Template:Proof/doc: Difference between revisions

(One intermediate revision by the same user not shown)

Line 1:

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:

Line 7:

Line 9:

| 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'' {{=}} ''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.

{{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>

Line 16:

Line 18:

| 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'' {{=}} ''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.

{{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.

}}

@@ Line 1: / Line 1: @@
 {{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:
@@ Line 7: / Line 9: @@
 | 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> {{=}} ''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.
+{{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>
@@ Line 16: / Line 18: @@
 | 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> {{=}} ''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.
+{{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.
 }}