https://artofproblemsolving.com/wiki/index.php?title=Ostrowski%27s_criterion&feed=atom&action=history
Ostrowski's criterion - Revision history
2024-03-28T09:20:31Z
Revision history for this page on the wiki
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https://artofproblemsolving.com/wiki/index.php?title=Ostrowski%27s_criterion&diff=156001&oldid=prev
Tigerzhang at 15:23, 15 June 2021
2021-06-15T15:23:35Z
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 15:23, 15 June 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l1" >Line 1:</td>
<td colspan="2" class="diff-lineno">Line 1:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Ostrowski's Criterion states that:</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Ostrowski's Criterion states that:</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del class="diffchange diffchange-inline">Left </del><math>f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\in \mathbb{Z}[x]</math>. If <math>a_0</math> is a prime and</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">Let </ins><math>f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\in \mathbb{Z}[x]</math>. If <math>a_0</math> is a prime and</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><cmath>|a_0|>|a_n|+|a_{n-1}|+\cdots+|a_1|</cmath></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><cmath>|a_0|>|a_n|+|a_{n-1}|+\cdots+|a_1|</cmath></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>then <math>f(x)</math> is irreducible.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>then <math>f(x)</math> is irreducible.</div></td></tr>
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Tigerzhang
https://artofproblemsolving.com/wiki/index.php?title=Ostrowski%27s_criterion&diff=97234&oldid=prev
Naman12 at 20:30, 14 August 2018
2018-08-14T20:30:32Z
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 20:30, 14 August 2018</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l5" >Line 5:</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>then <math>f(x)</math> is irreducible.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>then <math>f(x)</math> is irreducible.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Proof<del class="diffchange diffchange-inline">: </del>Let <math>\phi</math> be a root of <math>f(x)</math>. If <math>|\phi|\leq 1</math>, then</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">==</ins>Proof<ins class="diffchange diffchange-inline">==</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Let <math>\phi</math> be a root of <math>f(x)</math>. If <math>|\phi|\leq 1</math>, then</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><cmath>|a_0|=|a_1\phi+\cdots+a_n\phi^n|\leq |a_1|+\cdots+|a_n|</cmath></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><cmath>|a_0|=|a_1\phi+\cdots+a_n\phi^n|\leq |a_1|+\cdots+|a_n|</cmath></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>a contradiction. Therefore, <math>|\phi|>1</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>a contradiction. Therefore, <math>|\phi|>1</math>.</div></td></tr>
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Naman12
https://artofproblemsolving.com/wiki/index.php?title=Ostrowski%27s_criterion&diff=97233&oldid=prev
Naman12: Ostrowski's Criterion
2018-08-14T20:30:12Z
<p>Ostrowski's Criterion</p>
<p><b>New page</b></p><div>Ostrowski's Criterion states that:<br />
<br />
Left <math>f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\in \mathbb{Z}[x]</math>. If <math>a_0</math> is a prime and<br />
<cmath>|a_0|>|a_n|+|a_{n-1}|+\cdots+|a_1|</cmath><br />
then <math>f(x)</math> is irreducible.<br />
<br />
Proof: Let <math>\phi</math> be a root of <math>f(x)</math>. If <math>|\phi|\leq 1</math>, then<br />
<cmath>|a_0|=|a_1\phi+\cdots+a_n\phi^n|\leq |a_1|+\cdots+|a_n|</cmath><br />
a contradiction. Therefore, <math>|\phi|>1</math>.<br />
<br />
Suppose <math>f(x)=g(x)h(x)</math>. Since <math>f(0)=a_0</math>, one of <math>g(0)</math> and <math>h(0)</math> is 1. WLOG, assume <math>g(0)=1</math>. Then, let <math>g_n</math> be the leading coefficient of <math>g(x)</math>. If <math>\phi_1,\phi_2,\cdots,\phi_r</math> are the roots of <math>g(x)</math>, then <math>1<|\phi_1\phi_2\cdots \phi_n|=\frac{1}{|b|}\leq 1</math>. This is a contradiction, so <math>f(x)</math> is irreducible.<br />
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Naman12