means, all you need to do to get b is to add 1 to a.
     
  12. Oct 31, 2012 #11

    haruspex

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    If bonfire09 has been introduced to the floor function, fine, but that has not been established.
    Not in the context of a being set to floor(x). Besides, it only takes one line to complete the proof.
     
  13. Oct 31, 2012 #12

    HallsofIvy

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    You will need the Archimedean property: if x is any real number, there exist an integer n greater than x

    and the well-ordered property of the integers: if a non-empty set of integers has a lower bound, then it contains a smallest member.
     
    Last edited: Oct 31, 2012
  14. Oct 31, 2012 #13

    Mark44

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    Which is precisely why I wrote "by whatever means".
     
  15. Oct 31, 2012 #14
    I've seen the floor and ceiling function In a programming class but I have never seen it been used in my proofs class. I realize i tend to gloss over a detaail which messed me up. I tend to do this sometimes since proofs requiire a great deal of attention to details.
     
  16. Oct 31, 2012 #15

    SammyS

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    I suggest you look at HallsofIvy's post (Post #13) which I quote below.

     
  17. Nov 15, 2012 #16
    Instead of using the well ordering principle. Couldn't I just say ""When x is an integer then let a=x and b=x+1. x+1-x=1=b-a. When x is not an integer then a<x<b since a and b are integers. Thus a<x<a+1 when b=a+1. Hence there exists an integer greater than every real number x."
     
  18. Nov 15, 2012 #17

    Ray Vickson

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    I think the question is essentially asking for a proof of the existence of the floor function. If we know we can write x as n + r (, n = integer, 0 <= r < 1) then, of course, we can write floor(x) = n. Naturally, in many courses the floor function will be introduced and used, but without ever really proving it exists (which is intuitively obvious, anyway).

    RGV
     
  19. Nov 15, 2012 #18
    So in other wards like this "

    Case 1: Assume x is an integer then let a=x and b=x+1. x+1-x=1=b-a.

    Case:2 Assume x is not an integer then let a=⌊x⌋ and b=⌊x⌋+1. This becomes ⌊x⌋<x<⌊x⌋+1 and b-a=1.
     
    Last edited: Nov 15, 2012
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