MaplePrimes - comments on Post, Could the MRB constant be rational?

<rss xmlns:itunes="http://www.itunes.com/dtds/podcast-1.0.dtd" version="2.0">
  <channel>
    <title>MaplePrimes - comments on Post, Could the MRB constant be rational?</title>
    <link>http://www.mapleprimes.com/posts/101425-Could-The-MRB-Constant-Be-Rational</link>
    <language>en-us</language>
    <copyright>2026 Maplesoft, A Division of Waterloo Maple Inc.</copyright>
    <generator>Maplesoft Document System</generator>
    <lastBuildDate>Fri, 12 Jun 2026 01:54:24 GMT</lastBuildDate>
    <pubDate>Fri, 12 Jun 2026 01:54:24 GMT</pubDate>
    <itunes:subtitle />
    <itunes:summary />
    <description>The latest comments added to the Post, Could the MRB constant be rational?</description>
    <image>
      <url>http://www.mapleprimes.com/images/mapleprimeswhite.jpg</url>
      <title>MaplePrimes - comments on Post, Could the MRB constant be rational?</title>
      <link>http://www.mapleprimes.com/posts/101425-Could-The-MRB-Constant-Be-Rational</link>
    </image>
    <item>
      <title>&amp;nbsp;
</title>
      <link>http://www.mapleprimes.com/posts/101425-Could-The-MRB-Constant-Be-Rational?ref=Feed:MaplePrimes:Could the MRB constant be rational?:Comments#comment101539</link>
      <itunes:summary>&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;!--break--&gt;
&lt;p&gt;Note to self:&lt;/p&gt;
&lt;p&gt;Let x=q^(1/n) where n is the term of the power series.&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&lt;a href="http://marvinrayburns.com"&gt;marvinrayburns.com&lt;br&gt;&lt;/a&gt;&lt;/p&gt;</itunes:summary>
      <description>The latest comments added to the Post, Could the MRB constant be rational?</description>
      <guid>101539</guid>
      <pubDate>Wed, 09 Feb 2011 09:50:46 Z</pubDate>
      <itunes:author>Marvin Ray Burns</itunes:author>
      <author>Marvin Ray Burns</author>
    </item>
    <item>
      <title>Attempt 2</title>
      <link>http://www.mapleprimes.com/posts/101425-Could-The-MRB-Constant-Be-Rational?ref=Feed:MaplePrimes:Could the MRB constant be rational?:Comments#comment101556</link>
      <itunes:summary>&lt;p&gt;&lt;br&gt;p= q*sum((-1)^n*(n^(1/n)),n=1..infinity). By an associate law,&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;p=sum((-1)^n*q*(n^(1/n)),n=1..infinity) Expressing n and q by the same exponent we get&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;p=sum((-1)^n*(n*q^n)^(1/n),n=1..infinity). When we expand the series we see that&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;p=-1^(1/1)*(q^1)^(1/1)+2^(1/2)*(q^2)^(1/2)-3^(1/3)*(q^3)^(1/3)+4^(1/4)*(q^4)^(1/4)-... . Since (A^B)^C=(A^C)^B when they are all positive:&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;p=-1^(1/1)*(q^(1/1))^1+2^(1/2)*(q^(1/2))^2-3^(1/3)*(q^(1/3))^3+4^(1/4)*(q^(1/4))^4... . Accordingly&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;______________My mistake is still here.________________&lt;/p&gt;
&lt;p&gt;p=-1^(1/1)*a1^1+2^(1/2)*a2^2-3^(1/3)*a3^3+4^(1/4)*a4^4-..., where ax=(q^(1/x))^x., and assuming q and x are positive&amp;nbsp; (q^(1/x))^x =q.&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;Let x=q^(1/n) where n is the term of the power series&lt;/p&gt;
&lt;p&gt;&amp;nbsp;p=-1^(1/1)*x^1+2^(1/2)*x^2-3^(1/3)*x^3+4^(1/4)*x^4-&amp;hellip; . Since 1^(1/1) =1,&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;p=-1x^1+2^(1/2)*x^2-3^(1/3)*x^3+4^(1/4)*x^4-&amp;hellip;&amp;nbsp; . &amp;nbsp;Factoring out a x we get&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;p=x*(-1+2^(1/2)*x-3^(1/3)*x^2+4^(1/4)*x^3-&amp;hellip;).&amp;nbsp; Dividing both sides by x!=0 we get&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;p/x&amp;nbsp; =&amp;nbsp; -1+2^(1/2)*x-3^(1/3)*x^2+4^(1/4)*x^3-&amp;hellip; . Adding 1 to each side gives&lt;/p&gt;
&lt;p&gt;&amp;nbsp;p/x+1&amp;nbsp; =&amp;nbsp; +2^(1/2)*x-3^(1/3)*x^2+4^(1/4)*x^3-&amp;hellip; . Again, dividing both sides by x!=0 we get&lt;/p&gt;
&lt;p&gt;&amp;nbsp;(p/x+1)/x&amp;nbsp; =&amp;nbsp; +2^(1/2)-3^(1/3)*x+4^(1/4)*x^2-&amp;hellip; . Adding -2^(1/2) to each side gives&lt;/p&gt;
&lt;p&gt;&amp;nbsp;(p/x+1)/x-2^(1/2)&amp;nbsp; =&amp;nbsp; -3^(1/3)*x+4^(1/4)*x^2-&amp;hellip; . Repeating the previous two steps:&lt;/p&gt;
&lt;p&gt;&amp;nbsp;((p/x+1)/x-2^(1/2))/x&amp;nbsp; =&amp;nbsp; -3^(1/3)+4^(1/4)*x-&amp;hellip; ., and&lt;/p&gt;
&lt;p&gt;&amp;nbsp;((p/x+1)/x-2^(1/2))/x+3^(1/3)&amp;nbsp; =&amp;nbsp; +4^(1/4)*x-5^(1/5)+... .&lt;/p&gt;
&lt;p&gt;&amp;nbsp;((((((p/x+1)/x-2^(1/2))/x+3^(1/3))/x)-4^(1/4))/x+5^(1/5))/&amp;hellip;+&amp;hellip;=0. In pretty print to see the pattern that is forming:&lt;/p&gt;
&lt;form&gt; 
&lt;table width="576" align="center"&gt;
&lt;tbody&gt;
&lt;tr&gt;
&lt;td&gt;
&lt;table style="margin-left: 0px; margin-right: 0px;"&gt;
&lt;tbody&gt;
&lt;tr valign="baseline"&gt;
&lt;td&gt;&lt;span style="color: #ff0000; font-size: 100%; font-family: monospace,monospace; font-weight: bold; font-style: normal;"&gt;&amp;gt;&amp;nbsp;&lt;/span&gt;&lt;/td&gt;
&lt;td&gt;
&lt;p style="margin: 0 0 0 0; padding-top: 0px; padding-bottom: 0px;"&gt;&amp;nbsp;&lt;/p&gt;
&lt;/td&gt;
&lt;/tr&gt;
&lt;/tbody&gt;
&lt;/table&gt;
&lt;table style="margin-left: 0px; margin-right: 0px;"&gt;
&lt;tbody&gt;
&lt;tr valign="baseline"&gt;
&lt;td&gt;&lt;span style="color: #ff0000; font-size: 100%; font-family: monospace,monospace; font-weight: bold; font-style: normal;"&gt;&amp;gt;&amp;nbsp;&lt;/span&gt;&lt;/td&gt;
&lt;td&gt;
&lt;p style="margin: 0 0 0 0; padding-top: 0px; padding-bottom: 0px;"&gt;&lt;span style="color: #ff0000; font-size: 100%; font-family: monospace,monospace; font-weight: bold; font-style: normal;"&gt;&amp;nbsp; &lt;/span&gt;&lt;/p&gt;
&lt;/td&gt;
&lt;/tr&gt;
&lt;/tbody&gt;
&lt;/table&gt;
&lt;table style="margin-left: 0px; margin-right: 0px;"&gt;
&lt;tbody&gt;
&lt;tr valign="baseline"&gt;
&lt;td&gt;&lt;span style="color: #ff0000; font-size: 100%; font-family: monospace,monospace; font-weight: bold; font-style: normal;"&gt;&amp;gt;&amp;nbsp;&lt;/span&gt;&lt;/td&gt;
&lt;td&gt;
&lt;p style="margin: 0 0 0 0; padding-top: 0px; padding-bottom: 0px;"&gt;&lt;img style="vertical-align: -16;" src="/view.aspx?sf=101556/319535/bd47aea67b8c94503e69471542129288.gif" alt="" width="369" height="118"&gt;&lt;/p&gt;
&lt;/td&gt;
&lt;/tr&gt;
&lt;/tbody&gt;
&lt;/table&gt;
&lt;table style="margin-left: 0px; margin-right: 0px;"&gt;
&lt;tbody&gt;
&lt;tr valign="baseline"&gt;
&lt;td&gt;&lt;span style="color: #ff0000; font-size: 100%; font-family: monospace,monospace; font-weight: bold; font-style: normal;"&gt;&amp;gt;&amp;nbsp;&lt;/span&gt;&lt;/td&gt;
&lt;td&gt;
&lt;p style="margin: 0 0 0 0; padding-top: 0px; padding-bottom: 0px;"&gt;&amp;nbsp;&lt;/p&gt;
&lt;/td&gt;
&lt;/tr&gt;
&lt;/tbody&gt;
&lt;/table&gt;
&lt;p style="margin: 0 0 0 0; padding-top: 0px; padding-bottom: 0px;"&gt;&lt;span style="color: #000000; font-size: 100%; font-family: Times New Roman,serif; font-weight: normal; font-style: normal;"&gt;&amp;nbsp;&lt;/span&gt;&lt;/p&gt;
&lt;/td&gt;
&lt;/tr&gt;
&lt;/tbody&gt;
&lt;/table&gt;
&lt;br&gt;&lt;br&gt;&lt;a href="/view.aspx?sf=101556/319535/insert_Feb_09_201.mw"&gt;Download insert_Feb_09_201.mw&lt;/a&gt;&lt;/form&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;An inspection shows that, as n-&amp;gt;infinity, the LHS becomes independent of p.&lt;/p&gt;
&lt;p&gt;Now here do I dare say that p can be an integer even if q is?&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;</itunes:summary>
      <description>The latest comments added to the Post, Could the MRB constant be rational?</description>
      <guid>101556</guid>
      <pubDate>Thu, 10 Feb 2011 01:35:33 Z</pubDate>
      <itunes:author>Marvin Ray Burns</itunes:author>
      <author>Marvin Ray Burns</author>
    </item>
    <item>
      <title>Got nowhere</title>
      <link>http://www.mapleprimes.com/posts/101425-Could-The-MRB-Constant-Be-Rational?ref=Feed:MaplePrimes:Could the MRB constant be rational?:Comments#comment101593</link>
      <itunes:summary>&lt;p&gt;After correcting all of my mistakes, I see that I haven't got anywhere!&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;p= q*sum((-1)^n*(n^(1/n)),n=1..infinity). By an associate law,&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;p=sum((-1)^n*q*(n^(1/n)),n=1..infinity) Expressing n and q by the same exponent we get&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;p=sum((-1)^n*(n*q^n)^(1/n),n=1..infinity). When we expand the series we see that&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;p=-1^(1/1)*(q^1)^(1/1)+2^(1/2)*(q^2)^(1/2)-3^(1/3)*(q^3)^(1/3)+4^(1/4)*(q^4)^(1/4)-... . Since (A^B)^C=(A^C)^B when they are all positive:&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;p=-1^(1/1)*(q^(1/1))^1+2^(1/2)*(q^(1/2))^2-3^(1/3)*(q^(1/3))^3+4^(1/4)*(q^(1/4))^4... . Accordingly&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;p=-1^(1/1)*a1^1+2^(1/2)*a2^2-3^(1/3)*a3^3+4^(1/4)*a4^4-..., where ax=(q^(1/x))^x., and assuming q and x are positive&amp;nbsp; (q^(1/x))^x =q.&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; Let x=q^(1/n) where n is the term of the power series&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp; So to determine if it was rational you would have to have integers,p1,q1,p,q such that &lt;br&gt;&amp;nbsp;&lt;br&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; p1/q1=sum((-1)^n*(n^(1/n)-1),n=1..infinity), which implies that&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; p/q= sum((-1)^n*(n^(1/n)),n=1..infinity). Multiplying both sides by q we get&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; p= q*sum((-1)^n*(n^(1/n)),n=1..infinity). By an assosiate law,&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; p=sum((-1)^n*q*(n^(1/n)),n=1..infinity) Expressing n and q by the same exponent we get&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; p=sum((-1)^n*(n*q^n)^(1/n),n=1..infinity). When we expand the series we see that&lt;/p&gt;
&lt;p&gt;_____________________________A change made here._____________________________________&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; p=-q+sqrt(2)*sqrt(q^2)-3^(1/3)*(q^3)^(1/3)+4^(1/4)*(q^4)^(1/4)-5^(1/5)*(q^5)^(1/5)+6^(1/6)*(q^6)^(1/6)-...&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; p+q=sqrt(2)*sqrt(q^2)-3^(1/3)*(q^3)^(1/3)+4^(1/4)*(q^4)^(1/4)-5^(1/5)*(q^5)^(1/5)+6^(1/6)*(q^6)^(1/6)-... since sqrt(q^2)=q,&lt;br&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; &lt;br&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; p+q=sqrt(2)*q-3^(1/3)*(q^3)^(1/3)+4^(1/4)*(q^4)^(1/4)-5^(1/5)*(q^5)^(1/5)+6^(1/6)*(q^6)^(1/6)-...&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; p+q=q*(sqrt(2)-3^(1/3)+4^(1/4)-5^(1/5)+6^(1/6)-...)&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; (p+q)/q=(sqrt(2)-3^(1/3)+4^(1/4)-5^(1/5)+6^(1/6)-...). In series form we have&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; (p+q)/q= sum((-1)^n*(n^(1/n)),n=1..infinity)+1. Thus&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; (p+q)/q-1= sum((-1)^n*(n^(1/n)),n=1..infinity). But remember that&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; p/q= sum((-1)^n*(n^(1/n)),n=1..infinity). Since q!=0,&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; (p+q)/q-1 = p/q. Multiplying both sides by gives,&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; (p+q)-q = p. By association,&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; p+(q-q) = p. Since q-q=0,&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; p=p.&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;</itunes:summary>
      <description>The latest comments added to the Post, Could the MRB constant be rational?</description>
      <guid>101593</guid>
      <pubDate>Fri, 11 Feb 2011 01:29:38 Z</pubDate>
      <itunes:author>Marvin Ray Burns</itunes:author>
      <author>Marvin Ray Burns</author>
    </item>
    <item>
      <title>&amp;nbsp;p1/q1 =MRB constant. By definition</title>
      <link>http://www.mapleprimes.com/posts/101425-Could-The-MRB-Constant-Be-Rational?ref=Feed:MaplePrimes:Could the MRB constant be rational?:Comments#comment101598</link>
      <itunes:summary>&lt;p&gt;&lt;span&gt;&amp;nbsp;&lt;span&gt;p1/q1 =MRB constant. By definition then&lt;/span&gt;&lt;/span&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;&lt;span&gt;p1/q1= sum((-1)^n*(n^(1/n)-1),n=1..infinity) and since sum((-1)^n*(n^(1/n)-1),n=1..infinity)-1 = sum((-1)^n*(n^(1/n)-1),n=1..infinity)-q1/q1= sum((-1)^n*(n^(1/n)),n=1..infinity), and p1/q1-q1/q1=(p1-q1)/q1 =p/q1=p/q. Thus q=q1 and
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/span&gt;&lt;/p&gt;
&lt;p&gt;p/q= sum((-1)^n*(n^(1/n)),n=1..infinity). Multiplying by q we get
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;p= q*sum((-1)^n*(n^(1/n)),n=1..infinity). By an associate law,
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;&amp;nbsp;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;p=sum((-1)^n*q*(n^(1/n)),n=1..infinity) Expressing n and q by the same exponent we get
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;&amp;nbsp;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;p=sum((-1)^n*(n*q^n)^(1/n),n=1..infinity). When we expand the series we see that
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;&amp;nbsp;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;p=-1^(1/1)*(q^1)^(1/1)+2^(1/2)*(q^2)^(1/2)-3^(1/3)*(q^3)^(1/3)+4^(1/4)*(q^4)^(1/4)-... . Since (A^B)^C=(A^C)^B when they are all positive:
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;&amp;nbsp;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;p=-1^(1/1)*(q^(1/1))^1+2^(1/2)*(q^(1/2))^2-3^(1/3)*(q^(1/3))^3+4^(1/4)*(q^(1/4))^4... . Accordingly
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;&amp;nbsp;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;p=-1^(1/1)*a1^1+2^(1/2)*a2^2-3^(1/3)*a3^3+4^(1/4)*a4^4-..., where ax=(q^(1/x))^x., and assuming q and x are positive&amp;nbsp; (q^(1/x))^x =q.
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;&amp;nbsp;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; Let x=q^(1/n) where n is the term of the power series
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp; So to determine if it was rational you would have to have integers,p1,q1,p,q such that &lt;br&gt;&amp;nbsp;&lt;br&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; p1/q1=sum((-1)^n*(n^(1/n)-1),n=1..infinity), which implies that
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; p/q= sum((-1)^n*(n^(1/n)),n=1..infinity). Multiplying both sides by q we get&amp;nbsp;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; p= q*sum((-1)^n*(n^(1/n)),n=1..infinity). By an assosiate law,
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; p=sum((-1)^n*q*(n^(1/n)),n=1..infinity) Expressing n and q by the same exponent we get
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; p=sum((-1)^n*(n*q^n)^(1/n),n=1..infinity). When we expand the series we see that
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;_____________________________A change made here._____________________________________
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; p=-q+sqrt(2)*sqrt(q^2)-3^(1/3)*(q^3)^(1/3)+4^(1/4)*(q^4)^(1/4)-5^(1/5)*(q^5)^(1/5)+6^(1/6)*(q^6)^(1/6)-...
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; p+q=sqrt(2)*sqrt(q^2)-3^(1/3)*(q^3)^(1/3)+4^(1/4)*(q^4)^(1/4)-5^(1/5)*(q^5)^(1/5)+6^(1/6)*(q^6)^(1/6)-... since sqrt(q^2)=q,&lt;br&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; &lt;br&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; p+q=sqrt(2)*q-3^(1/3)*(q^3)^(1/3)+4^(1/4)*(q^4)^(1/4)-5^(1/5)*(q^5)^(1/5)+6^(1/6)*(q^6)^(1/6)-...
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; p+q=q*(sqrt(2)-3^(1/3)+4^(1/4)-5^(1/5)+6^(1/6)-...). Diving by q!=0,
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; (p+q)/q=(sqrt(2)-3^(1/3)+4^(1/4)-5^(1/5)+6^(1/6)-...). In series form we have
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;__________________________A new change made here.__________________________________
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; (p+q)/q = sum((-1)^n*n^(1/n), n = 2 .. infinity), and
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; sum((-1)^n*n^(1/n), n = 2 .. infinity)=MRB constant+1/2. Accordingly
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; (p+q)/q=MRB constant+1/2, But&amp;nbsp;&lt;span&gt;&amp;nbsp;&lt;span&gt;p1/q1 =MRB constant,&lt;span&gt;&lt;span&gt; so by subsitution and reflextivity,&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&amp;nbsp;(p+q)/q=p1/q1+1/2, Remembering that q=q1 we get&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&amp;nbsp;(p+q)/q=p1/q+1/2. Since 1/2 =(1/2*q)/q,&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&amp;nbsp;(p+q)/q=p1/q+(1/2*q)/q, Multiplying by q we get&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&amp;nbsp;(p+q)=p1+1/2*q.&amp;nbsp;Subtracting 1/2*q and by reflextivity&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;p1 =&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;(p+q)-1/2*q. Simplifying gives an end result of&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/p&gt;
&lt;p&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&lt;span&gt;&amp;nbsp;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&amp;nbsp;&lt;span&gt;&lt;img class="math" src="http://www.mapleprimes.com/MapleImage.ashx?f=2bab5a1dfe47d06b9f3d0956efae9633.gif" alt="p1=p+(1/2)*q"&gt;.&lt;/span&gt;&lt;/p&gt;
&lt;p&gt;&lt;span&gt;Now if p is an integer&amp;nbsp;and q is odd&amp;nbsp; then p1 is not an integer.&lt;/span&gt;&lt;/p&gt;
&lt;p&gt;&lt;span&gt;I'll have to look at this for a little bit. I doubt that I proved anything that easily.&lt;/span&gt;&lt;/p&gt;
&lt;p&gt;&lt;span&gt;&amp;nbsp;&lt;/span&gt;&lt;/p&gt;
&lt;p&gt;
&lt;p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;
&lt;p&gt;&amp;nbsp;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;/p&gt;
&lt;/p&gt;
&lt;/p&gt;
&lt;/p&gt;</itunes:summary>
      <description>The latest comments added to the Post, Could the MRB constant be rational?</description>
      <guid>101598</guid>
      <pubDate>Fri, 11 Feb 2011 07:43:47 Z</pubDate>
      <itunes:author>Marvin Ray Burns</itunes:author>
      <author>Marvin Ray Burns</author>
    </item>
    <item>
      <title>irrationality</title>
      <link>http://www.mapleprimes.com/posts/101425-Could-The-MRB-Constant-Be-Rational?ref=Feed:MaplePrimes:Could the MRB constant be rational?:Comments#comment101613</link>
      <itunes:summary>&lt;p&gt;Proving that a number is irrational is generally very difficult (with some exceptions, notably algebraic numbers).&amp;nbsp; This is not an area where an amateur is likely to succeed.&amp;nbsp;&lt;/p&gt;
&lt;p&gt;You wrote:&lt;/p&gt;
&lt;p&gt;&lt;em&gt;&lt;span class="mainBody document"&gt;&lt;span style="font-family: 'Verdana','sans-serif'; color: black; font-size: 7.5pt;"&gt;&lt;span style="font-family: 'Verdana','sans-serif'; color: black; font-size: 7.5pt;"&gt;For  instance, if you had an infinite number of irrational numbers and half  of them were the sqrt(2) and the other half were &amp;ndash;sqrt(2), their sum  would not have to&amp;nbsp;be irrational, would it?&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/em&gt;&lt;/p&gt;
&lt;p&gt;In that case, their sum would be neither rational nor irrational, it would simply not exist, i.e. it would be a divergent series.&lt;/p&gt;
&lt;p&gt;&lt;span class="mainBody document"&gt;&lt;span style="font-family: 'Verdana','sans-serif'; color: black; font-size: 7.5pt;"&gt;&lt;span style="font-family: 'Verdana','sans-serif'; color: black; font-size: 7.5pt;"&gt;&lt;br&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/p&gt;</itunes:summary>
      <description>The latest comments added to the Post, Could the MRB constant be rational?</description>
      <guid>101613</guid>
      <pubDate>Fri, 11 Feb 2011 13:00:56 Z</pubDate>
      <itunes:author>Robert Israel</itunes:author>
      <author>Robert Israel</author>
    </item>
    <item>
      <title>I might never</title>
      <link>http://www.mapleprimes.com/posts/101425-Could-The-MRB-Constant-Be-Rational?ref=Feed:MaplePrimes:Could the MRB constant be rational?:Comments#comment101629</link>
      <itunes:summary>&lt;p&gt;&amp;nbsp;@Robert&lt;/p&gt;
&lt;p&gt;I might never be able to prove the ir/rationality of the MRB constant, or I might only after 30 years of work and more education, or I just might get lucky and just happen to think of the right thing at the right time!&lt;/p&gt;
&lt;p&gt;I am&amp;nbsp;very scared at the thought of trying something so foolhardy; I might waste the rest of my life away.&lt;/p&gt;
&lt;p&gt;It is a great unknown into which I now tread; the giants are too busy with things that really matter, and I feel that if I don't try it will never be done.&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;For inspiration I look back at what I said on Jan 11, 1999:&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;I have started a search for a new&amp;nbsp;&lt;br&gt;mathematical constant!&lt;br&gt;Does anyone want to help me?&lt;br&gt;Consider, 1^(1/1)-2^(1/2)+3^(1/3)...&lt;br&gt;This may be a contradiction but I call rc&amp;nbsp;&lt;br&gt;a dual constant.&lt;br&gt;The root constant for even numbers&amp;nbsp;&lt;br&gt;(n+1=even number and rc is negative.)&lt;br&gt;The root constant for odd numbers&amp;nbsp;&lt;br&gt;(n+1=odd number and rc is positive.)&lt;br&gt;rc=1^(1/1)-2^(1/2)+3^(1/3)+4^(1/4)...&lt;br&gt;&lt;br&gt;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;And then&amp;nbsp;a few&amp;nbsp;days later I said of the constant and my search:&lt;/p&gt;
&lt;p&gt;It has already gone through hundreds of test.&amp;nbsp;&lt;br&gt;It matches no other known constant or limited&amp;nbsp;&lt;br&gt;combination.&amp;nbsp;I will be taking it apart and examining it "bit by bit".&amp;nbsp;It is my hope to find connections to all kinds of arithmetical manipulations.I realize I am out in "no man's land" but, I work best there! If anyone else is foolhardy enough to come along an offer advice&amp;nbsp;I welcome you.&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;So here I go. I don&amp;rsquo;t know if anyone will care to help; but if I don&amp;rsquo;t do it, it won&amp;rsquo;t get done.&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&lt;a href="http://marvinrayburns.com"&gt;marvinrayburns.com&lt;br&gt;&lt;/a&gt;&lt;/p&gt;</itunes:summary>
      <description>The latest comments added to the Post, Could the MRB constant be rational?</description>
      <guid>101629</guid>
      <pubDate>Sat, 12 Feb 2011 03:32:43 Z</pubDate>
      <itunes:author>Marvin Ray Burns</itunes:author>
      <author>Marvin Ray Burns</author>
    </item>
  </channel>
</rss>