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    Prove that there is a rational number in the open interval  (a,b), b>a, where  a, b \in \mathbb{R} .
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    http://homepages.warwick.ac.uk/~masdbl/w5.pdf

    Look at Assignment 1. See if that helps you.
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    (Original post by Ano123)
    Prove that there is a rational number in the open interval  (a,b), b>a, where  a, b \in \mathbb{R} .
    \frac{a+b}{2} is rational and within the range for any a,b \in \mathbb{R} thus statement is true. Not sure how to formally structure this sort of proof.
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    (Original post by RDKGames)
    \frac{a+b}{2} is rational and within the range for any a,b \in \mathbb{R} thus statement is true. Not sure how to formally structure this sort of proof.
    Let a=\sqrt{2} and b=1+\sqrt{2}

    \frac{a+b}{2} is not rational.
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    (Original post by Math12345)
    Let a=\sqrt{2} and b=1+\sqrt{2}

    (a+b)/2 is not rational.
    oh lol missed that where they don't say a and b are rationals, would be easy otherwise.
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    What have you tried?
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    If  b-a<1 , then  b-a=k, k\in \mathbb{R}^+ , k<1, if we multiply by some real constant  n so that  n(b-a)\geq 1, ( nk\geq 1 ) then we can say for sure that some integer  m lies in that interval which is of course rational. So from this we can deduce that  na<m<nb , dividing by  n we get  a<m/n<b .
    If  b-a>1 then an integer lies between the a and b which is obviously rational. This completes the proof.
 
 
 
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