# The Archimedean Principle | Proof by Contradiction

I am a first-semester BSc. Engineering student and I have been studying the Archimedean Principle as part of my coursework. Recently, I came across Jay Cummings' proof for this principle in his book 'Real Analysis - A Long Form Mathematics Textbook'. While studying the proof, I encountered a difficulty and would appreciate some clarification.

The Archimedean Principle states that,

'Iff a and b are real numbers with a > 0, then there exists a natural number n such that na > b.'

In 'Real Analysis'; the author proves the Archimedean Principle using a proof by contradiction.

A summarized version of his proof is as follows:

Since $$a, b \in \mathbb R;$$ then, $$a/ b \in \mathbb R$$,
Let $$x = a/ b ,(x \in \mathbb R)$$,
Assume
for contradiction, that $$x$$ is an upper bound of $$\mathbb N$$.

The author is attempting to prove that,
Statement: $$(\forall x \in \mathbb R)$$, $$(\exists n \in \mathbb Z)$$, such that $$n> x$$. (For any real number $$x$$, there exists an integer $$n$$ such that $$n> x$$.)

Since $$\mathbb N \in \mathbb R$$ and $$\mathbb R$$ is complete, $$sup(\mathbb N)$$ exists.
Let $$sup(\mathbb N)= \alpha$$,
Then, $$\alpha - 1$$ is
not an upper bound of $$\mathbb N$$.
That is, $$\exists m \in (\mathbb Z,\mathbb N)$$ such that, $$m> \alpha -1$$.
Consequently, $$m+ 1> \alpha$$.

#Contradiction $$(sup(\mathbb N)= \alpha$$. But also, $$m+ 1> \alpha$$ and $$(m+ 1)\in \mathbb N)$$

Therefore, we can conclude that $$x$$ is not an upper bound of $$\mathbb N$$.
By substituting $$b/ a$$ for $$x$$ in Statement we get,
$$\forall b/ a \in \mathbb R$$; $$\exists n \in \mathbb Z$$, such that $$n> b/ a$$.
Therefore it follows that, $$\exists n \in \mathbb N$$, such that na> b. (There exists a natural number n, such that na> b.
)

However, I have a question regarding the proof.

Since we are considering the set $$\mathbb N$$, where the last element is $$sup(\mathbb N)$$ and the element before that is $$sup(\mathbb N) - 1$$ all the numbers between these two are decimals and are not included in $$\mathbb N$$. On the other hand, $$m$$ belongs to $$\mathbb N$$, so $$m \le sup(\mathbb N)$$. Therefore, how can there exist an integer $$m$$ such that, $$m> sup(\mathbb N)- 1$$?

Doesn't this imply that there exists an integer $$m$$; such that, $$sup(\mathbb N)- 1< m \le sup(\mathbb N)$$, while $$sup(\mathbb N)- 1$$ and $$sup(\mathbb N)$$ are the two closest integers?

$$\mathbb N$$ = {$$1, 2, 3, \dots, sup(\mathbb N)- 2, sup(\mathbb N)- 1, sup(\mathbb N)$$}

• $\Bbb N$ is not an element of $\Bbb R$, $\Bbb N$ is a subset of $\Bbb R$. Commented Jul 16, 2023 at 3:40

## 1 Answer

"Doesn't this imply that there exists an integer $$m$$; such that, $$sup(\mathbb N)- 1< m \le sup(\mathbb N)$$, while $$sup(\mathbb N)- 1$$ and $$sup(\mathbb N)$$ are the two closest integers?"

Yes, in fact, since $$\alpha-1$$ is not an upper bound of $$\mathbb{N}$$ it means that there exists at least one number from $$\mathbb{N}$$ that is larger than it, let it be $$m$$ like shown in the proof. As you said $$m\le \sup{(\mathbb{N})},m\in \mathbb N$$ and this means that $$m=\sup{(\mathbb N)}$$.

I hope you understand the rest of the proof from here.