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Math Help - let a<b with a, b being real numbers and b-a>1. Show that a<m<b for a random m

  1. #1
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    let a<b with a, b being real numbers and b-a>1. Show that a<m<b for a random m

    Hi, I have the following three problems, but I only want to know what the first one is like... I am sure I will then be able to figure out the other two.

    The problem says:

    "In the following, a, b are real numbers.
    1) Let b>a+1. Show that there is an integer m such that a<m<b.
    2) Let a,b > 0. Show that there is an integer n such that n*a > b.
    3) Let a<b. Prove that there is an integer n such that n*b > n*a+1. Conclude that there is a rational number q such that a<q<b."

    I would really appreciate it if you could give me a hand with the first one so that I can see how to solve the other 2 problems. Thanks a lot!
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    Re: let a<b with a, b being real numbers and b-a>1. Show that a<m<b for a random m

    Let S be the set of all integers, x, with x< a. That set clearly is non-empty and has a as upper bound. By the "well ordered" property of the integers, any non-empty set of integers, having an upper bound, has a largest member. Let n be the largest member of this set. What can you say about n+1?
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  3. #3
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    Re: let a<b with a, b being real numbers and b-a>1. Show that a<m<b for a random m

    Quote Originally Posted by juanma101285 View Post
    The problem says:
    "In the following, a, b are real numbers.
    1) Let b>a+1. Show that there is an integer m such that a<m<b.
    I would really appreciate it if you could give me a hand with the first one so that I can see how to solve the other 2 problems.
    Theorm: If \alpha\in\mathbb{R} then \left( {\exists j \in \mathbb{Z}} \right)\left[ {j \leqslant \alpha  < j + 1} \right]
    The proof goes like this: Let A=\{n\in\mathbb{Z}:n\le\alpha\}.
    That set exists because the integers are not bounded below.
    Moreover the \sup(A)\in\mathbb{Z}. There is our j.

    That proves that the greatest integer \left\lfloor \alpha  \right\rfloor exists.

    Now for your question. Suppose that b>a+1.
    The theorem tells us that \left\lfloor a  \right\rfloor\le a<\left\lfloor a  \right\rfloor+1.
    So a<\left\lfloor a  \right\rfloor+1\le a+1<b

    EDIT this is a bit more. I did not see reply #2.
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