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I am confused on how the following proof works

given $ x

We may therefore suppose that $11$ for some integer $p$. Thus $1/p

if $\frac{k}{p}$ is the first rational number $>x$ then $\frac{k}{p}$ ought to be between $x$ and $y$, because $\frac{1}{p}

to be more precise, let $\mathbf{K}={\left \{ n | \frac{n}{p}>x \right \}}$

by the well ordering principle, $K$ has a least element $k$. Thus

$\frac{k}{p}>x$

But $\frac{k-1}{p}\leq x$

Therefore $\frac{k-1}{p} \leq x + \frac{1}{p} < x+e$

$\leq x+(y-x)$

Therefore $x\leq\frac{k}{p} \leq y$

Why is it ok to initially assume $x+n \leq r \leq y+n$ ? This is seems to be nearly assuming exactly what is trying to be proved. (I know this assumption is in P=>Q form but nontheless I do not understand how this helps.

How does $x \leq r'=r-n \leq y$ imply $1

Basically after that I get the jist of it but I am really confused on why the proof started the way it did.

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    good catch, its been been a long day lol – 2017-02-28

1 Answers 1

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The rationale as the following: You want to proof

If $x

The author first proves (as a lemma, if you prefer)

If $1

Once you have the lemma, you can prove the general theorem: There exists $n\in \Bbb N$ with $x+n>1$ (follows also from the Archimedean principle). Apply the lemma to $x'=x+n$, $y'=y+n$ and then let $r=r'-n$. You can then verify that $x'

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    perfect thanks for the help! – 2017-02-28
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    I feel a bit silly for asking this but since I'm learning this all myself I have no alternative. How does There exists $n\in \mathbb{N}$ with $x+n>1$ follow from For any $M$ and $e$ be any two positive integers, then there is a positive integer $n$ such that $ne>M$ – 2017-03-01
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    Here is how I think a proof for this would go. If $x\geq0$ let $n=2$ thus $x+2>1$ so $\exists n : x+n>1$. if $x<0$ let $n=-x+2$ so that $x+(2-x)>1$. Thus again $\exists n : x+n>1$. In any case $\exists n : x+n>1$. But this relies on $x$ being an integer and so I am not sure how Archimedes comes to the rescue here. – 2017-03-01