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Thread: Proof of irrationality of e using series

  1. #1
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    Proof of irrationality of e using series

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

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

    for some $\displaystyle z\in (0,1)$

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

    2) Suppose that:

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

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

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

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

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

    for some $\displaystyle z\in (0,1)$

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


    Yes.


    2) Suppose that:

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

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


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


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


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

    4) Conclude that $\displaystyle e$ 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; Sep 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) $\displaystyle \frac{p}{q}=e=1+\frac{1}{1!}+\frac{1}{2!}+...+\fra c{1}{k!}+\frac{e^z}{(k+1)!}$ for some $\displaystyle z\in(0,1)$

    $\displaystyle k!(e-s_k)=k!e-k!s_k$

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

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

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



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

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

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

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

    $\displaystyle \Rightarrow 0<(e-s_k)<1$



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

    This gives us a contradiction, therefore, $\displaystyle e$ is irrational
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