Mixing
Are there any other books which adopt this axiom of the real numbers $mathbb{R}$?
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I am reading "Calculus" by Takeshi Saito.
In this book, Saito adopts the following axiom of $mathbb{R}$.
I like this axiom.
I think it is easy to understand what this axiom is saying.
But I cannot find a book in which this axiom of $mathbb{R}$ is adopted.
Are there any other books which adopt this axiom of $mathbb{R}$?
Axiom 1.1.1:
1. If $a$ is a real number, then there exists an integer $n$ such that $n leq a leq n+1$.
2. If ${a_n}$ is a sequence such that $a_i in {0, 1}$ for all $i in {1, 2, cdots}$, then there exists a real number $b$ such that $$sum_{n=1}^m frac{a_n}{2^n} leq b leq sum_{n=1}^m frac{a_n}{2^n}+frac{1}{2^m}$$ for all $m in {0, 1, 2, cdots}$.
I will write how to use this axiom to prove an important theorem.
By Axiom 1.1.1.1, if $a$ is a real number, there exists a unique integer such that $m leq a < m+1$ and we define this $m$ as $[a]$. $[a]+1$ is the smallest integer which is greater than $a$.
Proposition 1.1.2:
Let $a, b$ be real numbers.
1. If $a < b$, then there exists $r in mathbb{Q}$ such that $a < r < b$.
2. If $|a-b|<frac{1}{n}$ for all $n in {1, 2, cdots}$, then $a=b$.
Proof:
1. Let $n:=[frac{1}{b-a}]+1$. Then, $n$ is the smallest integer such that $n > frac{1}{b-a} > 0$.
Let $m:=[na]+1$. Then, $na<mleq na+1 < nb$, so the rational number $r := frac{m}{n}$ satisfies $a < r < b$.
2. If $|a-b| > 0$, then, by Axiom 1.1.1.1, there exists $n in {1,2,cdots}$ such that $frac{1}{n} leq |a-b| < frac{1}{n}$. But this is a contradiction. So, $|a-b|=0$. So $a = b$.
Corollary 1.1.3:
If ${a_n}$ is a sequence such that $a_i in {0, 1}$ for all $i in {1, 2, cdots}$, then, there exists a unique real number $b$ such that $$sum_{n=1}^m frac{a_n}{2^n} leq b leq sum_{n=1}^m frac{a_n}{2^n}+frac{1}{2^m}$$ for all $m in {0, 1, 2, cdots}$.
Proof:
If $b$ and $c$ satisfy $$sum_{n=1}^m frac{a_n}{2^n} leq b leq sum_{n=1}^m frac{a_n}{2^n}+frac{1}{2^m}$$ $$sum_{n=1}^m frac{a_n}{2^n} leq c leq sum_{n=1}^m frac{a_n}{2^n}+frac{1}{2^m}$$ for all $m in {0, 1, 2, cdots}$, then $|b-c| leq frac{1}{2^m} < frac{1}{m}$. By Proposition 1.1.2.2, $b=c$.
We write this unique real number $b$ as $sum_{n=1}^{infty} frac{a_n}{2^n}$.
Theorem 1.1.4
Let $a, b$ be real numbers such that $a leq b$.
Let $A$ be a subset of $[a, b]$ which satisfies the following condition (D):
(D) If $x in A$, then $[a, x] subset A$.
Then, $A = [a, c]$ or $A = [a, c)$ for some $c in [a, b]$.
Proof:
First, we show this theorem when $a = 0, b = 1$.
By the condition (D), if $0 notin A$, then $A = emptyset = [0, 0)$.
By the condition (D), if $1 in A$, then $A = [0, 1]$.
So, we assume that $0 in A$ and $1 notin A$.
We define a sequence ${a_n}$ inductively as follows:
Let $s_0 := 0$
If $s_0 + frac{1}{2^{0+1}} in A$, then $a_1 := 1$.
If $s_0 + frac{1}{2^{0+1}} notin A$, then $a_1:=0$.
After $a_1, a_2, cdots, a_m$ are defined, let $s_m := sum_{n=1}^m frac{a_n}{2^n}$.
If $s_m + frac{1}{2^{m+1}} in A$, then $a_{m+1} := 1$.
If $s_m + frac{1}{2^{m+1}} notin A$, then $a_{m+1} := 0$.
By the definition of ${a_n}$ and by induction on $m$,
for all $m in {0, 1, 2, cdots }$, $s_m in A$ and $s_m + frac{1}{2^m} notin A$.
By Axiom 1.1.1.2, there exists a (unique) real number $c := sum_{n=1}^{infty} frac{a_n}{2^n}$ such that $$sum_{n=1}^m frac{a_n}{2^n} leq c leq sum_{n=1}^m frac{a_n}{2^n}+frac{1}{2^m}$$ for all $m in {0, 1, 2, cdots}$.
Now we prove that $A = [0, c)$ or $A = [0, c]$.
To prove this, we prove that $[0, c) subset A$ and $A cap (c, 1] = emptyset$.
If $x in [0, c)$, then, by Proposition 1.1.2.1, there exists an integer $m in {1, 2, cdots }$ such that $c - x geq frac{1}{m}$.
Then, $x leq c-frac{1}{m} leq c-frac{1}{2^m} leq s_m$ and $s_m in A$.
So, by the condition (D), $x in A$.
$therefore [0, c) subset A$.
If $c < x leq 1$, then by Proposition 1.1.2.1, there exists an integer $m in {1, 2, cdots }$ such that $x - c geq frac{1}{m}$.
Then, $x geq c+frac{1}{m} geq c+frac{1}{2^m} geq s_m+frac{1}{2^m}$ and $s_m+frac{1}{2^m} notin A$.
So, by the condition (D), $x notin A$.
$therefore A cap (c, 1] = emptyset$.
Now we prove the general case.
If $a = b$, then let $c = a = b$.
Then, $A = [a, c]$ or $A = emptyset = [a, c)$.
If $a < b$, then let $A' := {frac{x-a}{b-a} | x in A }$.
Then, $A' subset [0, 1]$ and $A'$ satisfies the condition (D).
Then, there exists $c'$ such that $A' = [0, c']$ or $A' = [0, c')$.
Then, $A = [a, a+(b-a)c']$ or $A = [a, a+(b-a)c')$.
calculus reference-request book-recommendation
asked Jan 29 at 13:38
tchappy hatchappy ha
788412
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show 1 more comment
$begingroup$
I am reading "Calculus" by Takeshi Saito.
In this book, Saito adopts the following axiom of $mathbb{R}$.
I like this axiom.
I think it is easy to understand what this axiom is saying.
But I cannot find a book in which this axiom of $mathbb{R}$ is adopted.
Are there any other books which adopt this axiom of $mathbb{R}$?
Axiom 1.1.1:
1. If $a$ is a real number, then there exists an integer $n$ such that $n leq a leq n+1$.
2. If ${a_n}$ is a sequence such that $a_i in {0, 1}$ for all $i in {1, 2, cdots}$, then there exists a real number $b$ such that $$sum_{n=1}^m frac{a_n}{2^n} leq b leq sum_{n=1}^m frac{a_n}{2^n}+frac{1}{2^m}$$ for all $m in {0, 1, 2, cdots}$.
I will write how to use this axiom to prove an important theorem.
By Axiom 1.1.1.1, if $a$ is a real number, there exists a unique integer such that $m leq a < m+1$ and we define this $m$ as $[a]$. $[a]+1$ is the smallest integer which is greater than $a$.
Proposition 1.1.2:
Let $a, b$ be real numbers.
1. If $a < b$, then there exists $r in mathbb{Q}$ such that $a < r < b$.
2. If $|a-b|<frac{1}{n}$ for all $n in {1, 2, cdots}$, then $a=b$.
Proof:
1. Let $n:=[frac{1}{b-a}]+1$. Then, $n$ is the smallest integer such that $n > frac{1}{b-a} > 0$.
Let $m:=[na]+1$. Then, $na<mleq na+1 < nb$, so the rational number $r := frac{m}{n}$ satisfies $a < r < b$.
2. If $|a-b| > 0$, then, by Axiom 1.1.1.1, there exists $n in {1,2,cdots}$ such that $frac{1}{n} leq |a-b| < frac{1}{n}$. But this is a contradiction. So, $|a-b|=0$. So $a = b$.
Corollary 1.1.3:
If ${a_n}$ is a sequence such that $a_i in {0, 1}$ for all $i in {1, 2, cdots}$, then, there exists a unique real number $b$ such that $$sum_{n=1}^m frac{a_n}{2^n} leq b leq sum_{n=1}^m frac{a_n}{2^n}+frac{1}{2^m}$$ for all $m in {0, 1, 2, cdots}$.
Proof:
If $b$ and $c$ satisfy $$sum_{n=1}^m frac{a_n}{2^n} leq b leq sum_{n=1}^m frac{a_n}{2^n}+frac{1}{2^m}$$ $$sum_{n=1}^m frac{a_n}{2^n} leq c leq sum_{n=1}^m frac{a_n}{2^n}+frac{1}{2^m}$$ for all $m in {0, 1, 2, cdots}$, then $|b-c| leq frac{1}{2^m} < frac{1}{m}$. By Proposition 1.1.2.2, $b=c$.
We write this unique real number $b$ as $sum_{n=1}^{infty} frac{a_n}{2^n}$.
Theorem 1.1.4
Let $a, b$ be real numbers such that $a leq b$.
Let $A$ be a subset of $[a, b]$ which satisfies the following condition (D):
(D) If $x in A$, then $[a, x] subset A$.
Then, $A = [a, c]$ or $A = [a, c)$ for some $c in [a, b]$.
Proof:
First, we show this theorem when $a = 0, b = 1$.
By the condition (D), if $0 notin A$, then $A = emptyset = [0, 0)$.
By the condition (D), if $1 in A$, then $A = [0, 1]$.
So, we assume that $0 in A$ and $1 notin A$.
We define a sequence ${a_n}$ inductively as follows:
Let $s_0 := 0$
If $s_0 + frac{1}{2^{0+1}} in A$, then $a_1 := 1$.
If $s_0 + frac{1}{2^{0+1}} notin A$, then $a_1:=0$.
After $a_1, a_2, cdots, a_m$ are defined, let $s_m := sum_{n=1}^m frac{a_n}{2^n}$.
If $s_m + frac{1}{2^{m+1}} in A$, then $a_{m+1} := 1$.
If $s_m + frac{1}{2^{m+1}} notin A$, then $a_{m+1} := 0$.
By the definition of ${a_n}$ and by induction on $m$,
for all $m in {0, 1, 2, cdots }$, $s_m in A$ and $s_m + frac{1}{2^m} notin A$.
By Axiom 1.1.1.2, there exists a (unique) real number $c := sum_{n=1}^{infty} frac{a_n}{2^n}$ such that $$sum_{n=1}^m frac{a_n}{2^n} leq c leq sum_{n=1}^m frac{a_n}{2^n}+frac{1}{2^m}$$ for all $m in {0, 1, 2, cdots}$.
Now we prove that $A = [0, c)$ or $A = [0, c]$.
To prove this, we prove that $[0, c) subset A$ and $A cap (c, 1] = emptyset$.
If $x in [0, c)$, then, by Proposition 1.1.2.1, there exists an integer $m in {1, 2, cdots }$ such that $c - x geq frac{1}{m}$.
Then, $x leq c-frac{1}{m} leq c-frac{1}{2^m} leq s_m$ and $s_m in A$.
So, by the condition (D), $x in A$.
$therefore [0, c) subset A$.
If $c < x leq 1$, then by Proposition 1.1.2.1, there exists an integer $m in {1, 2, cdots }$ such that $x - c geq frac{1}{m}$.
Then, $x geq c+frac{1}{m} geq c+frac{1}{2^m} geq s_m+frac{1}{2^m}$ and $s_m+frac{1}{2^m} notin A$.
So, by the condition (D), $x notin A$.
$therefore A cap (c, 1] = emptyset$.
Now we prove the general case.
If $a = b$, then let $c = a = b$.
Then, $A = [a, c]$ or $A = emptyset = [a, c)$.
If $a < b$, then let $A' := {frac{x-a}{b-a} | x in A }$.
Then, $A' subset [0, 1]$ and $A'$ satisfies the condition (D).
Then, there exists $c'$ such that $A' = [0, c']$ or $A' = [0, c')$.
Then, $A = [a, a+(b-a)c']$ or $A = [a, a+(b-a)c')$.
calculus reference-request book-recommendation
asked Jan 29 at 13:38
tchappy hatchappy ha
788412
$endgroup$
$begingroup$
seems like the Archimedean property
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– J. W. Tanner
Jan 29 at 14:49
1
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The axiom equates real numbers with infinite binary expansion. Another base such as decimal would be equivalent.
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– Somos
Jan 29 at 20:24
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Can you provide a link to that "Calculus" by Takeshi Saito?
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– user635162
Jan 30 at 0:27
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The question-part of your question ends with the statement of the axiom. What is the point of the Proposition-Corollary-Theorem and discussion that follow?
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– Gerry Myerson
Jan 30 at 0:39
1
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@palaeomathematician Calculus
$endgroup$
– tchappy ha
Jan 30 at 0:45
|
show 1 more comment
$begingroup$
I am reading "Calculus" by Takeshi Saito.
In this book, Saito adopts the following axiom of $mathbb{R}$.
I like this axiom.
I think it is easy to understand what this axiom is saying.
But I cannot find a book in which this axiom of $mathbb{R}$ is adopted.
Are there any other books which adopt this axiom of $mathbb{R}$?
Axiom 1.1.1:
1. If $a$ is a real number, then there exists an integer $n$ such that $n leq a leq n+1$.
2. If ${a_n}$ is a sequence such that $a_i in {0, 1}$ for all $i in {1, 2, cdots}$, then there exists a real number $b$ such that $$sum_{n=1}^m frac{a_n}{2^n} leq b leq sum_{n=1}^m frac{a_n}{2^n}+frac{1}{2^m}$$ for all $m in {0, 1, 2, cdots}$.
I will write how to use this axiom to prove an important theorem.
By Axiom 1.1.1.1, if $a$ is a real number, there exists a unique integer such that $m leq a < m+1$ and we define this $m$ as $[a]$. $[a]+1$ is the smallest integer which is greater than $a$.
Proposition 1.1.2:
Let $a, b$ be real numbers.
1. If $a < b$, then there exists $r in mathbb{Q}$ such that $a < r < b$.
2. If $|a-b|<frac{1}{n}$ for all $n in {1, 2, cdots}$, then $a=b$.
Proof:
1. Let $n:=[frac{1}{b-a}]+1$. Then, $n$ is the smallest integer such that $n > frac{1}{b-a} > 0$.
Let $m:=[na]+1$. Then, $na<mleq na+1 < nb$, so the rational number $r := frac{m}{n}$ satisfies $a < r < b$.
2. If $|a-b| > 0$, then, by Axiom 1.1.1.1, there exists $n in {1,2,cdots}$ such that $frac{1}{n} leq |a-b| < frac{1}{n}$. But this is a contradiction. So, $|a-b|=0$. So $a = b$.
Corollary 1.1.3:
If ${a_n}$ is a sequence such that $a_i in {0, 1}$ for all $i in {1, 2, cdots}$, then, there exists a unique real number $b$ such that $$sum_{n=1}^m frac{a_n}{2^n} leq b leq sum_{n=1}^m frac{a_n}{2^n}+frac{1}{2^m}$$ for all $m in {0, 1, 2, cdots}$.
Proof:
If $b$ and $c$ satisfy $$sum_{n=1}^m frac{a_n}{2^n} leq b leq sum_{n=1}^m frac{a_n}{2^n}+frac{1}{2^m}$$ $$sum_{n=1}^m frac{a_n}{2^n} leq c leq sum_{n=1}^m frac{a_n}{2^n}+frac{1}{2^m}$$ for all $m in {0, 1, 2, cdots}$, then $|b-c| leq frac{1}{2^m} < frac{1}{m}$. By Proposition 1.1.2.2, $b=c$.
We write this unique real number $b$ as $sum_{n=1}^{infty} frac{a_n}{2^n}$.
Theorem 1.1.4
Let $a, b$ be real numbers such that $a leq b$.
Let $A$ be a subset of $[a, b]$ which satisfies the following condition (D):
(D) If $x in A$, then $[a, x] subset A$.
Then, $A = [a, c]$ or $A = [a, c)$ for some $c in [a, b]$.
Proof:
First, we show this theorem when $a = 0, b = 1$.
By the condition (D), if $0 notin A$, then $A = emptyset = [0, 0)$.
By the condition (D), if $1 in A$, then $A = [0, 1]$.
So, we assume that $0 in A$ and $1 notin A$.
We define a sequence ${a_n}$ inductively as follows:
Let $s_0 := 0$
If $s_0 + frac{1}{2^{0+1}} in A$, then $a_1 := 1$.
If $s_0 + frac{1}{2^{0+1}} notin A$, then $a_1:=0$.
After $a_1, a_2, cdots, a_m$ are defined, let $s_m := sum_{n=1}^m frac{a_n}{2^n}$.
If $s_m + frac{1}{2^{m+1}} in A$, then $a_{m+1} := 1$.
If $s_m + frac{1}{2^{m+1}} notin A$, then $a_{m+1} := 0$.
By the definition of ${a_n}$ and by induction on $m$,
for all $m in {0, 1, 2, cdots }$, $s_m in A$ and $s_m + frac{1}{2^m} notin A$.
By Axiom 1.1.1.2, there exists a (unique) real number $c := sum_{n=1}^{infty} frac{a_n}{2^n}$ such that $$sum_{n=1}^m frac{a_n}{2^n} leq c leq sum_{n=1}^m frac{a_n}{2^n}+frac{1}{2^m}$$ for all $m in {0, 1, 2, cdots}$.
Now we prove that $A = [0, c)$ or $A = [0, c]$.
To prove this, we prove that $[0, c) subset A$ and $A cap (c, 1] = emptyset$.
If $x in [0, c)$, then, by Proposition 1.1.2.1, there exists an integer $m in {1, 2, cdots }$ such that $c - x geq frac{1}{m}$.
Then, $x leq c-frac{1}{m} leq c-frac{1}{2^m} leq s_m$ and $s_m in A$.
So, by the condition (D), $x in A$.
$therefore [0, c) subset A$.
If $c < x leq 1$, then by Proposition 1.1.2.1, there exists an integer $m in {1, 2, cdots }$ such that $x - c geq frac{1}{m}$.
Then, $x geq c+frac{1}{m} geq c+frac{1}{2^m} geq s_m+frac{1}{2^m}$ and $s_m+frac{1}{2^m} notin A$.
So, by the condition (D), $x notin A$.
$therefore A cap (c, 1] = emptyset$.
Now we prove the general case.
If $a = b$, then let $c = a = b$.
Then, $A = [a, c]$ or $A = emptyset = [a, c)$.
If $a < b$, then let $A' := {frac{x-a}{b-a} | x in A }$.
Then, $A' subset [0, 1]$ and $A'$ satisfies the condition (D).
Then, there exists $c'$ such that $A' = [0, c']$ or $A' = [0, c')$.
Then, $A = [a, a+(b-a)c']$ or $A = [a, a+(b-a)c')$.
calculus reference-request book-recommendation
asked Jan 29 at 13:38
tchappy hatchappy ha
788412
$endgroup$
I am reading "Calculus" by Takeshi Saito.
In this book, Saito adopts the following axiom of $mathbb{R}$.
I like this axiom.
I think it is easy to understand what this axiom is saying.
But I cannot find a book in which this axiom of $mathbb{R}$ is adopted.
Are there any other books which adopt this axiom of $mathbb{R}$?
Axiom 1.1.1:
1. If $a$ is a real number, then there exists an integer $n$ such that $n leq a leq n+1$.
2. If ${a_n}$ is a sequence such that $a_i in {0, 1}$ for all $i in {1, 2, cdots}$, then there exists a real number $b$ such that $$sum_{n=1}^m frac{a_n}{2^n} leq b leq sum_{n=1}^m frac{a_n}{2^n}+frac{1}{2^m}$$ for all $m in {0, 1, 2, cdots}$.
I will write how to use this axiom to prove an important theorem.
By Axiom 1.1.1.1, if $a$ is a real number, there exists a unique integer such that $m leq a < m+1$ and we define this $m$ as $[a]$. $[a]+1$ is the smallest integer which is greater than $a$.
Proposition 1.1.2:
Let $a, b$ be real numbers.
1. If $a < b$, then there exists $r in mathbb{Q}$ such that $a < r < b$.
2. If $|a-b|<frac{1}{n}$ for all $n in {1, 2, cdots}$, then $a=b$.
Proof:
1. Let $n:=[frac{1}{b-a}]+1$. Then, $n$ is the smallest integer such that $n > frac{1}{b-a} > 0$.
Let $m:=[na]+1$. Then, $na<mleq na+1 < nb$, so the rational number $r := frac{m}{n}$ satisfies $a < r < b$.
2. If $|a-b| > 0$, then, by Axiom 1.1.1.1, there exists $n in {1,2,cdots}$ such that $frac{1}{n} leq |a-b| < frac{1}{n}$. But this is a contradiction. So, $|a-b|=0$. So $a = b$.
Corollary 1.1.3:
If ${a_n}$ is a sequence such that $a_i in {0, 1}$ for all $i in {1, 2, cdots}$, then, there exists a unique real number $b$ such that $$sum_{n=1}^m frac{a_n}{2^n} leq b leq sum_{n=1}^m frac{a_n}{2^n}+frac{1}{2^m}$$ for all $m in {0, 1, 2, cdots}$.
Proof:
If $b$ and $c$ satisfy $$sum_{n=1}^m frac{a_n}{2^n} leq b leq sum_{n=1}^m frac{a_n}{2^n}+frac{1}{2^m}$$ $$sum_{n=1}^m frac{a_n}{2^n} leq c leq sum_{n=1}^m frac{a_n}{2^n}+frac{1}{2^m}$$ for all $m in {0, 1, 2, cdots}$, then $|b-c| leq frac{1}{2^m} < frac{1}{m}$. By Proposition 1.1.2.2, $b=c$.
We write this unique real number $b$ as $sum_{n=1}^{infty} frac{a_n}{2^n}$.
Theorem 1.1.4
Let $a, b$ be real numbers such that $a leq b$.
Let $A$ be a subset of $[a, b]$ which satisfies the following condition (D):
(D) If $x in A$, then $[a, x] subset A$.
Then, $A = [a, c]$ or $A = [a, c)$ for some $c in [a, b]$.
Proof:
First, we show this theorem when $a = 0, b = 1$.
By the condition (D), if $0 notin A$, then $A = emptyset = [0, 0)$.
By the condition (D), if $1 in A$, then $A = [0, 1]$.
So, we assume that $0 in A$ and $1 notin A$.
We define a sequence ${a_n}$ inductively as follows:
Let $s_0 := 0$
If $s_0 + frac{1}{2^{0+1}} in A$, then $a_1 := 1$.
If $s_0 + frac{1}{2^{0+1}} notin A$, then $a_1:=0$.
After $a_1, a_2, cdots, a_m$ are defined, let $s_m := sum_{n=1}^m frac{a_n}{2^n}$.
If $s_m + frac{1}{2^{m+1}} in A$, then $a_{m+1} := 1$.
If $s_m + frac{1}{2^{m+1}} notin A$, then $a_{m+1} := 0$.
By the definition of ${a_n}$ and by induction on $m$,
for all $m in {0, 1, 2, cdots }$, $s_m in A$ and $s_m + frac{1}{2^m} notin A$.
By Axiom 1.1.1.2, there exists a (unique) real number $c := sum_{n=1}^{infty} frac{a_n}{2^n}$ such that $$sum_{n=1}^m frac{a_n}{2^n} leq c leq sum_{n=1}^m frac{a_n}{2^n}+frac{1}{2^m}$$ for all $m in {0, 1, 2, cdots}$.
Now we prove that $A = [0, c)$ or $A = [0, c]$.
To prove this, we prove that $[0, c) subset A$ and $A cap (c, 1] = emptyset$.
If $x in [0, c)$, then, by Proposition 1.1.2.1, there exists an integer $m in {1, 2, cdots }$ such that $c - x geq frac{1}{m}$.
Then, $x leq c-frac{1}{m} leq c-frac{1}{2^m} leq s_m$ and $s_m in A$.
So, by the condition (D), $x in A$.
$therefore [0, c) subset A$.
If $c < x leq 1$, then by Proposition 1.1.2.1, there exists an integer $m in {1, 2, cdots }$ such that $x - c geq frac{1}{m}$.
Then, $x geq c+frac{1}{m} geq c+frac{1}{2^m} geq s_m+frac{1}{2^m}$ and $s_m+frac{1}{2^m} notin A$.
So, by the condition (D), $x notin A$.
$therefore A cap (c, 1] = emptyset$.
Now we prove the general case.
If $a = b$, then let $c = a = b$.
Then, $A = [a, c]$ or $A = emptyset = [a, c)$.
If $a < b$, then let $A' := {frac{x-a}{b-a} | x in A }$.
Then, $A' subset [0, 1]$ and $A'$ satisfies the condition (D).
Then, there exists $c'$ such that $A' = [0, c']$ or $A' = [0, c')$.
Then, $A = [a, a+(b-a)c']$ or $A = [a, a+(b-a)c')$.
calculus reference-request book-recommendation
calculus reference-request book-recommendation
asked Jan 29 at 13:38
tchappy hatchappy ha
788412
asked Jan 29 at 13:38
tchappy hatchappy ha
788412
edited Jan 30 at 0:50
tchappy ha
asked Jan 29 at 13:38
tchappy hatchappy ha
788412
asked Jan 29 at 13:38
tchappy hatchappy ha
788412
asked Jan 29 at 13:38
tchappy hatchappy ha
788412
788412
$begingroup$
seems like the Archimedean property
$endgroup$
– J. W. Tanner
Jan 29 at 14:49
1
$begingroup$
The axiom equates real numbers with infinite binary expansion. Another base such as decimal would be equivalent.
$endgroup$
– Somos
Jan 29 at 20:24
$begingroup$
Can you provide a link to that "Calculus" by Takeshi Saito?
$endgroup$
– user635162
Jan 30 at 0:27
$begingroup$
The question-part of your question ends with the statement of the axiom. What is the point of the Proposition-Corollary-Theorem and discussion that follow?
$endgroup$
– Gerry Myerson
Jan 30 at 0:39
1
$begingroup$
@palaeomathematician Calculus
$endgroup$
– tchappy ha
Jan 30 at 0:45
|
show 1 more comment
$begingroup$
seems like the Archimedean property
$endgroup$
– J. W. Tanner
Jan 29 at 14:49
1
$begingroup$
The axiom equates real numbers with infinite binary expansion. Another base such as decimal would be equivalent.
$endgroup$
– Somos
Jan 29 at 20:24
$begingroup$
Can you provide a link to that "Calculus" by Takeshi Saito?
$endgroup$
– user635162
Jan 30 at 0:27
$begingroup$
The question-part of your question ends with the statement of the axiom. What is the point of the Proposition-Corollary-Theorem and discussion that follow?
$endgroup$
– Gerry Myerson
Jan 30 at 0:39
1
$begingroup$
@palaeomathematician Calculus
$endgroup$
– tchappy ha
Jan 30 at 0:45
$begingroup$
seems like the Archimedean property
$endgroup$
– J. W. Tanner
Jan 29 at 14:49
$begingroup$
seems like the Archimedean property
$endgroup$
– J. W. Tanner
Jan 29 at 14:49
1
1
$begingroup$
The axiom equates real numbers with infinite binary expansion. Another base such as decimal would be equivalent.
$endgroup$
– Somos
Jan 29 at 20:24
$begingroup$
The axiom equates real numbers with infinite binary expansion. Another base such as decimal would be equivalent.
$endgroup$
– Somos
Jan 29 at 20:24
$begingroup$
Can you provide a link to that "Calculus" by Takeshi Saito?
$endgroup$
– user635162
Jan 30 at 0:27
$begingroup$
Can you provide a link to that "Calculus" by Takeshi Saito?
$endgroup$
– user635162
Jan 30 at 0:27
$begingroup$
The question-part of your question ends with the statement of the axiom. What is the point of the Proposition-Corollary-Theorem and discussion that follow?
$endgroup$
– Gerry Myerson
Jan 30 at 0:39
$begingroup$
The question-part of your question ends with the statement of the axiom. What is the point of the Proposition-Corollary-Theorem and discussion that follow?
$endgroup$
– Gerry Myerson
Jan 30 at 0:39
1
1
$begingroup$
@palaeomathematician Calculus
$endgroup$
– tchappy ha
Jan 30 at 0:45
$begingroup$
@palaeomathematician Calculus
$endgroup$
– tchappy ha
Jan 30 at 0:45
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
This axiom would be proved, ordinarily, by the archimidean property, as J.W.Tanner points out. It's actually not a great "axiom", because it includes things that can be proved from other axioms. Knowing that there's an $n+1$ that's greater than or equal to $x$ -- that's good. There's a least such number by well-ordering. For that $n+1$, we have that $n <= x$. So the left hand side of part 1 is redundant.
Part ii hardly seems to be an axiom at all, for it's easy to prove (assuming you've done the order axioms right) that between any two reals, there's another real, so between the left hand sum, and the LHS plus a little increment, there's a real.
Now you can have, as axioms, things that follow from other axioms -- it's not against the rules or anything. But it seems like a generally bad practice, and not a good approach to use as a model for beginning students.
I think I'd rather stick with the Archimedean principle, and prove the claims of the proposed "axiom" as small lemmas.
answered Jan 30 at 2:30
John HughesJohn Hughes
65.1k24292
$endgroup$
$begingroup$
Part ii is not as trivial as you suggest, because the same $b$ is supposed to work for all $m$'s.
$endgroup$
– Andreas Blass
Jan 30 at 3:52
$begingroup$
You're correct -- I missed that. I suspect that a least-upper-bound argument might work, but I could easily be wrong.
$endgroup$
– John Hughes
Jan 30 at 4:56
$begingroup$
Yes, a least upper bound argument works. The least upper bound of all the sums on the left sides of the inequalities (for all $m$) will serve as $b$, because the right sides of all those inequalities are among the upper bounds.
$endgroup$
– Andreas Blass
Jan 30 at 5:00
$begingroup$
This also reminds me of a nice exercise in the last chapter or so of Spivak, in which he has you generate "the algebra student's real numbers," which are decimals, with an equivalence relation to deal with the $0.999ldots = 1$ problem, observing that getting multiplication right is really ugly. :)
$endgroup$
– John Hughes
Jan 30 at 5:02
$begingroup$
Thank you very much, John Hughes and Andreas Blass.
$endgroup$
– tchappy ha
Jan 31 at 6:13
add a comment |
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$begingroup$
This axiom would be proved, ordinarily, by the archimidean property, as J.W.Tanner points out. It's actually not a great "axiom", because it includes things that can be proved from other axioms. Knowing that there's an $n+1$ that's greater than or equal to $x$ -- that's good. There's a least such number by well-ordering. For that $n+1$, we have that $n <= x$. So the left hand side of part 1 is redundant.
Part ii hardly seems to be an axiom at all, for it's easy to prove (assuming you've done the order axioms right) that between any two reals, there's another real, so between the left hand sum, and the LHS plus a little increment, there's a real.
Now you can have, as axioms, things that follow from other axioms -- it's not against the rules or anything. But it seems like a generally bad practice, and not a good approach to use as a model for beginning students.
I think I'd rather stick with the Archimedean principle, and prove the claims of the proposed "axiom" as small lemmas.
answered Jan 30 at 2:30
John HughesJohn Hughes
65.1k24292
$endgroup$
$begingroup$
Part ii is not as trivial as you suggest, because the same $b$ is supposed to work for all $m$'s.
$endgroup$
– Andreas Blass
Jan 30 at 3:52
$begingroup$
You're correct -- I missed that. I suspect that a least-upper-bound argument might work, but I could easily be wrong.
$endgroup$
– John Hughes
Jan 30 at 4:56
$begingroup$
Yes, a least upper bound argument works. The least upper bound of all the sums on the left sides of the inequalities (for all $m$) will serve as $b$, because the right sides of all those inequalities are among the upper bounds.
$endgroup$
– Andreas Blass
Jan 30 at 5:00
$begingroup$
This also reminds me of a nice exercise in the last chapter or so of Spivak, in which he has you generate "the algebra student's real numbers," which are decimals, with an equivalence relation to deal with the $0.999ldots = 1$ problem, observing that getting multiplication right is really ugly. :)
$endgroup$
– John Hughes
Jan 30 at 5:02
$begingroup$
Thank you very much, John Hughes and Andreas Blass.
$endgroup$
– tchappy ha
Jan 31 at 6:13
add a comment |
$begingroup$
This axiom would be proved, ordinarily, by the archimidean property, as J.W.Tanner points out. It's actually not a great "axiom", because it includes things that can be proved from other axioms. Knowing that there's an $n+1$ that's greater than or equal to $x$ -- that's good. There's a least such number by well-ordering. For that $n+1$, we have that $n <= x$. So the left hand side of part 1 is redundant.
Part ii hardly seems to be an axiom at all, for it's easy to prove (assuming you've done the order axioms right) that between any two reals, there's another real, so between the left hand sum, and the LHS plus a little increment, there's a real.
Now you can have, as axioms, things that follow from other axioms -- it's not against the rules or anything. But it seems like a generally bad practice, and not a good approach to use as a model for beginning students.
I think I'd rather stick with the Archimedean principle, and prove the claims of the proposed "axiom" as small lemmas.
answered Jan 30 at 2:30
John HughesJohn Hughes
65.1k24292
$endgroup$
$begingroup$
Part ii is not as trivial as you suggest, because the same $b$ is supposed to work for all $m$'s.
$endgroup$
– Andreas Blass
Jan 30 at 3:52
$begingroup$
You're correct -- I missed that. I suspect that a least-upper-bound argument might work, but I could easily be wrong.
$endgroup$
– John Hughes
Jan 30 at 4:56
$begingroup$
Yes, a least upper bound argument works. The least upper bound of all the sums on the left sides of the inequalities (for all $m$) will serve as $b$, because the right sides of all those inequalities are among the upper bounds.
$endgroup$
– Andreas Blass
Jan 30 at 5:00
$begingroup$
This also reminds me of a nice exercise in the last chapter or so of Spivak, in which he has you generate "the algebra student's real numbers," which are decimals, with an equivalence relation to deal with the $0.999ldots = 1$ problem, observing that getting multiplication right is really ugly. :)
$endgroup$
– John Hughes
Jan 30 at 5:02
$begingroup$
Thank you very much, John Hughes and Andreas Blass.
$endgroup$
– tchappy ha
Jan 31 at 6:13
add a comment |
$begingroup$
This axiom would be proved, ordinarily, by the archimidean property, as J.W.Tanner points out. It's actually not a great "axiom", because it includes things that can be proved from other axioms. Knowing that there's an $n+1$ that's greater than or equal to $x$ -- that's good. There's a least such number by well-ordering. For that $n+1$, we have that $n <= x$. So the left hand side of part 1 is redundant.
Part ii hardly seems to be an axiom at all, for it's easy to prove (assuming you've done the order axioms right) that between any two reals, there's another real, so between the left hand sum, and the LHS plus a little increment, there's a real.
Now you can have, as axioms, things that follow from other axioms -- it's not against the rules or anything. But it seems like a generally bad practice, and not a good approach to use as a model for beginning students.
I think I'd rather stick with the Archimedean principle, and prove the claims of the proposed "axiom" as small lemmas.
answered Jan 30 at 2:30
John HughesJohn Hughes
65.1k24292
$endgroup$
This axiom would be proved, ordinarily, by the archimidean property, as J.W.Tanner points out. It's actually not a great "axiom", because it includes things that can be proved from other axioms. Knowing that there's an $n+1$ that's greater than or equal to $x$ -- that's good. There's a least such number by well-ordering. For that $n+1$, we have that $n <= x$. So the left hand side of part 1 is redundant.
Part ii hardly seems to be an axiom at all, for it's easy to prove (assuming you've done the order axioms right) that between any two reals, there's another real, so between the left hand sum, and the LHS plus a little increment, there's a real.
Now you can have, as axioms, things that follow from other axioms -- it's not against the rules or anything. But it seems like a generally bad practice, and not a good approach to use as a model for beginning students.
I think I'd rather stick with the Archimedean principle, and prove the claims of the proposed "axiom" as small lemmas.
answered Jan 30 at 2:30
John HughesJohn Hughes
65.1k24292
answered Jan 30 at 2:30
John HughesJohn Hughes
65.1k24292
answered Jan 30 at 2:30
John HughesJohn Hughes
65.1k24292
answered Jan 30 at 2:30
John HughesJohn Hughes
65.1k24292
65.1k24292
$begingroup$
Part ii is not as trivial as you suggest, because the same $b$ is supposed to work for all $m$'s.
$endgroup$
– Andreas Blass
Jan 30 at 3:52
$begingroup$
You're correct -- I missed that. I suspect that a least-upper-bound argument might work, but I could easily be wrong.
$endgroup$
– John Hughes
Jan 30 at 4:56
$begingroup$
Yes, a least upper bound argument works. The least upper bound of all the sums on the left sides of the inequalities (for all $m$) will serve as $b$, because the right sides of all those inequalities are among the upper bounds.
$endgroup$
– Andreas Blass
Jan 30 at 5:00
$begingroup$
This also reminds me of a nice exercise in the last chapter or so of Spivak, in which he has you generate "the algebra student's real numbers," which are decimals, with an equivalence relation to deal with the $0.999ldots = 1$ problem, observing that getting multiplication right is really ugly. :)
$endgroup$
– John Hughes
Jan 30 at 5:02
$begingroup$
Thank you very much, John Hughes and Andreas Blass.
$endgroup$
– tchappy ha
Jan 31 at 6:13
add a comment |
$begingroup$
Part ii is not as trivial as you suggest, because the same $b$ is supposed to work for all $m$'s.
$endgroup$
– Andreas Blass
Jan 30 at 3:52
$begingroup$
You're correct -- I missed that. I suspect that a least-upper-bound argument might work, but I could easily be wrong.
$endgroup$
– John Hughes
Jan 30 at 4:56
$begingroup$
Yes, a least upper bound argument works. The least upper bound of all the sums on the left sides of the inequalities (for all $m$) will serve as $b$, because the right sides of all those inequalities are among the upper bounds.
$endgroup$
– Andreas Blass
Jan 30 at 5:00
$begingroup$
This also reminds me of a nice exercise in the last chapter or so of Spivak, in which he has you generate "the algebra student's real numbers," which are decimals, with an equivalence relation to deal with the $0.999ldots = 1$ problem, observing that getting multiplication right is really ugly. :)
$endgroup$
– John Hughes
Jan 30 at 5:02
$begingroup$
Thank you very much, John Hughes and Andreas Blass.
$endgroup$
– tchappy ha
Jan 31 at 6:13
$begingroup$
Part ii is not as trivial as you suggest, because the same $b$ is supposed to work for all $m$'s.
$endgroup$
– Andreas Blass
Jan 30 at 3:52
$begingroup$
Part ii is not as trivial as you suggest, because the same $b$ is supposed to work for all $m$'s.
$endgroup$
– Andreas Blass
Jan 30 at 3:52
$begingroup$
You're correct -- I missed that. I suspect that a least-upper-bound argument might work, but I could easily be wrong.
$endgroup$
– John Hughes
Jan 30 at 4:56
$begingroup$
You're correct -- I missed that. I suspect that a least-upper-bound argument might work, but I could easily be wrong.
$endgroup$
– John Hughes
Jan 30 at 4:56
$begingroup$
Yes, a least upper bound argument works. The least upper bound of all the sums on the left sides of the inequalities (for all $m$) will serve as $b$, because the right sides of all those inequalities are among the upper bounds.
$endgroup$
– Andreas Blass
Jan 30 at 5:00
$begingroup$
Yes, a least upper bound argument works. The least upper bound of all the sums on the left sides of the inequalities (for all $m$) will serve as $b$, because the right sides of all those inequalities are among the upper bounds.
$endgroup$
– Andreas Blass
Jan 30 at 5:00
$begingroup$
This also reminds me of a nice exercise in the last chapter or so of Spivak, in which he has you generate "the algebra student's real numbers," which are decimals, with an equivalence relation to deal with the $0.999ldots = 1$ problem, observing that getting multiplication right is really ugly. :)
$endgroup$
– John Hughes
Jan 30 at 5:02
$begingroup$
This also reminds me of a nice exercise in the last chapter or so of Spivak, in which he has you generate "the algebra student's real numbers," which are decimals, with an equivalence relation to deal with the $0.999ldots = 1$ problem, observing that getting multiplication right is really ugly. :)
$endgroup$
– John Hughes
Jan 30 at 5:02
$begingroup$
Thank you very much, John Hughes and Andreas Blass.
$endgroup$
– tchappy ha
Jan 31 at 6:13
$begingroup$
Thank you very much, John Hughes and Andreas Blass.
$endgroup$
– tchappy ha
Jan 31 at 6:13
add a comment |
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$begingroup$
seems like the Archimedean property
$endgroup$
– J. W. Tanner
Jan 29 at 14:49
1
$begingroup$
The axiom equates real numbers with infinite binary expansion. Another base such as decimal would be equivalent.
$endgroup$
– Somos
Jan 29 at 20:24
$begingroup$
Can you provide a link to that "Calculus" by Takeshi Saito?
$endgroup$
– user635162
Jan 30 at 0:27
$begingroup$
The question-part of your question ends with the statement of the axiom. What is the point of the Proposition-Corollary-Theorem and discussion that follow?
$endgroup$
– Gerry Myerson
Jan 30 at 0:39
1
$begingroup$
@palaeomathematician Calculus
$endgroup$
– tchappy ha
Jan 30 at 0:45