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rbeezer  committed 78e87c3

LaTeX/XML formatting in Solution LDS.T40 (Vladimir Yelkhimov)

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 Typo: Sage RLD, "nest" to "next" (Gavin Tranter)
 Typo: Theorem OSLI, "nozero" to "nonzero"  (Anna Dovzhik)
 Typo: Solution LDS.T55, B should be C at very end (Chris Spalding)
-                                                
+Typo: LaTeX/XML formatting in Solution LDS.T40 (Vladimir Yelkhimov)
                                                     
 v3.10 2013/08/20
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File src/section-LDS.xml

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 </problem>
 <solution contributor="robertbeezer">This is an equality of sets, so <acroref type="definition" acro="SE" /> applies.<br /><br />
 The <q>easy</q> half first.  Show that $X=\spn{\set{\vect{v}_1+\vect{v}_2,\,\vect{v}_1-\vect{v}_2}}\subseteq
-\spn{\set{\vect{v}_1,\,\vect{v}_2}}=Y$.\\
+\spn{\set{\vect{v}_1,\,\vect{v}_2}}=Y$.<br /><br />
 Choose $\vect{x}\in X$.  Then
 $\vect{x}=a_1(\vect{v}_1+\vect{v}_2)+a_2(\vect{v}_1-\vect{v}_2)$ for some scalars $a_1$ and $a_2$.  Then,
 <alignmath>
 <![CDATA[&=(a_1+a_2)\vect{v}_1+(a_1-a_2)\vect{v}_2]]>
 </alignmath>
 which qualifies $\vect{x}$ for membership in $Y$, as it is a linear combination of $\vect{v}_1,\,\vect{v}_2$.<br /><br />
-Now show the opposite inclusion, $Y=\spn{\set{\vect{v}_1,\,\vect{v}_2}}\subseteq\spn{\set{\vect{v}_1+\vect{v}_2,\,\vect{v}_1-\vect{v}_2}}=X$.\\
+Now show the opposite inclusion, $Y=\spn{\set{\vect{v}_1,\,\vect{v}_2}}\subseteq\spn{\set{\vect{v}_1+\vect{v}_2,\,\vect{v}_1-\vect{v}_2}}=X$.<br /><br />
 Choose $\vect{y}\in Y$.  Then there are scalars $b_1,\,b_2$ such that $ \vect{y}=b_1\vect{v}_1+b_2\vect{v}_2 $.  Rearranging, we obtain,
 <alignmath>
 \vect{y}