User:IssaRice/Formalizing adjustment of epsilons - Revision history

IssaRice: /* Tao's proof */

2020-05-24T06:59:02Z

Tao's proof

← Older revision		Revision as of 06:59, 24 May 2020
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	So now, use (3) to obtain some <math>\varepsilon</math>. By (1), we have <math>0 < \varepsilon < 1 \implies (x+\varepsilon)^2 \leq x^2 + 5\varepsilon</math>. Since <math>0 < \varepsilon < 1</math>, we can apply this implication to obtain <math>(x+\varepsilon)^2 \leq x^2 + 5\varepsilon</math>. By (3), we know <math>x^2 + 5\varepsilon < 2</math> so we have <math>(x+\varepsilon)^2 \leq x^2 + 5\varepsilon < 2</math> which is what we wanted.		So now, use (3) to obtain some <math>\varepsilon</math>. By (1), we have <math>0 < \varepsilon < 1 \implies (x+\varepsilon)^2 \leq x^2 + 5\varepsilon</math>. Since <math>0 < \varepsilon < 1</math>, we can apply this implication to obtain <math>(x+\varepsilon)^2 \leq x^2 + 5\varepsilon</math>. By (3), we know <math>x^2 + 5\varepsilon < 2</math> so we have <math>(x+\varepsilon)^2 \leq x^2 + 5\varepsilon < 2</math> which is what we wanted.

	So what was the issue? It was that in the first part of the proof, we were allowing <math>\varepsilon</math> to be arbitrary so that we could show the implication (1). Once we showed that, however, we want to close that context so that we could start a new context with some particular <math>\varepsilon</math>.		So what was the issue? It was that in the first part of the proof, we were allowing <math>\varepsilon</math> to be arbitrary so that we could show the implication (1). Once we showed that, however, we want to close that context so that we could start a new context with some particular <math>\varepsilon</math>. In first-order logic, these aren't the ''same'' <math>\varepsilon</math>, but when writing out proofs we aren't too careful about this.

	==The general case==		==The general case==

IssaRice: /* Tao's proof */

2020-05-24T06:53:04Z

Tao's proof

← Older revision		Revision as of 06:53, 24 May 2020
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	So now, use (3) to obtain some <math>\varepsilon</math>. By (1), we have <math>0 < \varepsilon < 1 \implies (x+\varepsilon)^2 \leq x^2 + 5\varepsilon</math>. Since <math>0 < \varepsilon < 1</math>, we can apply this implication to obtain <math>(x+\varepsilon)^2 \leq x^2 + 5\varepsilon</math>. By (3), we know <math>x^2 + 5\varepsilon < 2</math> so we have <math>(x+\varepsilon)^2 \leq x^2 + 5\varepsilon < 2</math> which is what we wanted.		So now, use (3) to obtain some <math>\varepsilon</math>. By (1), we have <math>0 < \varepsilon < 1 \implies (x+\varepsilon)^2 \leq x^2 + 5\varepsilon</math>. Since <math>0 < \varepsilon < 1</math>, we can apply this implication to obtain <math>(x+\varepsilon)^2 \leq x^2 + 5\varepsilon</math>. By (3), we know <math>x^2 + 5\varepsilon < 2</math> so we have <math>(x+\varepsilon)^2 \leq x^2 + 5\varepsilon < 2</math> which is what we wanted.

			So what was the issue? It was that in the first part of the proof, we were allowing <math>\varepsilon</math> to be arbitrary so that we could show the implication (1). Once we showed that, however, we want to close that context so that we could start a new context with some particular <math>\varepsilon</math>.

	==The general case==		==The general case==

IssaRice: /* Tao's proof */

2020-05-24T06:49:43Z

Tao's proof

← Older revision		Revision as of 06:49, 24 May 2020
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	To do this, use (2) to obtain <math>\varepsilon > 0</math> such that <math>x^2 + 5\varepsilon < 2</math>. If <math>\varepsilon < 1</math> then we are done. If <math>\varepsilon \geq 1</math>, then <math>1/2 < \varepsilon</math> so <math>x^2 + 5(1/2) < x^2 + 5\varepsilon < 2</math>, which means we can use 1/2. In either case we have found a number that works, so we have (3).		To do this, use (2) to obtain <math>\varepsilon > 0</math> such that <math>x^2 + 5\varepsilon < 2</math>. If <math>\varepsilon < 1</math> then we are done. If <math>\varepsilon \geq 1</math>, then <math>1/2 < \varepsilon</math> so <math>x^2 + 5(1/2) < x^2 + 5\varepsilon < 2</math>, which means we can use 1/2. In either case we have found a number that works, so we have (3).

	So now, use (3) to obtain some <math>\varepsilon</math>. By (1), we have <math>0 < \varepsilon < 1 \implies (x+\varepsilon)^2 \leq x^2 + 5\varepsilon</math>. Since <math>0 < \varepsilon < 1</math>, we can apply this implication to obtain <math>(x+\varepsilon)^2 \leq x^2 + 5\varepsilon</math>.		So now, use (3) to obtain some <math>\varepsilon</math>. By (1), we have <math>0 < \varepsilon < 1 \implies (x+\varepsilon)^2 \leq x^2 + 5\varepsilon</math>. Since <math>0 < \varepsilon < 1</math>, we can apply this implication to obtain <math>(x+\varepsilon)^2 \leq x^2 + 5\varepsilon</math>. By (3), we know <math>x^2 + 5\varepsilon < 2</math> so we have <math>(x+\varepsilon)^2 \leq x^2 + 5\varepsilon < 2</math> which is what we wanted.

	==The general case==		==The general case==

IssaRice: /* Tao's proof */

2020-05-24T06:48:32Z

Tao's proof

← Older revision		Revision as of 06:48, 24 May 2020
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	(3) <math>\exists \varepsilon (0 < \varepsilon < 1 \wedge x^2 + 5\varepsilon < 2)</math>		(3) <math>\exists \varepsilon (0 < \varepsilon < 1 \wedge x^2 + 5\varepsilon < 2)</math>

	To do this, use (2) to obtain <math>\varepsilon > 0</math> such that <math>x^2 + 5\varepsilon < 2</math>. If <math>\varepsilon < 1</math> then we are done. If <math>\varepsilon \geq 1</math>, then <math>1/2 < \varepsilon</math> so <math>x^2 + 5(1/2) < x^2 + 5\varepsilon < 2</math>, which means we can use 1/2.		To do this, use (2) to obtain <math>\varepsilon > 0</math> such that <math>x^2 + 5\varepsilon < 2</math>. If <math>\varepsilon < 1</math> then we are done. If <math>\varepsilon \geq 1</math>, then <math>1/2 < \varepsilon</math> so <math>x^2 + 5(1/2) < x^2 + 5\varepsilon < 2</math>, which means we can use 1/2. In either case we have found a number that works, so we have (3).

			So now, use (3) to obtain some <math>\varepsilon</math>. By (1), we have <math>0 < \varepsilon < 1 \implies (x+\varepsilon)^2 \leq x^2 + 5\varepsilon</math>. Since <math>0 < \varepsilon < 1</math>, we can apply this implication to obtain <math>(x+\varepsilon)^2 \leq x^2 + 5\varepsilon</math>.

	==The general case==		==The general case==

IssaRice: /* Tao's proof */

2020-05-24T06:46:38Z

Tao's proof

← Older revision		Revision as of 06:46, 24 May 2020
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	(3) <math>\exists \varepsilon (0 < \varepsilon < 1 \wedge x^2 + 5\varepsilon < 2)</math>		(3) <math>\exists \varepsilon (0 < \varepsilon < 1 \wedge x^2 + 5\varepsilon < 2)</math>

	To do this, use (2) to obtain <math>\varepsilon > 0</math> such that <math>x^2 + 5\varepsilon < 2</math>. If <math>\varepsilon < 1</math> then we are done. If <math>\varepsilon \geq 1</math>, then		To do this, use (2) to obtain <math>\varepsilon > 0</math> such that <math>x^2 + 5\varepsilon < 2</math>. If <math>\varepsilon < 1</math> then we are done. If <math>\varepsilon \geq 1</math>, then <math>1/2 < \varepsilon</math> so <math>x^2 + 5(1/2) < x^2 + 5\varepsilon < 2</math>, which means we can use 1/2.

	==The general case==		==The general case==

IssaRice: /* Tao's proof */

2020-05-24T06:45:04Z

Tao's proof

← Older revision		Revision as of 06:45, 24 May 2020
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	(3) <math>\exists \varepsilon (0 < \varepsilon < 1 \wedge x^2 + 5\varepsilon < 2)</math>		(3) <math>\exists \varepsilon (0 < \varepsilon < 1 \wedge x^2 + 5\varepsilon < 2)</math>

			To do this, use (2) to obtain <math>\varepsilon > 0</math> such that <math>x^2 + 5\varepsilon < 2</math>. If <math>\varepsilon < 1</math> then we are done. If <math>\varepsilon \geq 1</math>, then

	==The general case==		==The general case==

IssaRice: /* Tao's proof */

2020-05-24T06:43:10Z

Tao's proof

← Older revision		Revision as of 06:43, 24 May 2020
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	What we have are:		What we have are:
	* <math>\forall \varepsilon (0 < \varepsilon < 1 \implies (x+\varepsilon)^2 \leq x^2 + 5\varepsilon)</math>
	* <math>\exists \varepsilon (\varepsilon > 0 \wedge x^2 + 5\varepsilon < 2)</math>		(1) <math>\forall \varepsilon (0 < \varepsilon < 1 \implies (x+\varepsilon)^2 \leq x^2 + 5\varepsilon)</math><br>
			(2) <math>\exists \varepsilon (\varepsilon > 0 \wedge x^2 + 5\varepsilon < 2)</math>

			So now what we do is use (2) to obtain a particular <math>\varepsilon</math>. In order to use (1), we actually need to to show <math>0 < \varepsilon < 1</math>, but we only have <math>0 < \varepsilon</math>.

			So actually, we want to show

			(3) <math>\exists \varepsilon (0 < \varepsilon < 1 \wedge x^2 + 5\varepsilon < 2)</math>

	==The general case==		==The general case==

IssaRice: /* Tao's proof */

2020-05-24T06:35:29Z

Tao's proof

← Older revision		Revision as of 06:35, 24 May 2020
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	What we have are:		What we have are:
	* <math>\forall \varepsilon (0 < \varepsilon < 1 \implies x^2 + 5\varepsilon)</math>		* <math>\forall \varepsilon (0 < \varepsilon < 1 \implies (x+\varepsilon)^2 \leq x^2 + 5\varepsilon)</math>
	* <math>\exists \varepsilon (\varepsilon > 0 \wedge x^2 + 5\varepsilon < 2)</math>		* <math>\exists \varepsilon (\varepsilon > 0 \wedge x^2 + 5\varepsilon < 2)</math>

	==The general case==		==The general case==

IssaRice: /* Tao's proof */

2020-05-24T06:34:19Z

Tao's proof

← Older revision		Revision as of 06:34, 24 May 2020
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	We want to show <math>\exists \varepsilon (\varepsilon > 0 \wedge (x+\varepsilon)^2 < 2)</math>, because this will show that <math>x+\varepsilon \in E</math>, which will show that <math>x</math> isn't an upper bound of <math>E</math>.		We want to show <math>\exists \varepsilon (\varepsilon > 0 \wedge (x+\varepsilon)^2 < 2)</math>, because this will show that <math>x+\varepsilon \in E</math>, which will show that <math>x</math> isn't an upper bound of <math>E</math>.

			What we have are:
			* <math>\forall \varepsilon (0 < \varepsilon < 1 \implies x^2 + 5\varepsilon)</math>
			* <math>\exists \varepsilon (\varepsilon > 0 \wedge x^2 + 5\varepsilon < 2)</math>

	==The general case==		==The general case==

IssaRice: /* Tao's proof */

2020-05-24T06:32:40Z

Tao's proof

← Older revision		Revision as of 06:32, 24 May 2020
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	I'll use quantifiers so we can be really clear about what's bound and what's free.		I'll use quantifiers so we can be really clear about what's bound and what's free.

	We want to show <math>\exists \varepsilon > 0 ((x+\varepsilon)^2 < 2)</math>, because this will show that <math>x+\varepsilon \in E</math>, which will show that <math>x</math> isn't an upper bound of <math>E</math>.		We want to show <math>\exists \varepsilon (\varepsilon > 0 \wedge (x+\varepsilon)^2 < 2)</math>, because this will show that <math>x+\varepsilon \in E</math>, which will show that <math>x</math> isn't an upper bound of <math>E</math>.

	==The general case==		==The general case==