document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1204483>Question 1204483</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Prove that the set of all n-tuples of rational numbers {(q1, q2, . . . , qn)|qi ∈ Q} ⊂ R^n is NOT a subspace of R^n. </SPAN>\n" );
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document.write( "Review these rules to see what makes a subspace<BR>\n" );
document.write( "Scroll down to \"Definition  2.6.2: Subspace\"<BR>\n" );
document.write( "<a  rel=nofollow HREF= \"https://math.libretexts.org/Bookshelves/Linear_Algebra/Interactive_Linear_Algebra_(Margalit_and_Rabinoff)/02%3A_Systems_of_Linear_Equations-_Geometry/2.06%3A_Subspaces\">https://math.libretexts.org/Bookshelves/Linear_Algebra/Interactive_Linear_Algebra_(Margalit_and_Rabinoff)/02%3A_Systems_of_Linear_Equations-_Geometry/2.06%3A_Subspaces</a>\r<BR>\n" );
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document.write( "The zero vector is easy to prove it belongs in the set, since we can make all of the coordinates 0 and 0 is indeed rational. <BR>\n" );
document.write( "We have satisfied condition 1.\r<BR>\n" );
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document.write( "Furthermore, condition 2 works as well because any two n-tuples of the same size added together gets some other n-tuple in the set.<BR>\n" );
document.write( "Adding any two rational numbers yields a rational number.\r<BR>\n" );
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document.write( "Condition 3 is where things break down. If c and q were rational, then c*q would be rational as well. <BR>\n" );
document.write( "But if c is irrational while q is rational, then c*q would be irrational. <BR>\n" );
document.write( "We have escaped the set of rational numbers and therefore we don't have closure with multiplication. <BR>\n" );
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