Template:Proof/doc - Revision history

ArrowHead294 at 15:30, 25 January 2025

2025-01-25T15:30:43Z

← Older revision		Revision as of 15:30, 25 January 2025
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	{{dochead}}		{{dochead}}
			This is a box for proofs. This calls [[Template:Databox]] but it automatically appends [[Template:Qed]] in the end of the contents and adds the page to [[:Category:Pages with proofs]].

	=== Usage ===		=== Usage ===
	You type:		You type:

ArrowHead294 at 20:49, 10 December 2024

2024-12-10T20:49:41Z

← Older revision		Revision as of 20:49, 10 December 2024
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	\| title=Proof that the root of any integer is either an integer or irrational		\| title=Proof that the root of any integer is either an integer or irrational
	\| contents=Assume <math>\sqrt[n]{m}</math> is {{nowrap\|''p'' / ''q''}}, where {{nowrap\|''p'', ''q'' ∈ ℤ<sup>+</sup>}} and {{nowrap\|gcd(''p'', ''q'') {{=}} 1}}. Then		\| contents=Assume <math>\sqrt[n]{m}</math> is {{nowrap\|''p'' / ''q''}}, where {{nowrap\|''p'', ''q'' ∈ ℤ<sup>+</sup>}} and {{nowrap\|gcd(''p'', ''q'') {{=}} 1}}. Then
	{{nowrap\|''p<sup>n</sup>'' / ''q<sup>n</sup>'' {{=}} ''m''}}, and {{nowrap\|''p<sup>n''</sup> {{=}} ''mq<sup>n''</sup>}}. This means ''p'' is divisible by ''m''. Therefore, there exists some integer ''r'' such that {{nowrap\|''p'' {{=}} ''mr''}}, so we now have {{nowrap\|''m<sup>n</sup>r<sup>n</sup>'' {{=}} ''mq<sup>n</sup>''}}. Dividing both sides by ''m'' gives{{nowrap\|''m''<sup>{{nowrap\|''n'' − 1}}</sup>''r<sup>n''</sup> {{=}} ''q<sup>n</sup>''}}. This means that ''q'' must also be divisible by ''m'', which is a contradiction, since ''p'' and ''q'' were assumed to be relatively prime.		{{nowrap\|''p<sup>n</sup>'' / ''q<sup>n</sup>'' {{=}} ''m''}}, and {{nowrap\|''p<sup>n''</sup> {{=}} ''mq<sup>n''</sup>}}. This means ''p'' is divisible by ''m''. Therefore, there exists some integer ''r'' such that {{nowrap\|''p'' {{=}} ''mr''}}, so we now have {{nowrap\|''m<sup>n</sup>r<sup>n</sup>'' {{=}} ''mq<sup>n</sup>''}}. Dividing both sides by ''m'' gives {{nowrap\|''m''<sup>{{nowrap\|''n'' − 1}}</sup>''r<sup>n''</sup> {{=}} ''q<sup>n</sup>''}}. This means that ''q'' must also be divisible by ''m'', which is a contradiction, since ''p'' and ''q'' were assumed to be relatively prime.
	}}		}}
	</pre>		</pre>
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	\| title=Proof that the root of any integer is either an integer or irrational		\| title=Proof that the root of any integer is either an integer or irrational
	\| contents=Assume <math>\sqrt[n]{m}</math> is {{nowrap\|''p'' / ''q''}}, where {{nowrap\|''p'', ''q'' ∈ ℤ<sup>+</sup>}} and {{nowrap\|gcd(''p'', ''q'') {{=}} 1}}. Then		\| contents=Assume <math>\sqrt[n]{m}</math> is {{nowrap\|''p'' / ''q''}}, where {{nowrap\|''p'', ''q'' ∈ ℤ<sup>+</sup>}} and {{nowrap\|gcd(''p'', ''q'') {{=}} 1}}. Then
	{{nowrap\|''p<sup>n</sup>'' / ''q<sup>n</sup>'' {{=}} ''m''}}, and {{nowrap\|''p<sup>n''</sup> {{=}} ''mq<sup>n''</sup>}}. This means ''p'' is divisible by ''m''. Therefore, there exists some integer ''r'' such that {{nowrap\|''p'' {{=}} ''mr''}}, so we now have {{nowrap\|''m<sup>n</sup>r<sup>n</sup>'' {{=}} ''mq<sup>n</sup>''}}. Dividing both sides by ''m'' gives{{nowrap\|''m''<sup>{{nowrap\|''n'' − 1}}</sup>''r<sup>n''</sup> {{=}} ''q<sup>n</sup>''}}. This means that ''q'' must also be divisible by ''m'', which is a contradiction, since ''p'' and ''q'' were assumed to be relatively prime.		{{nowrap\|''p<sup>n</sup>'' / ''q<sup>n</sup>'' {{=}} ''m''}}, and {{nowrap\|''p<sup>n''</sup> {{=}} ''mq<sup>n''</sup>}}. This means ''p'' is divisible by ''m''. Therefore, there exists some integer ''r'' such that {{nowrap\|''p'' {{=}} ''mr''}}, so we now have {{nowrap\|''m<sup>n</sup>r<sup>n</sup>'' {{=}} ''mq<sup>n</sup>''}}. Dividing both sides by ''m'' gives {{nowrap\|''m''<sup>{{nowrap\|''n'' − 1}}</sup>''r<sup>n''</sup> {{=}} ''q<sup>n</sup>''}}. This means that ''q'' must also be divisible by ''m'', which is a contradiction, since ''p'' and ''q'' were assumed to be relatively prime.
	}}		}}

ArrowHead294 at 01:12, 7 December 2024

2024-12-07T01:12:10Z

← Older revision		Revision as of 01:12, 7 December 2024
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	{{nowrap\|''p<sup>n</sup>'' / ''q<sup>n</sup>'' {{=}} ''m''}}, and {{nowrap\|''p<sup>n''</sup> {{=}} ''mq<sup>n''</sup>}}. This means ''p'' is divisible by ''m''. Therefore, there exists some integer ''r'' such that {{nowrap\|''p'' {{=}} ''mr''}}, so we now have {{nowrap\|''m<sup>n</sup>r<sup>n</sup>'' {{=}} ''mq<sup>n</sup>''}}. Dividing both sides by ''m'' gives{{nowrap\|''m''<sup>{{nowrap\|''n'' − 1}}</sup>''r<sup>n''</sup> {{=}} ''q<sup>n</sup>''}}. This means that ''q'' must also be divisible by ''m'', which is a contradiction, since ''p'' and ''q'' were assumed to be relatively prime.		{{nowrap\|''p<sup>n</sup>'' / ''q<sup>n</sup>'' {{=}} ''m''}}, and {{nowrap\|''p<sup>n''</sup> {{=}} ''mq<sup>n''</sup>}}. This means ''p'' is divisible by ''m''. Therefore, there exists some integer ''r'' such that {{nowrap\|''p'' {{=}} ''mr''}}, so we now have {{nowrap\|''m<sup>n</sup>r<sup>n</sup>'' {{=}} ''mq<sup>n</sup>''}}. Dividing both sides by ''m'' gives{{nowrap\|''m''<sup>{{nowrap\|''n'' − 1}}</sup>''r<sup>n''</sup> {{=}} ''q<sup>n</sup>''}}. This means that ''q'' must also be divisible by ''m'', which is a contradiction, since ''p'' and ''q'' were assumed to be relatively prime.
	}}		}}


	{{escape notice}}		{{escape notice}}

ArrowHead294 at 15:52, 6 December 2024

2024-12-06T15:52:15Z

← Older revision		Revision as of 15:52, 6 December 2024
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	\| title=Proof that the root of any integer is either an integer or irrational		\| title=Proof that the root of any integer is either an integer or irrational
	\| contents=Assume <math>\sqrt[n]{m}</math> is {{nowrap\|''p'' / ''q''}}, where {{nowrap\|''p'', ''q'' ∈ ℤ<sup>+</sup>}} and {{nowrap\|gcd(''p'', ''q'') {{=}} 1}}. Then		\| contents=Assume <math>\sqrt[n]{m}</math> is {{nowrap\|''p'' / ''q''}}, where {{nowrap\|''p'', ''q'' ∈ ℤ<sup>+</sup>}} and {{nowrap\|gcd(''p'', ''q'') {{=}} 1}}. Then
	{{nowrap\|''p''<sup>n</sup> / ''q''<sup>n</sup> {{=}} ''m''}}, and {{nowrap\|''p''<sup>n</sup> {{=}} ~~m''q~~''<sup>n</sup>}}. This means ''p'' is divisible by ''m''. Therefore, there exists some integer ''r'' such that {{nowrap\|p {{=}} mr}}, so we now have {{nowrap\|''m''<sup>n</sup>''r''<sup>n</sup> {{=}} ~~m''q~~''<sup>n</sup>}}. Dividing both sides by ''m'' gives{{nowrap\|''m''<sup>{{nowrap\|n − 1}}</sup>''r''<sup>n</sup> {{=}} ''q''<sup>n</sup>}}. This means that ''q'' must also be divisible by ''m'', which is a contradiction, since ''p'' and ''q'' were assumed to be relatively prime.		{{nowrap\|''p<sup>n</sup>'' / ''q<sup>n</sup>'' {{=}} ''m''}}, and {{nowrap\|''p<sup>n''</sup> {{=}} ''mq<sup>n''</sup>}}. This means ''p'' is divisible by ''m''. Therefore, there exists some integer ''r'' such that {{nowrap\|''p'' {{=}} ''mr''}}, so we now have {{nowrap\|''m<sup>n</sup>r<sup>n</sup>'' {{=}} ''mq<sup>n</sup>''}}. Dividing both sides by ''m'' gives{{nowrap\|''m''<sup>{{nowrap\|''n'' − 1}}</sup>''r<sup>n''</sup> {{=}} ''q<sup>n</sup>''}}. This means that ''q'' must also be divisible by ''m'', which is a contradiction, since ''p'' and ''q'' were assumed to be relatively prime.
	}}		}}
	</pre>		</pre>
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	\| title=Proof that the root of any integer is either an integer or irrational		\| title=Proof that the root of any integer is either an integer or irrational
	\| contents=Assume <math>\sqrt[n]{m}</math> is {{nowrap\|''p'' / ''q''}}, where {{nowrap\|''p'', ''q'' ∈ ℤ<sup>+</sup>}} and {{nowrap\|gcd(''p'', ''q'') {{=}} 1}}. Then		\| contents=Assume <math>\sqrt[n]{m}</math> is {{nowrap\|''p'' / ''q''}}, where {{nowrap\|''p'', ''q'' ∈ ℤ<sup>+</sup>}} and {{nowrap\|gcd(''p'', ''q'') {{=}} 1}}. Then
	{{nowrap\|''p<sup>n</sup>'' / ''q<sup>n</sup>'' {{=}} ''m''}}, and {{nowrap\|''p<sup>n</sup>'' {{=}} m''q<sup>n</sup>''}}. This means ''p'' is divisible by ''m''. Therefore, there exists some integer ''r'' such that {{nowrap\|''p'' {{=}} ''mr''}}, so we now have {{nowrap\|''m<sup>n</sup>r<sup>n</sup>'' {{=}} ''mq<sup>n</sup>''}}. Dividing both sides by ''m'' gives {{nowrap\|''m<sup>n~~ ~~&minus~~;&nbsp~~;1</sup>r<sup>n</sup>'' {{=}} ''q<sup>n</sup>''}}. This means that ''q'' must also be divisible by ''m'', which is a contradiction, since ''p'' and ''q'' were assumed to be relatively prime.		{{nowrap\|''p<sup>n</sup>'' / ''q<sup>n</sup>'' {{=}} ''m''}}, and {{nowrap\|''p<sup>n''</sup> {{=}} ''mq<sup>n''</sup>}}. This means ''p'' is divisible by ''m''. Therefore, there exists some integer ''r'' such that {{nowrap\|''p'' {{=}} ''mr''}}, so we now have {{nowrap\|''m<sup>n</sup>r<sup>n</sup>'' {{=}} ''mq<sup>n</sup>''}}. Dividing both sides by ''m'' gives{{nowrap\|''m''<sup>{{nowrap\|''n'' − 1}}</sup>''r<sup>n''</sup> {{=}} ''q<sup>n</sup>''}}. This means that ''q'' must also be divisible by ''m'', which is a contradiction, since ''p'' and ''q'' were assumed to be relatively prime.
	}}		}}

ArrowHead294 at 14:52, 14 November 2024

2024-11-14T14:52:04Z

← Older revision		Revision as of 14:52, 14 November 2024
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	\| title=Proof that the root of any integer is either an integer or irrational		\| title=Proof that the root of any integer is either an integer or irrational
	\| contents=Assume <math>\sqrt[n]{m}</math> is {{nowrap\|''p'' / ''q''}}, where {{nowrap\|''p'', ''q'' ∈ ℤ<sup>+</sup>}} and {{nowrap\|gcd(''p'', ''q'') {{=}} 1}}. Then		\| contents=Assume <math>\sqrt[n]{m}</math> is {{nowrap\|''p'' / ''q''}}, where {{nowrap\|''p'', ''q'' ∈ ℤ<sup>+</sup>}} and {{nowrap\|gcd(''p'', ''q'') {{=}} 1}}. Then
	{{nowrap\|''p''<sup>n</sup> / ''q''<sup>n</sup> {{=}} ''m''}}, and {{nowrap\|''p''<sup>n</sup> {{=}} m''q''<sup>n</sup>}}. This means ''p'' is divisible by ''m''. Therefore, there exists some integer ''r'' such that {{nowrap\|p {{=}} mr}}, so we now have {{nowrap\|''m''<sup>n</sup>''r''<sup>n</sup> {{=}} m''q''<sup>n</sup>}}. Dividing both sides by ''m'' gives{{nowrap\|''m''<sup>{{nowrap\|n − 1}}</sup>''r''<sup>n</sup> {{=}} ''q''<sup>n</sup>}}. This means that ''q'' must also be divisible by ''m'', which is a contradiction, since ''p'' and ''q'' were assumed to be relatively prime. ~~{{qed}}~~		{{nowrap\|''p''<sup>n</sup> / ''q''<sup>n</sup> {{=}} ''m''}}, and {{nowrap\|''p''<sup>n</sup> {{=}} m''q''<sup>n</sup>}}. This means ''p'' is divisible by ''m''. Therefore, there exists some integer ''r'' such that {{nowrap\|p {{=}} mr}}, so we now have {{nowrap\|''m''<sup>n</sup>''r''<sup>n</sup> {{=}} m''q''<sup>n</sup>}}. Dividing both sides by ''m'' gives{{nowrap\|''m''<sup>{{nowrap\|n − 1}}</sup>''r''<sup>n</sup> {{=}} ''q''<sup>n</sup>}}. This means that ''q'' must also be divisible by ''m'', which is a contradiction, since ''p'' and ''q'' were assumed to be relatively prime.
	}}		}}
	</pre>		</pre>

ArrowHead294 at 02:33, 31 August 2024

2024-08-31T02:33:47Z

← Older revision		Revision as of 02:33, 31 August 2024
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	You get:		You get:

	{{{{~~ROOTPAGENAMEE~~}}		{{{{ROOTPAGENAME}}
	\| title=Proof that the root of any integer is either an integer or irrational		\| title=Proof that the root of any integer is either an integer or irrational
	\| contents=Assume <math>\sqrt[n]{m}</math> is {{nowrap\|''p'' / ''q''}}, where {{nowrap\|''p'', ''q'' ∈ ℤ<sup>+</sup>}} and {{nowrap\|gcd(''p'', ''q'') {{=}} 1}}. Then		\| contents=Assume <math>\sqrt[n]{m}</math> is {{nowrap\|''p'' / ''q''}}, where {{nowrap\|''p'', ''q'' ∈ ℤ<sup>+</sup>}} and {{nowrap\|gcd(''p'', ''q'') {{=}} 1}}. Then
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	}}		}}

	~~Note: To include the pipe (<code>~~{{!}}</code>) character, it must be escaped as <code>{{((}}!{{))}}</code>, unless it appears inside preformatted text (such as a <code>{{^(}}syntaxhighlight{{)^}}</code> or <code>{{^(}}math{{)^}}</code> tag). Additionally, to include the equals sign, it must be escaped as <code>{{((}}={{))}}</code>.		{{escape notice}}

	=== See also ===		=== See also ===
	* [[Template:Theorem]]		* [[Template:Theorem]]

ArrowHead294 at 14:12, 29 August 2024

2024-08-29T14:12:31Z

← Older revision		Revision as of 14:12, 29 August 2024
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	You get:		You get:

	{{~~Databox~~		{{{{ROOTPAGENAMEE}}
	\| 1=Proof that the root of any integer is either an integer or irrational		\| title=Proof that the root of any integer is either an integer or irrational
	\| 2=Assume <math>\sqrt[n]{m}</math> is {{nowrap\|''p'' / ''q''}}, where {{nowrap\|''p'', ''q'' ∈ ℤ<sup>+</sup>}} and {{nowrap\|gcd(''p'', ''q'') {{=}} 1}}. Then		\| contents=Assume <math>\sqrt[n]{m}</math> is {{nowrap\|''p'' / ''q''}}, where {{nowrap\|''p'', ''q'' ∈ ℤ<sup>+</sup>}} and {{nowrap\|gcd(''p'', ''q'') {{=}} 1}}. Then
	{{nowrap\|''p<sup>n</sup>'' / ''q<sup>n</sup>'' {{=}} ''m''}}, and {{nowrap\|''p<sup>n</sup>'' {{=}} m''q<sup>n</sup>''}}. This means ''p'' is divisible by ''m''. Therefore, there exists some integer ''r'' such that {{nowrap\|''p'' {{=}} ''mr''}}, so we now have {{nowrap\|''m<sup>n</sup>r<sup>n</sup>'' {{=}} ''mq<sup>n</sup>''}}. Dividing both sides by ''m'' gives {{nowrap\|''m<sup>n − 1</sup>r<sup>n</sup>'' {{=}} ''q<sup>n</sup>''}}. This means that ''q'' must also be divisible by ''m'', which is a contradiction, since ''p'' and ''q'' were assumed to be relatively prime. ~~{{qed}}~~		{{nowrap\|''p<sup>n</sup>'' / ''q<sup>n</sup>'' {{=}} ''m''}}, and {{nowrap\|''p<sup>n</sup>'' {{=}} m''q<sup>n</sup>''}}. This means ''p'' is divisible by ''m''. Therefore, there exists some integer ''r'' such that {{nowrap\|''p'' {{=}} ''mr''}}, so we now have {{nowrap\|''m<sup>n</sup>r<sup>n</sup>'' {{=}} ''mq<sup>n</sup>''}}. Dividing both sides by ''m'' gives {{nowrap\|''m<sup>n − 1</sup>r<sup>n</sup>'' {{=}} ''q<sup>n</sup>''}}. This means that ''q'' must also be divisible by ''m'', which is a contradiction, since ''p'' and ''q'' were assumed to be relatively prime.
	}}		}}

	Note: To include the pipe (<code>\|</code>) character, it must be escaped as <code>{{((}}!{{))}}</code>, unless it appears inside preformatted text (such as a <~~nowiki~~><syntaxhighlight></~~nowiki~~> or <~~nowiki~~><math></~~nowiki~~> tag). Additionally, to include the equals sign, it must be escaped as <code>{{((}}={{))}}</code>.		Note: To include the pipe (<code>{{!}}</code>) character, it must be escaped as <code>{{((}}!{{))}}</code>, unless it appears inside preformatted text (such as a <code>{{^(}}syntaxhighlight{{)^}}</code> or <code>{{^(}}math{{)^}}</code> tag). Additionally, to include the equals sign, it must be escaped as <code>{{((}}={{))}}</code>.

	=== See also ===		=== See also ===
	* [[Template:Theorem]]		* [[Template:Theorem]]

ArrowHead294: Generalised proof

2024-07-09T18:47:35Z

Generalised proof

← Older revision		Revision as of 18:47, 9 July 2024
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	<pre>		<pre>
	{{Proof\|title=Proof that ~~<math>\sqrt{2}</math>~~ is irrational\|		{{Proof
	contents=Assume <math>\sqrt{2}</math> is ''p''/''q'', where ''p'', ''q'' ∈ ℤ<~~sub~~>>0</~~sub~~> and gcd(''p'', ''q'') = 1. Then ''p''<sup>2</sup>/''q''<sup>2</sup> = 2, and ''p''<sup>2</sup> = 2''q''<sup>2</sup>. ~~Since~~ ''p''~~<sup>2</sup>~~ is ~~even, and gcd(~~''p'', ''q'') = 1, ~~gcd(~~''p''<sup>2</sup>, ''q''<sup>2</sup>) = 1, ''q''<sup>2</sup> ~~must be odd~~. ~~Hence~~ ''p''<sup>2</sup> ~~≡ 2 (mod 4)~~, which is ~~impossible because the square of an integer must always~~ be ~~congruent to 0 or 1 (mod 4)~~.		\| title=Proof that the root of any integer is either an integer or irrational
			\| contents=Assume <math>\sqrt[n]{m}</math> is {{nowrap\|''p'' / ''q''}}, where {{nowrap\|''p'', ''q'' ∈ ℤ<sup>+</sup>}} and {{nowrap\|gcd(''p'', ''q'') {{=}} 1}}. Then
			{{nowrap\|''p''<sup>n</sup> / ''q''<sup>n</sup> {{=}} ''m''}}, and {{nowrap\|''p''<sup>n</sup> {{=}} m''q''<sup>n</sup>}}. This means ''p'' is divisible by ''m''. Therefore, there exists some integer ''r'' such that {{nowrap\|p {{=}} mr}}, so we now have {{nowrap\|''m''<sup>n</sup>''r''<sup>n</sup> {{=}} m''q''<sup>n</sup>}}. Dividing both sides by ''m'' gives{{nowrap\|''m''<sup>{{nowrap\|n − 1}}</sup>''r''<sup>n</sup> {{=}} ''q''<sup>n</sup>}}. This means that ''q'' must also be divisible by ''m'', which is a contradiction, since ''p'' and ''q'' were assumed to be relatively prime. {{qed}}
	}}		}}
	</pre>		</pre>
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	{{Databox		{{Databox
	\| 1=Proof that ~~<math>\sqrt{2}</math>~~ is irrational		\| 1=Proof that the root of any integer is either an integer or irrational
	\| 2=Assume <math>\sqrt{2}</math> is ''p''/''q'', where ''p'', ''q'' ∈ ℤ<~~sub>~~>0</~~sub~~> and gcd(''p'', ''q'') = 1. Then ''p''<sup>2</sup>/''q''<sup>2</sup> = 2, and ''p''<sup>2</sup> = 2''q''<sup>2</sup>. ~~Since~~ ''p''~~<sup>2</sup>~~ is ~~even~~, ~~and gcd(~~''p'', ''q'') = ~~1, gcd(~~''p''<sup>2</sup>, ''q''<sup>2</sup>~~) = 1,~~ ''q''<sup>2</sup> ~~must be odd. Hence~~ ''p''<sup>2</sup> ~~≡ 2 (mod 4)~~, which is ~~impossible because the square of an integer must always~~ be ~~congruent to 0 or 1 (mod 4)~~. {{qed}}		\| 2=Assume <math>\sqrt[n]{m}</math> is {{nowrap\|''p'' / ''q''}}, where {{nowrap\|''p'', ''q'' ∈ ℤ<sup>+</sup>}} and {{nowrap\|gcd(''p'', ''q'') {{=}} 1}}. Then
			{{nowrap\|''p<sup>n</sup>'' / ''q<sup>n</sup>'' {{=}} ''m''}}, and {{nowrap\|''p<sup>n</sup>'' {{=}} m''q<sup>n</sup>''}}. This means ''p'' is divisible by ''m''. Therefore, there exists some integer ''r'' such that {{nowrap\|''p'' {{=}} ''mr''}}, so we now have {{nowrap\|''m<sup>n</sup>r<sup>n</sup>'' {{=}} ''mq<sup>n</sup>''}}. Dividing both sides by ''m'' gives {{nowrap\|''m<sup>n − 1</sup>r<sup>n</sup>'' {{=}} ''q<sup>n</sup>''}}. This means that ''q'' must also be divisible by ''m'', which is a contradiction, since ''p'' and ''q'' were assumed to be relatively prime. {{qed}}
	}}		}}

	Note: To include the ~~character~~ <code>\|</code>, it must be escaped as <code>{{((}}!{{))}}</code>, unless it appears inside preformatted text (such as a <nowiki><syntaxhighlight></nowiki> or <nowiki><math></nowiki> tag).		Note: To include the pipe (<code>\|</code>) character, it must be escaped as <code>{{((}}!{{))}}</code>, unless it appears inside preformatted text (such as a <nowiki><syntaxhighlight></nowiki> or <nowiki><math></nowiki> tag). Additionally, to include the equals sign, it must be escaped as <code>{{((}}={{))}}</code>.

	=== See also ===		=== See also ===
	* [[Template:Theorem]]		* [[Template:Theorem]]

ArrowHead294: Docpage

2024-05-22T02:31:56Z

Docpage

New page

{{dochead}}
=== Usage ===
You type:

<pre>
{{Proof|title=Proof that <math>\sqrt{2}</math> is irrational|
contents=Assume <math>\sqrt{2}</math> is ''p''/''q'', where ''p'', ''q'' ∈ ℤ>0 and gcd(''p'', ''q'') = 1. Then ''p''2/''q''2 = 2, and ''p''2 = 2''q''2. Since ''p''2 is even, and gcd(''p'', ''q'') = 1, gcd(''p''2, ''q''2) = 1, ''q''2 must be odd. Hence ''p''2 ≡ 2 (mod 4), which is impossible because the square of an integer must always be congruent to 0 or 1 (mod 4).
}}
</pre>

You get:

{{Databox
| 1=Proof that <math>\sqrt{2}</math> is irrational
| 2=Assume <math>\sqrt{2}</math> is ''p''/''q'', where ''p'', ''q'' ∈ ℤ>0 and gcd(''p'', ''q'') = 1. Then ''p''2/''q''2 = 2, and ''p''2 = 2''q''2. Since ''p''2 is even, and gcd(''p'', ''q'') = 1, gcd(''p''2, ''q''2) = 1, ''q''2 must be odd. Hence ''p''2 ≡ 2 (mod 4), which is impossible because the square of an integer must always be congruent to 0 or 1 (mod 4). {{qed}}
}}

Note: To include the character <code>|</code>, it must be escaped as <code>{{((}}!{{))}}</code>, unless it appears inside preformatted text (such as a <nowiki><syntaxhighlight></nowiki> or <nowiki><math></nowiki> tag).

=== See also ===
* [[Template:Theorem]]