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    Made a big error in my post to this thread about the irrationality of rt(2)+e.[missed a 2 from one line to the next which gave a false proof]
    http://www.thestudentroom.co.uk/showthread.php?t=203073
    Just to redeem myself,and earn the rep i received for it, here is a solution :


    result 1) e^2 is irrational.
    pf

    assume not so we have

    e^2=a/b

    choose m large and set
    p=2^(m)
    q=p-m
    we have
    2^q divides p!
    proof
    highest power of 2 dividing p is given by
    [p/2]+[p/2^2]+...[p/2^(2m)]
    =2^(2m)-1 + 2^(2m-2)+.....+1
    =(2^(2^2m)-1)/1
    =2^p-1
    but
    2^q=2^(p-m)
    and p-m is less than 2^(p)-1 so result follows.

    we have

    p! 2^n
    ------ is an integer for n<p
    n! 2^q

    proof
    we need what power of 2 divides n!
    let r be the highest power of 2 such that [n/2^r] is non-zero
    the required power of 2 is
    [n/2]+[n/2^2]+..[n/2^r]<n{1/2+1/4+....1/2^r}
    =n(1-2(-r))
    <n
    so highest power of 2 dividing n! is bounded by 2^n
    cancelling these out with the 2^n leaves 2^(k) with k positive.
    The p! has enough powers of 2 to rid us of 2^q and the remaining
    numbers of n! since n<p when we have factored out the powers of 2.
    This leaves us integers on the numerator so the expression is an
    integer.

    now consider the
    p! 2^n
    the sum of the terms ------ for n=p+1 to oo
    n! 2^q
    =
    2^(p+1-q) ... 2^(p+2-q)
    --------- + --------- +.....
    ( p+1) ..... (p+1)(p+2)

    but q=p-m so the sum becomes

    2^(m+1) ... 2^(m+2)...
    --------- + --------- +...
    p+1 ......... (p+1)(p+2)

    <
    2^m[ 2 ...... 2^2
    ..... [ ---- + -----
    ...... [ p...... p^2
    =
    2^(m+1)
    -------
    ( p-2)
    which is less than 1 since p=2^2m

    Since the sum of terms with n > p, is not even close to 1, we can multiply by b*p!/2^q (instead of just p!/2^q) and then we just have to choose m to be large enough that the tail is *still* smaller than 1 and result(1) follows.



    result 2) 1,e,e^2 are linearly independent over Q.

    pf.

    assume not so there exist p,q,r in Q not all 0 s.t
    p+qe+re^2=0
    multiply to rid us of the denominators to get
    q'e+r'e^2+p'=0 for p',q',r' integers
    ie q'e+r'e^2 =p'' for some integer d
    but then
    q'+r'e=p''e^(-1)
    so there exists integers a,b st
    ae+be^(-1) is an integer
    but

    ae+be^(-1)=
    2a+sum (a+b) + .. (a-b)
    ........... ----- ... -----
    ............ (2k)! .... (2k+1)!

    let p>a+b+1

    for k<=p

    (2p+1)!(a+b) + ... (a-b)
    .......... ----- .... ------ is an integer.
    ......... (2k)! .... (2k+1)!

    for k>=p+1 we are left with the sum of terms
    (a+b) +.. (a-b) ............ (a+b)
    ----- ... ---------- +..... --------------------- +......
    (2p+2) .. (2p+2)(2p+3) .... (2p+2)(2p+3)(2p+4)

    < a+b . a+b
    ...--- + --- + ....
    ...p.......p^2

    =
    a+b [ 1 + 1 + 1
    ---..[......-- . ---
    ..p...[......p.. p^2

    =
    a+b
    ----
    p-1

    which is less than 1 by our choice of p.

    hence ae+be^(-1) is not integral.
    Therefore 1,e,e^2 are linearly independent over Q.

    Finally we have

    result (3) rt(2)+e is irrational

    pf
    assume not so
    rt(2)+e=a/b
    brt(2)=a-be
    2b^2=a^2+b^2e^-2abe
    b^2e^2-2abe+(a^2-2b^2).1=0
    this gives a linear dependence between 1,e,e^2 contradicting resul(2) hence
    rt(2)+e is irrational.

    Note also have rt(p)+e is irrational for all rt(p) irrational
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    now edited so it all lines up.
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    Wow... i guess that is redemption.
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    Will read it all soon. Excellent work!
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    (Original post by evariste)
    rt(2)+e=a/b
    brt(2)=a+be
    2b^2=a^2+b^2e^+2abe
    Surely thats brt(2)=a-be ??
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    (Original post by Matt_2K)
    Surely thats brt(2)=a-be ??
    Yes, but fortunately that doesn't affect the proof. We still have a dependece relation. Plus everything else in the proof is fine, as far as I can tell.
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    (Original post by Matt_2K)
    Surely thats brt(2)=a-be ??
    Yes you are correct, its now corrected.
 
 
 
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