# Unclear ideas in the proof of the Archimedean Principle

Here is how the principle is laid out in the text:

Given real numbers $$a$$ and $$b$$, with $$a \gt 0$$, there is an integer $$n \in \mathbb{N}$$ such that $$b \lt na$$

First off what is important about finding an $$n$$ that satisfies $$b \lt na$$

It talks about the strategy behind the proof and goes on to say that any non-empty subset of integers that is bounded above has "a largest integer". If $$k_0$$ is the largest integer that satisfies $$k_0a \leq b$$, then $$n=(k_0+1)$$ must satisfy $$na \gt b$$. In order to justify this application of the Completeness axiom, we have two details. 1) Is the set $$E:= \{k \space \in \mathbb{N} \space : \space ka \leq b\}$$ bounded above and 2) Is $$E$$ non-empty? This answer depends on whether $$b \lt a$$ or not.

What I don't understand is why is $$k_0$$ introduced? and why does it matter if $$k_0a \leq b$$? Where did the idea of $$k_0a \leq b$$ come from and why is it important? Why does $$E$$ being non-empty depend on whether $$b \lt a$$ or not? How does it follow that n = $$k_0 + 1$$ must satisfy $$b \lt na$$?

Then the proof is laid out:

If $$b \lt a$$ set $$n=1$$. If. If $$a \leq b$$, consider the set $$E:= \{k \space \in \mathbb{N} \space : \space ka \leq b\}$$. $$E$$ is nonempty since $$1 \in E$$. Let $$k \in E$$. Since $$a \gt 0$$ it follows from the first multiplicative property that $$k \leq \frac{b}{a}$$. This proves $$E$$ is bounded by $$\frac{b}{a}$$. Thus by the completeness axiom $$E$$ has a finite supremum $$s$$ that belongs to $$E$$. set $$n= s+1$$. Then $$n \in \mathbb{N}$$, n cannot belong to $$E$$ thus $$na \gt b$$.

Whats the significance of if $$b \lt a$$ then set $$n=1$$? I thought we were dealing with the set $$E$$. How is $$E$$ considered non-empty when $$1 \in E$$, by that I mean where was the $$1$$ pulled from? Just the fact that we are dealing with the set $$\mathbb{N}$$? I'm confused. What justifies setting $$n = k_0 + 1$$?

• ...by that I mean where was the 1 pulled from? Given $a\le b$ and defining $E:= \{k \space \in \mathbb{N} \space : \space ka \leq b\}$, it is obvious $1*a \le b$. So $k = 1\in E$ – rsadhvika Jul 15 at 6:01

First off what is important about finding an n that satisfies b < na

Nothing is important about finding the $$n$$. What's important is to understand that is you have real positive number $$a$$, no matter how small, and another positive real number $$b$$, no matter how big, you will be able to add up some number of the teeny-tiny $$a$$ and end up something bigger than the huge ginormous gargantuan $$b$$.

This is important. It means: The integers are not bounded. It means no number so large that in can not be surpassed by adding a smaller value enough times. It means by corrolary that no positive number is so small we can't find a smaller positive number. And it means, also as a corollary that two numbers, no matter how close they are together will always have a number between them.

Now these observations may seem obvious but that are not givens. They have to be proven.

What I don't understand is why is k0 introduced?

We need to rule out that we can't just keep adding $$a$$s together and never get above $$b$$. If the set of all integers $$k$$ where $$ak \le b$$ is bounded above then it must have a greatest element, $$k_0$$. If it has a greatest element then we are done; we just take the next element $$k_0 +1$$ it is not in the set of all such integers. So $$a(k_0 + 1) \not \le b$$ so $$a(k_0 + 1) > b$$.

So we would be done with our proof... IF we know the set was bounded above.

Why does E being non-empty depend on whether b

If $$b < a$$ then $$b < 1*a$$ and for any $$n \in \mathbb N$$ we have $$n \ge 1$$ so $$an > b$$ and there are no natural numbers where $$an \le b$$. So $$E$$ is empty.

If $$b\ge a$$ then $$a*1 \le b$$ and $$1 \in E$$ and $$E$$ is not empty.

How does it follow that n = k0+1 must satisfy b

$$k_0$$ is the largest natural number in $$E$$. $$n= k_0 + 1 > k_0$$. So $$n$$ is larger than then largest number in $$E$$. So $$n$$ is NOT in $$E$$. So it is not true that $$an \le b$$. So $$an > b$$.

Whats the significance of if b < a then set n=1?

There are two different proofs. There is one proof where $$a > b$$.

There is another if $$a\le b$$.

The proof if $$a > b$$ doesn't deal with $$E$$ at all. (For one thing, if $$a > b$$ then $$E$$ is empty.) For another if $$a > b$$ the proof is trivial.

I thought we were dealing with the set E. How is E considered non-empty when 1∈E

If $$a > b$$ then $$1\not \in E$$. That's the problem. If $$a > b$$ then we can't do the proof as it is presented If $$a > b$$ then $$E$$ is empty. So we have to do another proof. The other proof is one line long.

If $$a > b$$ then $$1*a > b$$ and therefore there an $$n$$ where $$na > b$$. Because $$1$$ can be that $$n$$.

But if $$a \le b$$ then we have to do the "real" proof.

Let $$E$$ but the set of all natural numbers, $$k$$ so that $$ak \le b$$.

If $$a\le b$$ then $$1*a \le b$$ so $$1 \in E$$. (IF $$a \le b$$). So $$E$$ is non-empty.

Let $$k \in E$$. That means $$ka \le b$$. So $$k \le \frac ba$$. So $$E$$ is bounded above by $$\frac ba$$. (IF $$a \le b$$... if $$a > b$$ then $$E$$ is empty and none of this matters...)

Since $$E$$ is bounded above it has a largest element. Call the largest number ..... you know what, DON'T call it $$k_0$$. That is tripping you up somehow. Call $$LOOKATMEIMTHEBIGGESTNUMBERINALLOFE$$. $$LOOKATMEIMTHEBIGGESTNUMBERINALLOFE$$ is the biggest number in all of $$E$$.

Now let $$HEYIMEVENBIGGERTHANLOOKATMEIMTHEBIGGESTNUMBERINALLOFE = LOOKATMEIMTHEBIGGESTNUMBERINALLOFE + 1$$. Now $$HEYIMEVENBIGGERTHANLOOKATMEIMTHEBIGGESTNUMBERINALLOFE$$ is even bigger than $$LOOKATMEIMTHEBIGGESTNUMBERINALLOFE$$ which is the biggest number in $$E$$. So $$HEYIMEVENBIGGERTHANLOOKATMEIMTHEBIGGESTNUMBERINALLOFE$$ is even bigger than the biggest number in $$E$$ so it is too big to be in $$E$$.

So $$HEYIMEVENBIGGERTHANLOOKATMEIMTHEBIGGESTNUMBERINALLOFE \not \in E$$ which means $$HEYIMEVENBIGGERTHANLOOKATMEIMTHEBIGGESTNUMBERINALLOFE\cdot a \not \le b$$ and $$HEYIMEVENBIGGERTHANLOOKATMEIMTHEBIGGESTNUMBERINALLOFE\cdot a> b$$

And that's the proof.

• Is this a deliberate attempt to stretch the powers of MathJax beyond its reasonable capabilities? – Asaf Karagila Jul 15 at 13:53
• If it's funny I'll claim it was my attempt. I was attempting to make it explicitely clear what the meaning of $k_0$ and $n= k_0+1$ were for. – fleablood Jul 15 at 15:29

Note that if $$b < a$$, it follows that $$b < 1 \cdot a$$, so the theorem is proved right away: There is a positive integer $$n$$, specifically $$1$$, such that $$b < n \cdot a.$$

If $$a \leq b,$$ we do not have the result instantaneously, so we resort to completeness axiom. We wish to find some positive integer $$n$$ such that $$b < n \cdot a.$$ We realize that if some positive integer $$x$$ has the desired property, any integer $$y>x$$ must have the property as well because $$b < x \cdot a < y \cdot a$$ since $$a, x, y > 0$$. What we have to determine then, is if there exists such $$x$$. Hence, we wish to know if there exists some boundary that separates positive integers that work and those that do NOT work. This makes us consider the set of all positive integers $$k$$ that do NOT satisfy the given property and see if this set is bounded above, i.e., contains the maximum.

Once you prove that the maximum exists, denoted $$k_0$$ in your proof, it follows that $$k_0+1$$ has to satisfy our desired property.