Template:Proof/doc: Difference between revisions

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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 ===

You type:

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

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

}}

</pre>

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| 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+}} 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~~ ~~&minus~~;&nbsp~~;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.

{{nowrap|''pn'' / ''qn'' {{=}} ''m''}}, and {{nowrap|''pn'' {{=}} ''mqn''}}. 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|''m''{{nowrap|''n'' − 1}}''rn'' {{=}} ''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.

}}

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 {{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 ===
 You type:
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 | 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 &minus; 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> {{=}} ''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'' &minus; 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>
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 | 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>'' {{=}} ''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.
+{{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'' &minus; 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}}