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 $n = k_0 + 1$?

  • $\begingroup$ ...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 $k = 1\in E$ $\endgroup$ – 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 $E$.



And that's the proof.

  • 1
    $\begingroup$ Is this a deliberate attempt to stretch the powers of MathJax beyond its reasonable capabilities? $\endgroup$ – Asaf Karagila Jul 15 at 13:53
  • $\begingroup$ 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. $\endgroup$ – 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.


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