Template:Proof/doc: Difference between revisions

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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|''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.

}}

</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> {{=}} 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> {{=}} 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.
 }}
 </pre>