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Help with two proofs

  1. Mar 11, 2008 #1
    Hi there everyone, I have the basic idea of what to do, its just trying to show the cases work is where the problems occurs. Anyways for the first one:

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

    Prove that if x is any positive integer, then ⌈x/2⌉ ≤ (x + 1)/2. (Here, for
    any real number r, ⌈r⌉ is the smallest integer greater than or equal to r. Thus, ⌈2.3⌉ = 3,
    ⌈2⌉ = 2, etc.) Do a proof by cases.

    2. Relevant equations

    None that I know of.

    3. The attempt at a solution

    Since it's a proof by cases, and it only deals with positive integers, I think my two cases should be if x is an even integer, and an odd integer. This is the point where I get confused, do I basically plug in 2k and 2k+1 for x, and then solve? If that is true? What exactly am I solving for?

    Anyways for the next proof:

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

    Show that if r is an irrational number, there is a unique integer n such that the distance between r and n is less than 1/2. Be sure to argue the uniqueness of n.

    2. Relevant equations

    None that I know of.

    3. The attempt at a solution

    I have absolutely no idea on where to even start for this problem. I all know is that since r is irrational, it can be said that it could lie in between two integers n and n +1, much like n < r < n + 1. So I have to solve for n < r and r < n + 1. But again, what exactly am I solving for, and what do I have to do to show that? I guess my main problem in these types of proofs is just verifying they the cases work, I really just dont see how any of the cases could lead to a conclusion that I could use to help prove the problems. Anyways, any help is appreciated, thanks in advance!
     
  2. jcsd
  3. Mar 12, 2008 #2

    HallsofIvy

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    If n is even, say x= 2k, what is ⌈x/2⌉? What is (x+1)/2?
    If n is odd, say x= 2k+1, what is ⌈x/2⌉? What is (x+1)/2?

    What do you have to work with? How you would prove something depends strongly on what prior theorems you know. If I were trying to prove that I would use: 1) The Archimedean property: given any real number r, there exist an integer larger than r; and 2) the "well ordered property" of the positive integers: any set of positive integers contains a smallest integer.
    But I don't know if you have those to work with.
     
  4. Mar 12, 2008 #3
    So for the first one I know that if x = 2k, then I have ⌈x/2⌉ = ⌈2k / 2⌉ which is equal to k right? then (x + 1)/2 would equal (2k + 1) / 2, which can be written as (2k)/2 + 1/2 which can also be written as k + 1/2? so I would have ⌈k⌉ ≤ k + 1/2? Would that solve the even case?
    Working the odd case I would have ⌈(2k + 1) /2⌉ which equals ⌈2k / 2 + 1 /2⌉ which can equal ⌈k + 1/2⌉ and since ⌈r⌉ is the smallest integer greater than or equal to r I would end up having:
    ⌈k + 1⌉ ≤ (2k + 1) + 1 / 2 = (2k + 2) /2 = 2k/2 + 2/2 = k + 1 which is equal to the left hand side so it still works right?

    And for the second one, I apologize, I am not too versed with those properties, but Ill keep thinking about the problem, and how the play into the overall proof

    EDIT: Actually I think I'm starting to see what you mean by those properties, since an irrational number is a real number, by the Archimedean property, there exists a interger n such that r < n. Also by the well ordered property, we see that within the set of two integers, one exists as the smallest integer. But the only inequality that I see is n < r < n + 1, unless you are referring to the less than 1/2 side. Haha, nevermind, I think I just lost it.
     
    Last edited: Mar 12, 2008
  5. Mar 13, 2008 #4

    HallsofIvy

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    First, your analysis of the first problem is correct.

    For the second, you are trying to prove "Show that if r is an irrational number, there is a unique integer n such that the distance between r and n is less than 1/2."

    There is NOT a set of two integers because you have not yet proved that there is an "n" and "n+1".

    As you said, by the Archimedean property, there exists integers larger than r. Let "A" be the set of all integers larger than r. By the "well ordered" property, that set has a smallest integer, n. Then n is the smallest integer larger than A. Since n-1 is also an integer, and smaller than n, it cannot be larger than r: n-1< r (why can't it be equal to r?) If r- n< 1/2, we are done! (with existence). If r-n> 1/2 (note: you will need to prove that because r is irrational r-n cannot equal 1/2) then look at n-1< r. |n-(n-1)|[itex]\le[/itex]|r-(n-1)|+ |n-r| (triangle inequality). But n-(n-1)= 1. Use that to prove that the distance from r to n-1 is less than 1/2.

    And, of course, "Be sure to argue the uniqueness of n." Suppose there were two integers, n and m, whose distance from r were less than 1/2. What is the distance from n to m (triangle inequality again).
     
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