Apostol - Mathematical Analysis - Theorem 1.1
$begingroup$
I realize there are several questions dealing with Theorem 1.1 from Apostol's "Mathematical Analysis", but those questions are mostly about how to prove the result or the geometrical intuition around it. My question is about the application of the theorem. The theorem states:
If $aleq b+varepsilon$
for all $varepsilon>0$, then
$aleq b$.
Take an example.
$a=2$, $b=1$ and $e = 3$; $e := varepsilon$ Then, $2 leq 1 + 3$;
$2 leq 4$
These meet the conditions required in the theorem.
But, now I get the absurd result: $2 leq 1$
What is wrong in my usage of the theorem?
analysis
edited Dec 15 '18 at 16:54
Henrik
6,03792030
asked Dec 14 '18 at 6:58
USAUSA
1
$endgroup$
add a comment |
$begingroup$
I realize there are several questions dealing with Theorem 1.1 from Apostol's "Mathematical Analysis", but those questions are mostly about how to prove the result or the geometrical intuition around it. My question is about the application of the theorem. The theorem states:
If $aleq b+varepsilon$
for all $varepsilon>0$, then
$aleq b$.
Take an example.
$a=2$, $b=1$ and $e = 3$; $e := varepsilon$ Then, $2 leq 1 + 3$;
$2 leq 4$
These meet the conditions required in the theorem.
But, now I get the absurd result: $2 leq 1$
What is wrong in my usage of the theorem?
analysis
edited Dec 15 '18 at 16:54
Henrik
6,03792030
asked Dec 14 '18 at 6:58
USAUSA
1
$endgroup$
2
$begingroup$
In "for all ε>0", you are missing "for all".
$endgroup$
– Did
Dec 14 '18 at 7:02
add a comment |
$begingroup$
I realize there are several questions dealing with Theorem 1.1 from Apostol's "Mathematical Analysis", but those questions are mostly about how to prove the result or the geometrical intuition around it. My question is about the application of the theorem. The theorem states:
If $aleq b+varepsilon$
for all $varepsilon>0$, then
$aleq b$.
Take an example.
$a=2$, $b=1$ and $e = 3$; $e := varepsilon$ Then, $2 leq 1 + 3$;
$2 leq 4$
These meet the conditions required in the theorem.
But, now I get the absurd result: $2 leq 1$
What is wrong in my usage of the theorem?
analysis
edited Dec 15 '18 at 16:54
Henrik
6,03792030
asked Dec 14 '18 at 6:58
USAUSA
1
$endgroup$
I realize there are several questions dealing with Theorem 1.1 from Apostol's "Mathematical Analysis", but those questions are mostly about how to prove the result or the geometrical intuition around it. My question is about the application of the theorem. The theorem states:
If $aleq b+varepsilon$
for all $varepsilon>0$, then
$aleq b$.
Take an example.
$a=2$, $b=1$ and $e = 3$; $e := varepsilon$ Then, $2 leq 1 + 3$;
$2 leq 4$
These meet the conditions required in the theorem.
But, now I get the absurd result: $2 leq 1$
What is wrong in my usage of the theorem?
analysis
analysis
edited Dec 15 '18 at 16:54
Henrik
6,03792030
asked Dec 14 '18 at 6:58
USAUSA
1
edited Dec 15 '18 at 16:54
Henrik
6,03792030
asked Dec 14 '18 at 6:58
USAUSA
1
edited Dec 15 '18 at 16:54
Henrik
6,03792030
edited Dec 15 '18 at 16:54
Henrik
6,03792030
edited Dec 15 '18 at 16:54
Henrik
6,03792030
6,03792030
asked Dec 14 '18 at 6:58
USAUSA
1
asked Dec 14 '18 at 6:58
USAUSA
1
asked Dec 14 '18 at 6:58
USAUSA
1
1
2
$begingroup$
In "for all ε>0", you are missing "for all".
$endgroup$
– Did
Dec 14 '18 at 7:02
add a comment |
2
$begingroup$
In "for all ε>0", you are missing "for all".
$endgroup$
– Did
Dec 14 '18 at 7:02
2
2
$begingroup$
In "for all ε>0", you are missing "for all".
$endgroup$
– Did
Dec 14 '18 at 7:02
$begingroup$
In "for all ε>0", you are missing "for all".
$endgroup$
– Did
Dec 14 '18 at 7:02
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The premise says for all $epsilon>0.$ You have only shown the inequality holds for one $epsilon$ (namely $epsilon=3$). But it fails for $epsilon = 1/2,$ for example, so the premise does not hold and you cannot draw the conclusion.
answered Dec 14 '18 at 7:03
spaceisdarkgreenspaceisdarkgreen
32.9k21753
$endgroup$
$begingroup$
Thanks. Live to learn and fight again.
$endgroup$
– USA
Dec 15 '18 at 5:01
add a comment |
$begingroup$
It must hold for all $varepsilon > 0$ that $a leq b + varepsilon$.
In your example, you consider $a = 2, b = 1$. Then that gives us $2 leq 1 + varepsilon$. However, notice, if you solve for $varepsilon$, this requires $varepsilon geq 1$.
Since your example does not hold for all $varepsilon>0$ - namely, for all $varepsilon in (0,1)$ - then the conclusion doesn't follow, because the theorem itself doesn't apply.
In other words, if there exists $varepsilon > 0$ for which $a > b + varepsilon$, then the theorem doesn't apply.
answered Dec 14 '18 at 7:04
Eevee TrainerEevee Trainer
6,0631936
$endgroup$
$begingroup$
Many many thanks. It is a universal quantifier.
$endgroup$
– USA
Dec 15 '18 at 5:01
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The premise says for all $epsilon>0.$ You have only shown the inequality holds for one $epsilon$ (namely $epsilon=3$). But it fails for $epsilon = 1/2,$ for example, so the premise does not hold and you cannot draw the conclusion.
answered Dec 14 '18 at 7:03
spaceisdarkgreenspaceisdarkgreen
32.9k21753
$endgroup$
$begingroup$
Thanks. Live to learn and fight again.
$endgroup$
– USA
Dec 15 '18 at 5:01
add a comment |
$begingroup$
The premise says for all $epsilon>0.$ You have only shown the inequality holds for one $epsilon$ (namely $epsilon=3$). But it fails for $epsilon = 1/2,$ for example, so the premise does not hold and you cannot draw the conclusion.
answered Dec 14 '18 at 7:03
spaceisdarkgreenspaceisdarkgreen
32.9k21753
$endgroup$
$begingroup$
Thanks. Live to learn and fight again.
$endgroup$
– USA
Dec 15 '18 at 5:01
add a comment |
$begingroup$
The premise says for all $epsilon>0.$ You have only shown the inequality holds for one $epsilon$ (namely $epsilon=3$). But it fails for $epsilon = 1/2,$ for example, so the premise does not hold and you cannot draw the conclusion.
answered Dec 14 '18 at 7:03
spaceisdarkgreenspaceisdarkgreen
32.9k21753
$endgroup$
The premise says for all $epsilon>0.$ You have only shown the inequality holds for one $epsilon$ (namely $epsilon=3$). But it fails for $epsilon = 1/2,$ for example, so the premise does not hold and you cannot draw the conclusion.
answered Dec 14 '18 at 7:03
spaceisdarkgreenspaceisdarkgreen
32.9k21753
answered Dec 14 '18 at 7:03
spaceisdarkgreenspaceisdarkgreen
32.9k21753
answered Dec 14 '18 at 7:03
spaceisdarkgreenspaceisdarkgreen
32.9k21753
answered Dec 14 '18 at 7:03
spaceisdarkgreenspaceisdarkgreen
32.9k21753
32.9k21753
$begingroup$
Thanks. Live to learn and fight again.
$endgroup$
– USA
Dec 15 '18 at 5:01
add a comment |
$begingroup$
Thanks. Live to learn and fight again.
$endgroup$
– USA
Dec 15 '18 at 5:01
$begingroup$
Thanks. Live to learn and fight again.
$endgroup$
– USA
Dec 15 '18 at 5:01
$begingroup$
Thanks. Live to learn and fight again.
$endgroup$
– USA
Dec 15 '18 at 5:01
add a comment |
$begingroup$
It must hold for all $varepsilon > 0$ that $a leq b + varepsilon$.
In your example, you consider $a = 2, b = 1$. Then that gives us $2 leq 1 + varepsilon$. However, notice, if you solve for $varepsilon$, this requires $varepsilon geq 1$.
Since your example does not hold for all $varepsilon>0$ - namely, for all $varepsilon in (0,1)$ - then the conclusion doesn't follow, because the theorem itself doesn't apply.
In other words, if there exists $varepsilon > 0$ for which $a > b + varepsilon$, then the theorem doesn't apply.
answered Dec 14 '18 at 7:04
Eevee TrainerEevee Trainer
6,0631936
$endgroup$
$begingroup$
Many many thanks. It is a universal quantifier.
$endgroup$
– USA
Dec 15 '18 at 5:01
add a comment |
$begingroup$
It must hold for all $varepsilon > 0$ that $a leq b + varepsilon$.
In your example, you consider $a = 2, b = 1$. Then that gives us $2 leq 1 + varepsilon$. However, notice, if you solve for $varepsilon$, this requires $varepsilon geq 1$.
Since your example does not hold for all $varepsilon>0$ - namely, for all $varepsilon in (0,1)$ - then the conclusion doesn't follow, because the theorem itself doesn't apply.
In other words, if there exists $varepsilon > 0$ for which $a > b + varepsilon$, then the theorem doesn't apply.
answered Dec 14 '18 at 7:04
Eevee TrainerEevee Trainer
6,0631936
$endgroup$
$begingroup$
Many many thanks. It is a universal quantifier.
$endgroup$
– USA
Dec 15 '18 at 5:01
add a comment |
$begingroup$
It must hold for all $varepsilon > 0$ that $a leq b + varepsilon$.
In your example, you consider $a = 2, b = 1$. Then that gives us $2 leq 1 + varepsilon$. However, notice, if you solve for $varepsilon$, this requires $varepsilon geq 1$.
Since your example does not hold for all $varepsilon>0$ - namely, for all $varepsilon in (0,1)$ - then the conclusion doesn't follow, because the theorem itself doesn't apply.
In other words, if there exists $varepsilon > 0$ for which $a > b + varepsilon$, then the theorem doesn't apply.
answered Dec 14 '18 at 7:04
Eevee TrainerEevee Trainer
6,0631936
$endgroup$
It must hold for all $varepsilon > 0$ that $a leq b + varepsilon$.
In your example, you consider $a = 2, b = 1$. Then that gives us $2 leq 1 + varepsilon$. However, notice, if you solve for $varepsilon$, this requires $varepsilon geq 1$.
Since your example does not hold for all $varepsilon>0$ - namely, for all $varepsilon in (0,1)$ - then the conclusion doesn't follow, because the theorem itself doesn't apply.
In other words, if there exists $varepsilon > 0$ for which $a > b + varepsilon$, then the theorem doesn't apply.
answered Dec 14 '18 at 7:04
Eevee TrainerEevee Trainer
6,0631936
answered Dec 14 '18 at 7:04
Eevee TrainerEevee Trainer
6,0631936
answered Dec 14 '18 at 7:04
Eevee TrainerEevee Trainer
6,0631936
answered Dec 14 '18 at 7:04
Eevee TrainerEevee Trainer
6,0631936
6,0631936
$begingroup$
Many many thanks. It is a universal quantifier.
$endgroup$
– USA
Dec 15 '18 at 5:01
add a comment |
$begingroup$
Many many thanks. It is a universal quantifier.
$endgroup$
– USA
Dec 15 '18 at 5:01
$begingroup$
Many many thanks. It is a universal quantifier.
$endgroup$
– USA
Dec 15 '18 at 5:01
$begingroup$
Many many thanks. It is a universal quantifier.
$endgroup$
– USA
Dec 15 '18 at 5:01
add a comment |
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2
$begingroup$
In "for all ε>0", you are missing "for all".
$endgroup$
– Did
Dec 14 '18 at 7:02