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アルキメデスの原理（性質）Archimedean property
For any real number a, b> 0
There exists n ∈ N such that na> b ... ①
It is that.
I will prove it below (simplified version).
証明 Proof：
(* Use contradiction.)
In other words, from ① above,
it is said that b is not the upper bound of
na,
so b should be regarded as the upper bound of na to lead to a
contradiction.
If c is the supremum of na, then na ≤ c… ②.
Therefore, c−a <c, and this c−a is not the upper bound of c.
Therefore, there is something where c−a <na, and c <(n + 1) a.
However, since n + 1 ∈ N, this contradicts ②.
This is complete.
稠密性とは dense
For any real number a, b (where a <b), there exists x ∈ Q that satisfies
a <x <b.
It is that.
I will prove it below.
証明 Proof：
* Use the above Archimedean property.
Since a <b, there exists n ∈ N such that b−a> 0 and n (b−a)> 1.
Also, using the properties of Archimedean,
there exists $m_{1}$,$ m_{2}$ ∈ N
that satisfies $m_{1}$> na and $m_{2}$> −na.
Therefore,$ -m_{2}$ <na <$m_{1}$.
Therefore, it is assumed that there exists m that satisfies m-1 ≤ na <m,
and this m satisfies $−m_{2}$ ≤ m ≤ $m_{1}$.
Therefore, na <m ≤ 1 + na <nb.
* Finally, n (b−a)> 1 was used.
Now divide by n to get a <m / n <b.
If x = m / n, then a <x <b.
This is complete.
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