Template:Proof/doc: Difference between revisions

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{{~~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+}} and {{nowrap|gcd(''p'', ''q'') {{=}} 1}}. Then

| contents=Assume <math>\sqrt[n]{m}</math> is {{nowrap|''p'' / ''q''}}, where {{nowrap|''p'', ''q'' ∈ ℤ+}} and {{nowrap|gcd(''p'', ''q'') {{=}} 1}}. Then

{{nowrap|''pn'' / ''qn'' {{=}} ''m''}}, and {{nowrap|''pn'' {{=}} m''qn''}}. This means ''p'' is divisible by ''m''. Therefore, there exists some integer ''r'' such that {{nowrap|''p'' {{=}} ''mr''}}, so we now have {{nowrap|''mnrn'' {{=}} ''mqn''}}. Dividing both sides by ''m'' gives {{nowrap|''mn − 1rn'' {{=}} ''qn''}}. 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|''pn'' / ''qn'' {{=}} ''m''}}, and {{nowrap|''pn'' {{=}} m''qn''}}. This means ''p'' is divisible by ''m''. Therefore, there exists some integer ''r'' such that {{nowrap|''p'' {{=}} ''mr''}}, so we now have {{nowrap|''mnrn'' {{=}} ''mqn''}}. Dividing both sides by ''m'' gives {{nowrap|''mn − 1rn'' {{=}} ''qn''}}. 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 ===

* [[Template:Theorem]]

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 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&nbsp;&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. {{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&nbsp;&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.
 }}
-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 ===
 * [[Template:Theorem]]