document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1017551>Question 1017551</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Is there any other way to prove that <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+sqrt%283%29+ ALIGN=MIDDLE  ALT=\"+sqrt%283%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" > is irrational?\r<BR>\n" );
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document.write( "for e.g. you could prove that sum of 2 even number <BR>\n" );
document.write( "2n+2n = 4n\r<BR>\n" );
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document.write( "Can you do something like this to prove that <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+sqrt%283%29+ ALIGN=MIDDLE  ALT=\"+sqrt%283%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "I already know how to prove sqrt(3) is irrational if i set <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+sqrt%283%29+=+p%2F1+ ALIGN=MIDDLE  ALT=\"+sqrt%283%29+=+p%2F1+\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "any other ways than this? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #633870 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=633870\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I can think of 2 ways<BR>\n" );
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document.write( "1) You can do a proof by contradiction<BR>\n" );
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document.write( "assume that there are integers p and q such that<BR>\n" );
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document.write( "square root(3) = p/q which implies that<BR>\n" );
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document.write( "3 = p^2 / q^2 and p/q in lowest terms(no common factor except 1)<BR>\n" );
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document.write( "3q^2 = p^2<BR>\n" );
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document.write( "so p^2 is divisible by 3 and p is as well<BR>\n" );
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document.write( "this means that there exists an integer k, such that, p = 3k, therefore<BR>\n" );
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document.write( "3q^2 = p^2 implies 3q^2 = (3k)^2 implies 3q^2 = 9k^2<BR>\n" );
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document.write( "divide both sides by 3 and we get<BR>\n" );
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document.write( "q^2 = 3k^2<BR>\n" );
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document.write( "which implies q is divisible by 3<BR>\n" );
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document.write( "but we also have p is divisible by 3 and<BR>\n" );
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document.write( "this means that p and q are divisible by 3, so they were not in lowest terms - the contradiction that p and q only have 1 as a divisor<BR>\n" );
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document.write( "2) consider x^2 - p = 0 where p is prime<BR>\n" );
document.write( "then apply the Rational Root Theorem  </SPAN>\n" );
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