description/proof of that for \(2\) distinct non-negative real numbers and natural number larger than \(1\), there are 2nd and 3rd natural numbers s.t. 3rd number divided by number to power of 2nd number is exactly between real numbers
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of real numbers set.
Target Context
- The reader will have a description and a proof of the proposition that for any \(2\) distinct non-negative real numbers and any natural number larger than \(1\), there are some 2nd and 3rd natural numbers such that the 3rd number divided by the number to the power of the 2nd number is exactly between the real numbers.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(\{r_1, r_2\}\): \(\subseteq \mathbb{R}\), such that \(0 \le r_1, r_2 \land r_1 \lt r_2\)
\(n\): \(\in \mathbb{N} \setminus \{0, 1\}\)
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Statements:
\(\exists m, j \in \mathbb{N} (r_1 \lt j / n^m \lt r_2)\)
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2: Note
\(n \lt 1\) is required, because if \(n = 1\), \(n^m = 1\) for whatever \(m\) and \(j / n^m = j / 1 = j\) could not satisfy \(r_1 \lt j / n^m \lt r_2\) for \(r_1, r_2 \lt 1\).
3: Proof
Whole Strategy: Step 1: take an \(m\) such that \(1 \lt n^m (r_2 - r_1)\) and take a \(j\) such that \(n^m r_1 \lt j \lt n^m r_2\).
Step 1:
As \(0 \lt r_2 - r_1\) and \(n^m\) can be any larger by choosing a large \(m\) (because \(1 \lt n\)), let us choose an \(m \in \mathbb{N}\) such that \(1 \lt n^m (r_2 - r_1) = n^m r_2 - n^m r_1\).
There is a \(j \in \mathbb{N}\) such that \(n^m r_1 \lt j \lt n^m r_2\): when \(n^m r_1\) is a natural number, \(j := n^m r_1 + 1 \lt n^m r_2\) will do; otherwise, \(j\) can be taken to be the smallest natural number larger than \(n^m r_1\), then, \(j \lt n^m r_2\).
Then, \(r_1 \lt j / n^m \lt r_2\).
