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Thread: One Question about Proof on Irrationality of e

  1. #1
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    One Question about Proof on Irrationality of e

    Dear all,

    I'm looking at this particular proof that $\displaystyle e $ is irrational. Please find it beneath my writing.

    This proof by contradiction assumes that $\displaystyle p,q \in \mathbb{Z} $. I understand this.

    However, while proving that $\displaystyle q! \cdot b \in \mathbb{Z}$ is the contradiction, the proof uses the second boxed inequality
    where

    $\displaystyle \frac{1}{q + 1} > \frac{1}{2} \Longleftrightarrow q + 1 > 2 \Longleftrightarrow q > 1 $

    and $\displaystyle q > 1 \Longleftrightarrow q + 2 > 3 > 2 \Longleftrightarrow q + 2 > 2 \Longleftrightarrow \frac{1}{q + 2} < \frac{1}{2}$

    and so on...

    However, although $\displaystyle e > 0$, it is entirely possible for $\displaystyle p \in \mathbb{R^-} $ AND $\displaystyle q \in \mathbb{R^-}$ since $\displaystyle p,q \in \mathbb{Z} $.

    In this case then the $\displaystyle q \ngtr 1 $, so the above steps and the proof itself fail?

    Thank you very much for your assistance.

    Proof:

    Last edited by scherz0; Apr 13th 2012 at 07:26 PM.
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  2. #2
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    Re: One Question about Proof on Irrationality of e

    Quote Originally Posted by scherz0 View Post
    Dear all,

    I'm looking at this particular proof that $\displaystyle e $ is irrational. Please find it beneath my writing.

    This proof by contradiction assumes that $\displaystyle p,q \in \mathbb{Z} $. I understand this.

    However, while proving that $\displaystyle q! \cdot b \in \mathbb{Z}$ is the contradiction, the proof uses the second boxed inequality
    where

    $\displaystyle \frac{1}{q + 1} > \frac{1}{2} \Longleftrightarrow q + 1 > 2 \Longleftrightarrow q > 1 $

    and $\displaystyle q > 1 \Longleftrightarrow q + 2 > 3 > 2 \Longleftrightarrow q + 2 > 2 \Longleftrightarrow \frac{1}{q + 2} < \frac{1}{2}$

    and so on...

    However, although $\displaystyle e > 0$, it is entirely possible for $\displaystyle p \in \mathbb{R^-} $ AND $\displaystyle q \in \mathbb{R^-}$ since $\displaystyle p,q \in \mathbb{Z} $.

    In this case then the $\displaystyle q \ngtr 1 $, so the above steps and the proof itself fail?

    Thank you very much for your assistance.

    Proof:

    It should be pretty obvious that q is a positive integer, because of how q is defined in a and b.

    And since the proof involves showing that e = p/q, and it can easily be seen that e > 0, since q > 0, then so must be p.
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