# Proof verification of archimedean property through an intuition using the number line

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!

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.