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Math Help - Proof of irrationality of e using series

  1. #1
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    Proof of irrationality of e using series

    1) If e were rational, it would be of the form e=\frac{p}{q}, where p and q are positive integers. Select an integer k such that k\geq 3 and k\geq q. Use Taylor's theorem to show that

    \frac{p}{q}=e=1+\frac{1}{1!}+\frac{1}{2!}+...+\fra  c{1}{k!}+\frac{e^z}{(k+1)!}

    for some z\in (0,1)

    Is this just the Taylor series about 0. and the end part is the Lagrange formula for the remainder?

    2) Suppose that:

    s_k=1+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{k!}

    Show that k!(e-s_k) is an integer

    3) Show that 0<k!(e-s_k)<1

    4) Conclude that e is irrational



    NOTE: I've posted my answer below. I just need someone to check whether I've done it correct. Thanks
    Last edited by acevipa; September 16th 2010 at 09:35 PM.
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  2. #2
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    Quote Originally Posted by acevipa View Post
    1) If e were rational, it would be of the form e=\frac{p}{q}, where p and q are positive integers. Select an integer k such that k\geq 3 and k\geq q. Use Taylor's theorem to show that

    \frac{p}{q}=e=1+\frac{1}{1!}+\frac{1}{2!}+...+\fra  c{1}{k!}+\frac{e^z}{(k+1)!}

    for some z\in (0,1)

    Is this just the Taylor series about 0. and the end part is the Lagrange formula for the remainder?


    Yes.


    2) Suppose that:

    s_k=1+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{k!}

    Show that k!(e-s_k) is an integer


    Check separatedly \,\,k!e\,,\,k!s_k


    3) Show that 0<k!(e-s_k)<1


    For this I think you need to already know that e<4 and do some simple algebra.

    4) Conclude that e is irrational


    Obvious from the above.

    Tonio




    .
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  3. #3
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    Thanks... I think I kind of understand it now.
    Last edited by acevipa; September 16th 2010 at 07:50 AM.
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  4. #4
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    So can someone verify my working, to make sure I haven't missed anything.

    2) \frac{p}{q}=e=1+\frac{1}{1!}+\frac{1}{2!}+...+\fra  c{1}{k!}+\frac{e^z}{(k+1)!} for some z\in(0,1)

    k!(e-s_k)=k!e-k!s_k

    =k!\left(\frac{p}{q}\right)-k!\left(1+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{k  !}\right)

    \Rightarrow\left(\frac{k!}{q}\right)p-\left(k!+\frac{k!}{1!}+\frac{k!}{2!}+...+\frac{k!}  {k!}\right)

    Since k\geq q\Rightarrow \frac{k!}{q} is an integer. Therefore since all the terms are integers, it follows that k!(e-s_k) must also be an integer



    3) e-s_k=\frac{e^z}{(k+1)!} for some z\in (0,1) from (1) (Taylor's theorem)

    \Rightarrow 0<k!(e-s_k)<\frac{e}{(k+1)}

    \Rightarrow 0<(e-s_k)<\frac{3}{(k+1)}

    Since k\geq 3\Rightarrow \frac{3}{(k+1)}<1

    \Rightarrow 0<(e-s_k)<1



    4) From (2), since k!(e-s_k) is an integer, but since 0<k!(e-s_k)<1.

    This gives us a contradiction, therefore, e is irrational
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