About Walter Rudin's “Principles of Mathematical Analysis Third Edition” pp.18-19 STEP4 (A5)
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I am reading "Principles of Mathematical Analysis Third Edition" by Walter Rudin.
At p.19 in this book, there is the following statement:
Let $w$ be a positive rational number, and $alpha$ be a cut. Then, there is an integer $n$ such that $n w in alpha$ but $(n + 1) w notin alpha$. (Note that this depends on the fact that $mathbb{Q}$ has the archimedean property!)
The definition of cuts is the following:
The members of $mathbb{R}$ will be certain subsets of $mathbb{Q}$, called cuts. A cut is, by definition, any set $alpha subset mathbb{Q}$ with the following properties.
$alpha$ is not empty, $alpha neq mathbb{Q}$.
- If $p in alpha$, $q in mathbb{Q}$, and $q < p$, then $q in alpha$.
- If $p in alpha$, then $p < r$ for some $r in alpha$.
I proved the above statement, but I'm not sure my proof is OK or not.
Is the following proof OK or not?
By definition of a cut, there exist $p$, $q$ such that $p in alpha$, $q notin alpha$, $q in mathbb{Q}$.
By archimedean property of $mathbb{Q}$, there exists $m in mathbb{Z}^{+}$ such that $-p < m w$. So $p > (-m) w$. By the property 2 of cuts, $(-m) w in alpha$.
By archimedean property of $mathbb{Q}$, there exists $n in mathbb{Z}^{+}$ such that $q < n w$. By the property 2 of cuts, $n w notin alpha$.
So, $S:={m in mathbb{Z} | m w in alpha}$ is not empty.
So, $T:={m in mathbb{Z} | m w notin alpha}$ is not empty.
If $m_1 in S$ and $m_2 in T$, then $m_1 w < m_2 w$ by the property 2 of cuts.
$therefore m_1 < m_2$.
$therefore m_1$ is a lower bound of $T$.
$therefore$ There exists $min T$.
Let $n := min T - 1$.
Then, $(n + 1) w notin alpha$, and $n w in alpha$.
calculus proof-verification real-numbers
asked Dec 2 '18 at 23:05
tchappy hatchappy ha
583311
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add a comment |
$begingroup$
I am reading "Principles of Mathematical Analysis Third Edition" by Walter Rudin.
At p.19 in this book, there is the following statement:
Let $w$ be a positive rational number, and $alpha$ be a cut. Then, there is an integer $n$ such that $n w in alpha$ but $(n + 1) w notin alpha$. (Note that this depends on the fact that $mathbb{Q}$ has the archimedean property!)
The definition of cuts is the following:
The members of $mathbb{R}$ will be certain subsets of $mathbb{Q}$, called cuts. A cut is, by definition, any set $alpha subset mathbb{Q}$ with the following properties.
$alpha$ is not empty, $alpha neq mathbb{Q}$.
- If $p in alpha$, $q in mathbb{Q}$, and $q < p$, then $q in alpha$.
- If $p in alpha$, then $p < r$ for some $r in alpha$.
I proved the above statement, but I'm not sure my proof is OK or not.
Is the following proof OK or not?
By definition of a cut, there exist $p$, $q$ such that $p in alpha$, $q notin alpha$, $q in mathbb{Q}$.
By archimedean property of $mathbb{Q}$, there exists $m in mathbb{Z}^{+}$ such that $-p < m w$. So $p > (-m) w$. By the property 2 of cuts, $(-m) w in alpha$.
By archimedean property of $mathbb{Q}$, there exists $n in mathbb{Z}^{+}$ such that $q < n w$. By the property 2 of cuts, $n w notin alpha$.
So, $S:={m in mathbb{Z} | m w in alpha}$ is not empty.
So, $T:={m in mathbb{Z} | m w notin alpha}$ is not empty.
If $m_1 in S$ and $m_2 in T$, then $m_1 w < m_2 w$ by the property 2 of cuts.
$therefore m_1 < m_2$.
$therefore m_1$ is a lower bound of $T$.
$therefore$ There exists $min T$.
Let $n := min T - 1$.
Then, $(n + 1) w notin alpha$, and $n w in alpha$.
calculus proof-verification real-numbers
asked Dec 2 '18 at 23:05
tchappy hatchappy ha
583311
$endgroup$
$begingroup$
I'm going to be really fussy. Why do you know a set of integers which is bounded below has a minimum? (It could be that I am being much too fussy then is justified.)
$endgroup$
– fleablood
Dec 3 '18 at 1:26
$begingroup$
In Rudin's book, we can use the familiar properties of $mathbb{N}, mathbb{Z}, mathbb{Q}$ freely without proofs. I guess we can show the property of $mathbb{Z}$ that I used above without the theory of $mathbb{R}$.
$endgroup$
– tchappy ha
Dec 3 '18 at 2:11
$begingroup$
But is that a familiar enough property? Still your proof seems good. I think I'd use the archemedian principal to prove there is an $n$ so that $nw > q - p > 0$ and as $0*w = 0$ then $n > w$ and ... yeah... you're right. That's pretty clear that there most be a smallest such $n$...
$endgroup$
– fleablood
Dec 3 '18 at 2:22
add a comment |
$begingroup$
I am reading "Principles of Mathematical Analysis Third Edition" by Walter Rudin.
At p.19 in this book, there is the following statement:
Let $w$ be a positive rational number, and $alpha$ be a cut. Then, there is an integer $n$ such that $n w in alpha$ but $(n + 1) w notin alpha$. (Note that this depends on the fact that $mathbb{Q}$ has the archimedean property!)
The definition of cuts is the following:
The members of $mathbb{R}$ will be certain subsets of $mathbb{Q}$, called cuts. A cut is, by definition, any set $alpha subset mathbb{Q}$ with the following properties.
$alpha$ is not empty, $alpha neq mathbb{Q}$.
- If $p in alpha$, $q in mathbb{Q}$, and $q < p$, then $q in alpha$.
- If $p in alpha$, then $p < r$ for some $r in alpha$.
I proved the above statement, but I'm not sure my proof is OK or not.
Is the following proof OK or not?
By definition of a cut, there exist $p$, $q$ such that $p in alpha$, $q notin alpha$, $q in mathbb{Q}$.
By archimedean property of $mathbb{Q}$, there exists $m in mathbb{Z}^{+}$ such that $-p < m w$. So $p > (-m) w$. By the property 2 of cuts, $(-m) w in alpha$.
By archimedean property of $mathbb{Q}$, there exists $n in mathbb{Z}^{+}$ such that $q < n w$. By the property 2 of cuts, $n w notin alpha$.
So, $S:={m in mathbb{Z} | m w in alpha}$ is not empty.
So, $T:={m in mathbb{Z} | m w notin alpha}$ is not empty.
If $m_1 in S$ and $m_2 in T$, then $m_1 w < m_2 w$ by the property 2 of cuts.
$therefore m_1 < m_2$.
$therefore m_1$ is a lower bound of $T$.
$therefore$ There exists $min T$.
Let $n := min T - 1$.
Then, $(n + 1) w notin alpha$, and $n w in alpha$.
calculus proof-verification real-numbers
asked Dec 2 '18 at 23:05
tchappy hatchappy ha
583311
$endgroup$
I am reading "Principles of Mathematical Analysis Third Edition" by Walter Rudin.
At p.19 in this book, there is the following statement:
Let $w$ be a positive rational number, and $alpha$ be a cut. Then, there is an integer $n$ such that $n w in alpha$ but $(n + 1) w notin alpha$. (Note that this depends on the fact that $mathbb{Q}$ has the archimedean property!)
The definition of cuts is the following:
The members of $mathbb{R}$ will be certain subsets of $mathbb{Q}$, called cuts. A cut is, by definition, any set $alpha subset mathbb{Q}$ with the following properties.
$alpha$ is not empty, $alpha neq mathbb{Q}$.
- If $p in alpha$, $q in mathbb{Q}$, and $q < p$, then $q in alpha$.
- If $p in alpha$, then $p < r$ for some $r in alpha$.
I proved the above statement, but I'm not sure my proof is OK or not.
Is the following proof OK or not?
By definition of a cut, there exist $p$, $q$ such that $p in alpha$, $q notin alpha$, $q in mathbb{Q}$.
By archimedean property of $mathbb{Q}$, there exists $m in mathbb{Z}^{+}$ such that $-p < m w$. So $p > (-m) w$. By the property 2 of cuts, $(-m) w in alpha$.
By archimedean property of $mathbb{Q}$, there exists $n in mathbb{Z}^{+}$ such that $q < n w$. By the property 2 of cuts, $n w notin alpha$.
So, $S:={m in mathbb{Z} | m w in alpha}$ is not empty.
So, $T:={m in mathbb{Z} | m w notin alpha}$ is not empty.
If $m_1 in S$ and $m_2 in T$, then $m_1 w < m_2 w$ by the property 2 of cuts.
$therefore m_1 < m_2$.
$therefore m_1$ is a lower bound of $T$.
$therefore$ There exists $min T$.
Let $n := min T - 1$.
Then, $(n + 1) w notin alpha$, and $n w in alpha$.
calculus proof-verification real-numbers
calculus proof-verification real-numbers
asked Dec 2 '18 at 23:05
tchappy hatchappy ha
583311
asked Dec 2 '18 at 23:05
tchappy hatchappy ha
583311
asked Dec 2 '18 at 23:05
tchappy hatchappy ha
583311
asked Dec 2 '18 at 23:05
tchappy hatchappy ha
583311
asked Dec 2 '18 at 23:05
tchappy hatchappy ha
583311
583311
$begingroup$
I'm going to be really fussy. Why do you know a set of integers which is bounded below has a minimum? (It could be that I am being much too fussy then is justified.)
$endgroup$
– fleablood
Dec 3 '18 at 1:26
$begingroup$
In Rudin's book, we can use the familiar properties of $mathbb{N}, mathbb{Z}, mathbb{Q}$ freely without proofs. I guess we can show the property of $mathbb{Z}$ that I used above without the theory of $mathbb{R}$.
$endgroup$
– tchappy ha
Dec 3 '18 at 2:11
$begingroup$
But is that a familiar enough property? Still your proof seems good. I think I'd use the archemedian principal to prove there is an $n$ so that $nw > q - p > 0$ and as $0*w = 0$ then $n > w$ and ... yeah... you're right. That's pretty clear that there most be a smallest such $n$...
$endgroup$
– fleablood
Dec 3 '18 at 2:22
add a comment |
$begingroup$
I'm going to be really fussy. Why do you know a set of integers which is bounded below has a minimum? (It could be that I am being much too fussy then is justified.)
$endgroup$
– fleablood
Dec 3 '18 at 1:26
$begingroup$
In Rudin's book, we can use the familiar properties of $mathbb{N}, mathbb{Z}, mathbb{Q}$ freely without proofs. I guess we can show the property of $mathbb{Z}$ that I used above without the theory of $mathbb{R}$.
$endgroup$
– tchappy ha
Dec 3 '18 at 2:11
$begingroup$
But is that a familiar enough property? Still your proof seems good. I think I'd use the archemedian principal to prove there is an $n$ so that $nw > q - p > 0$ and as $0*w = 0$ then $n > w$ and ... yeah... you're right. That's pretty clear that there most be a smallest such $n$...
$endgroup$
– fleablood
Dec 3 '18 at 2:22
$begingroup$
I'm going to be really fussy. Why do you know a set of integers which is bounded below has a minimum? (It could be that I am being much too fussy then is justified.)
$endgroup$
– fleablood
Dec 3 '18 at 1:26
$begingroup$
I'm going to be really fussy. Why do you know a set of integers which is bounded below has a minimum? (It could be that I am being much too fussy then is justified.)
$endgroup$
– fleablood
Dec 3 '18 at 1:26
$begingroup$
In Rudin's book, we can use the familiar properties of $mathbb{N}, mathbb{Z}, mathbb{Q}$ freely without proofs. I guess we can show the property of $mathbb{Z}$ that I used above without the theory of $mathbb{R}$.
$endgroup$
– tchappy ha
Dec 3 '18 at 2:11
$begingroup$
In Rudin's book, we can use the familiar properties of $mathbb{N}, mathbb{Z}, mathbb{Q}$ freely without proofs. I guess we can show the property of $mathbb{Z}$ that I used above without the theory of $mathbb{R}$.
$endgroup$
– tchappy ha
Dec 3 '18 at 2:11
$begingroup$
But is that a familiar enough property? Still your proof seems good. I think I'd use the archemedian principal to prove there is an $n$ so that $nw > q - p > 0$ and as $0*w = 0$ then $n > w$ and ... yeah... you're right. That's pretty clear that there most be a smallest such $n$...
$endgroup$
– fleablood
Dec 3 '18 at 2:22
$begingroup$
But is that a familiar enough property? Still your proof seems good. I think I'd use the archemedian principal to prove there is an $n$ so that $nw > q - p > 0$ and as $0*w = 0$ then $n > w$ and ... yeah... you're right. That's pretty clear that there most be a smallest such $n$...
$endgroup$
– fleablood
Dec 3 '18 at 2:22
add a comment |
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$begingroup$
I'm going to be really fussy. Why do you know a set of integers which is bounded below has a minimum? (It could be that I am being much too fussy then is justified.)
$endgroup$
– fleablood
Dec 3 '18 at 1:26
$begingroup$
In Rudin's book, we can use the familiar properties of $mathbb{N}, mathbb{Z}, mathbb{Q}$ freely without proofs. I guess we can show the property of $mathbb{Z}$ that I used above without the theory of $mathbb{R}$.
$endgroup$
– tchappy ha
Dec 3 '18 at 2:11
$begingroup$
But is that a familiar enough property? Still your proof seems good. I think I'd use the archemedian principal to prove there is an $n$ so that $nw > q - p > 0$ and as $0*w = 0$ then $n > w$ and ... yeah... you're right. That's pretty clear that there most be a smallest such $n$...
$endgroup$
– fleablood
Dec 3 '18 at 2:22