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Apostol's Calculus I. 3.12 - Verify Solution.

  1. Feb 8, 2017 #1
    1. If x and y are arbitrary real numbers with x < y, prove that there is at least one real z satisfying
    x<z<y.



    2. I'll be using this theorem:
    T 1.32 Let h be a given positive number and let S be a set of real numbers. (a) If S has a supremum, then for some x in S we have x > sup S - h.



    3. The attempt at a solution
    1. Let S be the set of all real x satisfying x < y for a given real y.
    2. y = sup S.
    3. z > y - h where z is in S. ( because of T 1.32 )
    4. Let h = y - x then z > x.
    5. Because z is a member of S, then z must also be less than y.
    Therefore I have shown there exists a real number z, such that x < z < y.

    Is my logic sound? I am a little thrown off by the phrase, "If x and y are arbitrary reals.." I am not sure how to stick that construction into a set S, hence the wordiness of step 1. If so, how can I better formalize the proof, I would prefer some type of set/formal logic notation, at least for step 1. Thank you for your help.
     
  2. jcsd
  3. Feb 8, 2017 #2

    fresh_42

    Staff: Mentor

    I don't see any loophole. I would write ##S_y## instead of ##S## for it depends on ##y##, but this doesn't affect the logic. As ##x,y## are given and as such fixed, you're free to define the special ##S_y\,.## Maybe it would also help to define ##S_y = \{r \in \mathbb{R}\,\vert \,r < y\}## to avoid the double meaning of ##x##.

    In general it always helps to track down dependencies. E.g. in the ##\varepsilon## definition of a limit, the choice of ##N \in \mathbb{N}## is actually depending on the choice of ##\varepsilon## and a notation ##N_\varepsilon## or ##N(\varepsilon)## is more precise. For the same reason I'd define ##S## as ##S_y## in the above.
    The double usage of variables - such as ##x## in your case as given number and as variable notation for elements of ##S_y## - is also problematic, because it is a potential (and unnecessary) risk of confusion at best and errors at worst.
    As a minor hint, you could also use ##h := y-x## with the double point in front of the equality sign to indicate that it is a definition, an allocation in contrast to a derived formula. But this isn't really necessary in the sense that many authors don't use it. In my opinion it is simply easier to read with than without.

    If you want to be very petty, you should define ##sup \, S_y =: \overline{y}## and ##h:= \overline{y} - x## because you haven't shown, that actually ##y## is the supremum of ##S_y\,.## You don't need to show ##\overline{y} = y##, but in case you use both, you also should explain ##h > 0## as needed for the theorem.
     
    Last edited: Feb 8, 2017
  4. Feb 8, 2017 #3

    Ray Vickson

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    Your proof looks basically OK, but if all you want is to prove the statement about the existence of some ##z##, then looking at ##z = (x+y)/2## is a lot easier. (In fact, you can show the existence of infinitely many ##z## between ##x## and ##y## by looking at ##z = (1-t) x + t y##, where ##0 < t < 1.##)
     
  5. Feb 8, 2017 #4

    fresh_42

    Staff: Mentor

    How do you make sure, that ##z## exists in this case?
     
  6. Feb 8, 2017 #5

    Ray Vickson

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    Well,certainly ##z## exists; the only issue is whether or not ##x < z < y##, and that is an exercise in elementary inequality manipulation. Of course, it all depends on what properties the OP is allowed to use as known background information.
     
  7. Feb 8, 2017 #6

    fresh_42

    Staff: Mentor

    To me it appeared as an exercise about Dedekind cuts or whichever way the reals are constructed in this case. So "certainly it exists" might not be allowed to assume. The normal tools of calculus might not be available at this stage as I understood it.
     
  8. Feb 8, 2017 #7

    Ray Vickson

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    If we are allowed to know/use the facts that a sum (or difference) of two numbers is a number, and the product of two numbers is a number, then ##z## does, indeed exist. If we are not allowed to use those properties, then, of course, everything changes.

    As for the inequalities, if we know that a multiplying both sides of an inequality by a positive number preserves the direction of the inequality, and the sum of two same-direction inequalities is another same-direction inequality, then the rest of the result follows as well. But again, if we are not allowed to know/use those properties, life becomes more difficult.
     
  9. Feb 9, 2017 #8
    Apostol's Calculus p. 17

    Huh, so I think Ray Vickson's solution is totally valid, but I think Apostol wanted the reader to solve it using the idea of a supremum. It seems the difference between the two solutions is one where we're explicitly constructing a z that is in between y and x and the other solution just proves there exists a z in between y and x but doesn't tell what it is. I am guessing Apostol is going to be using that generality to insert the irrationals into the real number system in the proceeding section.

    fresh_42, thank you very much for that wonderful post!

    For what's it worth to anyone going through Apostol's book, I think I polished the proof a bit:
    x, y ∈ ℝ such that x < y
    S = { z < y | z ∈ ℝ }
    Because S is bounded above by y, by the completeness axiom, S has a supremum such that sup S = y.
    Let h = y - x which is positive.
    There is a z > y - h by T 1.32, that is z > x.
    Hence there is a real z such that x < z < y.
     
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