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Math Help - Proof , Real system

  1. #1
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    Proof , Real system

    Hello guys I'm new here and I hope I'm asking this in the right place.
    I'm sturting to study calculus by myself and I'm using apostol book.
    Im still in the introduction part and I'm having some troubles by doing some proofs.

    If x and y are arbitrary real numbers with x < y , prove that there is at least one real z satisfying x < z < y

    The exercise is in the chapter of supremum and infimum, archimedean property of the real-number system

    I appreciate any help
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  2. #2
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    Quote Originally Posted by yannick View Post
    Hello guys I'm new here and I hope I'm asking this in the right place.
    I'm sturting to study calculus by myself and I'm using apostol book.
    Im still in the introduction part and I'm having some troubles by doing some proofs.

    If x and y are arbitrary real numbers with x < y , prove that there is at least one real z satisfying x < z < y

    The exercise is in the chapter of supremum and infimum, archimedean property of the real-number system

    I appreciate any help
    If x < y, then x < \frac{x + y}{2} < y ....
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  3. #3
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    but how to proove this?
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  4. #4
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    let z = (x + y)/2 then 2z - y = x and 2z - x = y
    x < y
    2z - y < y
    2z < 2y
    z < y

    x< y
    x < 2z - x
    2x < 2z
    x < z

    so we have

    x < z < y


    really thx!
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  5. #5
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    I know that I should create a new topic for this but is quite the same problem.
    After some other exercizes I found that the same exercise but asking to prove that there is at least a rational number z and another exercise that ask to prove that exist at least one irrational number z
    any suggestion?
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by yannick View Post
    I know that I should create a new topic for this but is quite the same problem.
    After some other exercizes I found that the same exercise but asking to prove that there is at least a rational number z and another exercise that ask to prove that exist at least one irrational number z
    any suggestion?
    i suppose you want to prove the denseness of the rationals and the denseness of the irrationals (can't find a link for the irrationals yet)
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  7. #7
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    In more general terms if (X,d) is a metric space we say a subset Y\subset X is dense in X when \bar Y = X (the closure of Y). Note, \bar Y = Y \cup \partial Y. Thus, if X = \mathbb{R} and Y = \mathbb{Q} then \widehat{\mathbb{Q}} = \mathbb{Q} \cup \partial \mathbb{Q} but \partial \mathbb{Q} = \mathbb{R} because any point x we chose in \mathbb{R} if we draw a small interval around it (x-\epsilon,x+\epsilon) is shall contain both rational and irrational points. Thus, any real point is a boundary point of a rational number. And thus the rationals are dense in the reals.
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  8. #8
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by yannick View Post
    I know that I should create a new topic for this but is quite the same problem.
    After some other exercizes I found that the same exercise but asking to prove that there is at least a rational number z and another exercise that ask to prove that exist at least one irrational number z
    any suggestion?
    Pick an integer b such that \frac{1}{y-x}<b. Then b is positive and by-bx>1. Since the difference between by and bx is greater than 1, there must be an integer a between them*, i.e. bx<a<by. Hence x<\frac{a}{b}<y, showing that there is a rational number between x and y.

    *The set of integers \{n\in\mathbb{Z}:n\leq bx\} is nonempty (by the Archimedean principle) and bounded above, so it has a maximum member c. It follows that bx<c+1<by.

    For irrational, choose b>\frac{j}{y-x}, where j is any irrational number greater than 1 (e.g. \sqrt{2}). This should lead to x<\frac{aj}{b}<y.
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  9. #9
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    Quote Originally Posted by JaneBennet View Post
    Pick an integer b such that \frac{1}{y-x}<b. Then b is positive and by-bx>1. Since the difference between by and bx is greater than 1, there must be an integer a between them*, i.e. bx<a<by. Hence x<\frac{a}{b}<y, showing that there is a rational number between x and y.

    *The set of integers \{n\in\mathbb{Z}:n\leq bx\} is nonempty (by the Archimedean principle) and bounded above, so it has a maximum member c. It follows that bx<c+1<by.

    For irrational, choose b>\frac{j}{y-x}, where j is any irrational number greater than 1 (e.g. \sqrt{2}). This should lead to x<\frac{aj}{b}<y.
    Well thank u so much.

    and thx to everyone who helped me
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  10. #10
    Senior Member JaneBennet's Avatar
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    For irrational, choose b>\frac{j}{y-x}, where j is any irrational number greater than 1 (e.g. \sqrt{2}). This should lead to x<\frac{aj}{b}<y.
    I’ve just realized that this would not work if a=0. This would only occur when x<0<y; in all other cases, it shouldn’t be a problem.

    Well, if x<0<y, then we just choose a positive integer b such that b>\frac{1}{y\sqrt{2}}. That will do.
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