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I am currently going through a textbook called Elementary Real Analysis, and I've been trying to get down an intuitive proof of the archimedean property, stated as follows:

The set of natural numbers $\mathbb{N}$ has no upper bounds.

I have come across some other proof, but had a hard time getting an intuitive grasp of it so I attempted to work it out in this other way. My intuition goes as follows, in an attempt to prove by contradiction: supposing there is a number $b$ on a number line, by virtue of its place it will be greater than any natural number coming before it. Now, if we take a certain number, $a \in\mathbb{N}$, and sum it to itself such that $a*n$ is the last sum of $a$ before $b$ and $a(n + 1)$ ends up to the right of $b$, we'll have proven that there can be a natural number beyond $b$ and made a contradiction, since $\mathbb{N}$ is closed under multiplication.

Now obviously, this in its current form can't work because if $a = 1$, then we would have to let $n$ be greater than $b$, but if we let $n \in\mathbb{N}$ then it must be limited by $b$.

Based on this idea of summing $a$ to cross beyond $b$, I attempted to get something that behaves similarly to the a(n+1) that would cross beyond, could someone please check this proof?

Proof. Let $\mathbb{N}$ have an upper bound. Then it has a least upper bound. Let $b$ be this number. Let some number $α$ be s.t. $α > b$ and $α - a < b$. Let $n\in \mathbb{N}$ and $ß$ be some number s.t. $α = a*n + ß$ and $a*n + ß \leq a*n + a$. Then: $$α > b$$ $$a*n + ß > b$$ $$a*n + a > b$$

Because $a, b \in \mathbb{N}$, and $\mathbb{N}$ is closed under addition and multiplication, $a*n + a$ is a natural number and it is greater than b, which is a contradiction. $\square$

So, because of the way we've defined $α$, we can think of $ß$ as a small non-natural number added before the sum $an$, and we can think of both $b$ and $an + ß$ as lying between $an$ and $an + a$. And if $a*n + ß \leq a*n + a$, then $ß$ is smaller or equal to $a$, meaning that if we replace the little $ß$ at the beginning of our sum $a*n$, then $a*n + a$ is sure to cross beyond $b$.

If this doesn't work, any guidance is highly appreciated! I'm not used to writing proofs, so if this isn't how it's done, please let me know!

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As in your attempt, we start from the least upper bound property of the reals and let $b=\sup\Bbb N$. Then $b-1$ cannot be an upper bound for $\Bbb N$, hence there exists $n\in \Bbb N$ with $n>b-1$. But then $n+1>b$, constradiction.

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  • $\begingroup$ I came across this proof, but had a very hard time visualizing it so I tried to get something more intuitive. What I have a hard time understanding is, how can n > b - 1 be anything else than b in this case? Yes the algebra ends up making n + 1 > b, but I have no clue how it ended up there. Thanks for the contribution! $\endgroup$
    – shintuku
    Feb 27, 2020 at 22:38
  • $\begingroup$ I finally understood the proof you posted. In n > b - 1, since n cannot go beyond b, n is indeed equal to b. That is why n + 1 goes beyond b, which we write n + 1 > b. $\endgroup$
    – shintuku
    Feb 28, 2020 at 0:06

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