document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=32844>Question 32844</A>: <I><SPAN STYLE=\"word-break: break-all;\"> (a) Prove that the set S of rational numbers (in lowest term) with odd denominators is<BR>\n" );
document.write( "a subring of Q.<BR>\n" );
document.write( "(b) Let I be the set of elements of S with even numerators. Prove that I is an ideal in<BR>\n" );
document.write( "S.<BR>\n" );
document.write( "(c) Show that S/I consists of exactly two elements.\r<BR>\n" );
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document.write( "Thank You </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #19331 by </B><A HREF=/tutors/aboutme.mpl?userid=khwang><B>khwang(438)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=khwang><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=19331\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Prove that the set S of rational numbers (in lowest term) with odd denominators is<BR>\n" );
document.write( "a subring of Q.<BR>\n" );
document.write( "(b) Let I be the set of elements of S with even numerators. Prove that I is an ideal in S.<BR>\n" );
document.write( "(c) Show that S/I consists of exactly two elements.\r<BR>\n" );
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document.write( " proof: a) S = { p/q| p, q in Z,(p,q) = 1 & q is odd}<BR>\n" );
document.write( "   p1/q1, p2/q2 in S --> p1/q1- p2/q2 = (p1q2 - p2q1)/q1 q2 is in S<BR>\n" );
document.write( "  and p1/q1* p2/q2 = p1p2/q1q2 is in S<BR>\n" );
document.write( " So, S is a subring of Q.<BR>\n" );
document.write( " <BR>\n" );
document.write( " b) I = {p/q in S| p even}<BR>\n" );
document.write( "  As in a) test I is a subring of S (for you).\r<BR>\n" );
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document.write( "  for r in S, p/q in I, so p is even and so<BR>\n" );
document.write( "  r * pq = pp'/qq' if r = p'/q' is in I since pp' is even.<BR>\n" );
document.write( "  Hence, rI < I  and so I is an ideof Q.\r<BR>\n" );
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document.write( " c) S/I ={ s + I | s in S}<BR>\n" );
document.write( "   for any s in S, let s = p/q <BR>\n" );
document.write( " s is in I if p is even.(i.e s+I = I)<BR>\n" );
document.write( " if p is odd, then s - 1/3 = p/q - 1/3 = (3p-q)/3q belonging to I<BR>\n" );
document.write( "  (since 3p-q is even)<BR>\n" );
document.write( " Hence, s + I = 1/3 for any s = p/q (p odd) in I.<BR>\n" );
document.write( " This shows S/I  = {I, 1/3 + I} with two elements.\r<BR>\n" );
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document.write( " Kenny\r<BR>\n" );
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document.write( "   </SPAN>\n" );
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