description/proof of a way to systematically choose rational number that is larger than real number and is equal to or smaller than another real number
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of real numbers set.
- The reader knows a definition of rational numbers set.
Target Context
- The reader will have a description and a proof of a way for systematically choosing a rational number that is larger than any real number and is equal to or smaller than another any real number.
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:
\(r_1\): \(\in \mathbb{R}\)
\(r_2\): \(\in \mathbb{R}\), such that \(r_1 \lt r_2\)
//
Statements:
choose \(q\) as follows:
let \(r_1\) and \(r_2\) be expressed as the decimals without any trailing '999...'
when \(0 \le r_2\), when \(r_1 \lt 0\), take \(q = 0\), otherwise, take the 1st digit on which \(r_1\) and \(r_2\) disagree and take \(q\) as \(r_2\) with with the subsequent digits cut off
when \(r_2 \lt 0\), take the 1st digit on which \(r_1\) and \(r_2\) disagree, take the 1st digit of \(r_2\) after that that is not '9', and take \(q\) as \(r_2\) with the digit incremented by \(1\) and the subsequent digits cut off
\(\implies\)
\(q\) has been systematically chosen satisfying \(q \in \mathbb{Q} \land r_1 \lt q \le r_2\)
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2: Note
We cannot do like \(Max (\{q \in \mathbb{Q} \vert q \le r_2\})\), because such the maximum may not exist: what is the maximum when \(r_2 = \sqrt{2}\)? We cannot do like \(Sup (\{q \in \mathbb{Q} \vert q \le r_2\})\), because such the supremum may not exist in \(\mathbb{Q}\) but in \(\mathbb{R}\): what is the supremum in \(\mathbb{Q}\) when \(r_2 = \sqrt{2}\)?
When we have only some finite number of pairs, \((r_{1, 1}, r_{1, 2}), ..., (r_{n, 1}, r_{n, 2})\), we can just claim that we can choose some rational numbers, \(r_{1, 1} \lt q_1 \le r_{1, 2}, ..., r_{n, 1} \lt q_n \le r_{n, 2}\). But when we have (possibly uncountably) infinite number of pairs, \(\{(r_{j, 1}, r_{j, 2}) \vert j \in J\}\), there can be an objection against our just claiming that we can choose a \(q_j \in \mathbb{Q}\) such that \(r_{j, 1} \lt q_j \le r_{j, 2}\) for each \(j \in J\). The axiom of choice could be employed, but let us do without it. If there is a systematic way of choosing a \(q\) for any \((r_1, r_2)\), we can claim that we can choose a \(q_j \in \mathbb{Q}\) such that \(r_{j, 1} \lt q_j \le r_{j, 2}\) for each \(j \in J\) by that way.
Of course, there can be many other ways, but presenting a way is enough for our purpose of claiming that we can choose a \(q_j \in \mathbb{Q}\) for each \(j \in J\).
\(q\) can be also systematically chosen satisfying \(q \in \mathbb{Q} \land r_1 \lt q \lt r_2\), because we can take \((r_1 + r_2) / 2\) and take \(r_1 \lt q \le (r_1 + r_2) / 2 \lt r_2\), applying this proposition.
3: Proof
Whole Strategy: Step 1: let \(r_1\) and \(r_2\) be expressed as the decimals without any trailing '999...'; Step 2: when \(0 \le r_2\), take \(q\) as is mentioned in Statements, and see that \(q \in \mathbb{Q}\) and \(r_1 \lt q \le r_2\); Step 3: when \(r_2 \lt 0\), take \(q\) as is mentioned in Statements, and see that \(q \in \mathbb{Q}\) and \(r_1 \lt q \le r_2\); Step 4: conclude the proposition.
Step 1:
Let \(r_1\) and \(r_2\) be expressed as the decimals without any trailing '999...' (for example, '12.3' instead of '12.2999...'), which makes the expressions unique. When a decimal is finite, let the decimal have the trailing '000...'.
Step 2:
Let us suppose that \(0 \le r_2\).
\(r_1 \lt 0\) or \(0 \le r_1\).
When \(r_1 \lt 0\), let \(q = 0\).
Then, \(q \in \mathbb{Q}\) and \(r_0 \lt q \le r_2\).
Let us suppose that \(0 \le r_1\).
There is the 1st digit on which \(r_1\) and \(r_2\) disagree.
Let \(q\) be \(r_2\) with the subsequent digits cut off.
Then, \(q \in \mathbb{Q}\), because it has the finite decimal, and \(r_1 \lt q \le r_2\), obviously.
An example is \(r_1 = 12.344678..., r_2 = 12.345678...\), and \(q = 12.345\).
Step 3:
Let us suppose that \(r_2 \lt 0\).
There is the 1st digit on which \(r_1\) and \(r_2\) disagree.
There is the 1st digit of \(r_2\) after that that is not '9' (because the expression is without any trailing '999...').
Let \(q\) be \(r_2\) with the digit incremented by \(1\) and the subsequent digits cut off.
Then, \(q \in \mathbb{Q}\), because it has the finite decimal, and \(r_1 \lt q \le r_2\), because as the disagreeing digit is not changed, \(r_1 \lt q\) holds.
An example is \(r_1 = - 12.346678..., r_2 = - 12.3459678...\), and \(q = - 12.34597\).
Step 4:
\(q\) has been chosen without any arbitrariness satisfying \(q \in \mathbb{Q} \land r_1 \lt q \le r_2\).
