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 \(j \in J\).
According to the way of this proposition, \(q\) can be expressed as \(m / 10^n\), where \(n \in \mathbb{N}\) and \(m \in \mathbb{Z}\), which is because we have used the decimal expressions of \(r_1\) and \(r_2\).
In fact, the expressions do not really need to be the decimals but can be the binaries, the hexadecimals, or the expressions with the base as any natural number larger than \(1\), \(b\), because Proof can be modified for any base with "\(9\)" replaced by \(b - 1\): for the binary case, \(1\), for the hexadecimal case, \(15\), then \(q\) can be expressed as \(m / b^n\).
\(q\) 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 \(q \in \mathbb{Q} \land r_1 \lt q \le r_2\).
