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pretraining/mathematica/linear_algebra/lp_norm_w_steps/1025.txt

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+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -3 \\
+ 8 \\
+ 0 \\
+ -3 \\
+ -7 \\
+ -8 \\
+ 3 \\
+ -3 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, (-3,8,0,-3,-7,-8,3,-3)\, : \\
+ \| \, (-3,8,0,-3,-7,-8,3,-3)\, \|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{8-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{8-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2+v_8^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, (-3,8,0,-3,-7,-8,3,-3)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \| \, (-3,8,0,-3,-7,-8,3,-3)\, \| =\sqrt{(-3)^2+8^2+0^2+(-3)^2+(-7)^2+(-8)^2+3^2+(-3)^2}=\sqrt{213} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/1029.txt

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+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -\frac{9}{2} \\
+ \frac{11}{10} \\
+ \frac{39}{10} \\
+ -\frac{3}{2} \\
+ \frac{39}{10} \\
+ -\frac{5}{2} \\
+ 5 \\
+ \frac{33}{5} \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, \left(-\frac{9}{2},\frac{11}{10},\frac{39}{10},-\frac{3}{2},\frac{39}{10},-\frac{5}{2},5,\frac{33}{5}\right)\, : \\
+ \left\| \, \left(-\frac{9}{2},\frac{11}{10},\frac{39}{10},-\frac{3}{2},\frac{39}{10},-\frac{5}{2},5,\frac{33}{5}\right)\, \right\|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{8-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{8-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2+v_8^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, \left(-\frac{9}{2},\frac{11}{10},\frac{39}{10},-\frac{3}{2},\frac{39}{10},-\frac{5}{2},5,\frac{33}{5}\right)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \left\| \, \left(-\frac{9}{2},\frac{11}{10},\frac{39}{10},-\frac{3}{2},\frac{39}{10},-\frac{5}{2},5,\frac{33}{5}\right)\, \right\| =\sqrt{\left(\frac{-9}{2}\right)^2+\left(\frac{11}{10}\right)^2+\left(\frac{39}{10}\right)^2+\left(\frac{-3}{2}\right)^2+\left(\frac{39}{10}\right)^2+\left(\frac{-5}{2}\right)^2+5^2+\left(\frac{33}{5}\right)^2}=\frac{\sqrt{\frac{6447}{2}}}{5} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/1188.txt

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+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ 8 \\
+ \frac{7}{2} \\
+ 9 \\
+ 8 \\
+ 1 \\
+ -5 \\
+ \frac{3}{2} \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, \left(8,\frac{7}{2},9,8,1,-5,\frac{3}{2}\right)\, : \\
+ \left\| \, \left(8,\frac{7}{2},9,8,1,-5,\frac{3}{2}\right)\, \right\|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{7-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{7-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, \left(8,\frac{7}{2},9,8,1,-5,\frac{3}{2}\right)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \left\| \, \left(8,\frac{7}{2},9,8,1,-5,\frac{3}{2}\right)\, \right\| =\sqrt{8^2+\left(\frac{7}{2}\right)^2+9^2+8^2+1^2+(-5)^2+\left(\frac{3}{2}\right)^2}=\sqrt{\frac{499}{2}} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/1218.txt

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+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ 10 \\
+ -6 \\
+ -4 \\
+ 8 \\
+ 7 \\
+ -5 \\
+ 5 \\
+ 8 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, (10,-6,-4,8,7,-5,5,8)\, : \\
+ \| \, (10,-6,-4,8,7,-5,5,8)\, \|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{8-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{8-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2+v_8^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, (10,-6,-4,8,7,-5,5,8)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \| \, (10,-6,-4,8,7,-5,5,8)\, \| =\sqrt{10^2+(-6)^2+(-4)^2+8^2+7^2+(-5)^2+5^2+8^2}=\sqrt{379} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/1387.txt

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+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -1 \\
+ -4 \\
+ 0 \\
+ -6 \\
+ -7 \\
+ 3 \\
+ 3 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, (-1,-4,0,-6,-7,3,3)\, : \\
+ \| \, (-1,-4,0,-6,-7,3,3)\, \|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{7-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{7-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, (-1,-4,0,-6,-7,3,3)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \| \, (-1,-4,0,-6,-7,3,3)\, \| =\sqrt{(-1)^2+(-4)^2+0^2+(-6)^2+(-7)^2+3^2+3^2}=2 \sqrt{30} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/1428.txt

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+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -9 \\
+ -4 \\
+ -1 \\
+ 1 \\
+ -10 \\
+ 9 \\
+ -7 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, (-9,-4,-1,1,-10,9,-7)\, : \\
+ \| \, (-9,-4,-1,1,-10,9,-7)\, \|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{7-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{7-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, (-9,-4,-1,1,-10,9,-7)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \| \, (-9,-4,-1,1,-10,9,-7)\, \| =\sqrt{(-9)^2+(-4)^2+(-1)^2+1^2+(-10)^2+9^2+(-7)^2}=\sqrt{329} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/149.txt

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+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -7 \\
+ 3 \\
+ 3 \\
+ 1 \\
+ 5 \\
+ 2 \\
+ 2 \\
+ -9 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, (-7,3,3,1,5,2,2,-9)\, : \\
+ \| \, (-7,3,3,1,5,2,2,-9)\, \|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{8-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{8-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2+v_8^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, (-7,3,3,1,5,2,2,-9)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \| \, (-7,3,3,1,5,2,2,-9)\, \| =\sqrt{(-7)^2+3^2+3^2+1^2+5^2+2^2+2^2+(-9)^2}=\sqrt{182} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/1510.txt

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+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -1 \\
+ -3 \\
+ 4 \\
+ -5 \\
+ -3 \\
+ -4 \\
+ 2 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, (-1,-3,4,-5,-3,-4,2)\, : \\
+ \| \, (-1,-3,4,-5,-3,-4,2)\, \|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{7-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{7-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, (-1,-3,4,-5,-3,-4,2)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \| \, (-1,-3,4,-5,-3,-4,2)\, \| =\sqrt{(-1)^2+(-3)^2+4^2+(-5)^2+(-3)^2+(-4)^2+2^2}=4 \sqrt{5} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/1516.txt

@@ -0,0 +1,41 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -\frac{49}{5} \\
+ \frac{4}{5} \\
+ 4 \\
+ -\frac{67}{10} \\
+ \frac{63}{10} \\
+ -\frac{49}{5} \\
+ \frac{14}{5} \\
+ \frac{16}{5} \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, \left(-\frac{49}{5},\frac{4}{5},4,-\frac{67}{10},\frac{63}{10},-\frac{49}{5},\frac{14}{5},\frac{16}{5}\right)\, : \\
+ \left\| \, \left(-\frac{49}{5},\frac{4}{5},4,-\frac{67}{10},\frac{63}{10},-\frac{49}{5},\frac{14}{5},\frac{16}{5}\right)\, \right\|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{8-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{8-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2+v_8^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, \left(-\frac{49}{5},\frac{4}{5},4,-\frac{67}{10},\frac{63}{10},-\frac{49}{5},\frac{14}{5},\frac{16}{5}\right)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \left\| \, \left(-\frac{49}{5},\frac{4}{5},4,-\frac{67}{10},\frac{63}{10},-\frac{49}{5},\frac{14}{5},\frac{16}{5}\right)\, \right\| =\sqrt{\left(\frac{-49}{5}\right)^2+\left(\frac{4}{5}\right)^2+4^2+\left(\frac{-67}{10}\right)^2+\left(\frac{63}{10}\right)^2+\left(\frac{-49}{5}\right)^2+\left(\frac{14}{5}\right)^2+\left(\frac{16}{5}\right)^2}=\frac{\sqrt{\frac{15569}{2}}}{5} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/1545.txt

@@ -0,0 +1,40 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ \frac{5}{4} \\
+ \frac{17}{4} \\
+ \frac{15}{4} \\
+ 6 \\
+ \frac{5}{4} \\
+ -\frac{5}{4} \\
+ 6 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, \left(\frac{5}{4},\frac{17}{4},\frac{15}{4},6,\frac{5}{4},-\frac{5}{4},6\right)\, : \\
+ \left\| \, \left(\frac{5}{4},\frac{17}{4},\frac{15}{4},6,\frac{5}{4},-\frac{5}{4},6\right)\, \right\|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{7-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{7-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, \left(\frac{5}{4},\frac{17}{4},\frac{15}{4},6,\frac{5}{4},-\frac{5}{4},6\right)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \left\| \, \left(\frac{5}{4},\frac{17}{4},\frac{15}{4},6,\frac{5}{4},-\frac{5}{4},6\right)\, \right\| =\sqrt{\left(\frac{5}{4}\right)^2+\left(\frac{17}{4}\right)^2+\left(\frac{15}{4}\right)^2+6^2+\left(\frac{5}{4}\right)^2+\left(\frac{-5}{4}\right)^2+6^2}=\frac{\sqrt{1741}}{4} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/1574.txt

@@ -0,0 +1,40 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ 5 \\
+ 9 \\
+ 2 \\
+ 1 \\
+ 1 \\
+ 2 \\
+ 5 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, (5,9,2,1,1,2,5)\, : \\
+ \| \, (5,9,2,1,1,2,5)\, \|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{7-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{7-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, (5,9,2,1,1,2,5)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \| \, (5,9,2,1,1,2,5)\, \| =\sqrt{5^2+9^2+2^2+1^2+1^2+2^2+5^2}=\sqrt{141} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/168.txt

@@ -0,0 +1,40 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ 7 \\
+ -5 \\
+ -4 \\
+ 10 \\
+ -1 \\
+ -6 \\
+ -8 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, (7,-5,-4,10,-1,-6,-8)\, : \\
+ \| \, (7,-5,-4,10,-1,-6,-8)\, \|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{7-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{7-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, (7,-5,-4,10,-1,-6,-8)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \| \, (7,-5,-4,10,-1,-6,-8)\, \| =\sqrt{7^2+(-5)^2+(-4)^2+10^2+(-1)^2+(-6)^2+(-8)^2}=\sqrt{291} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/1737.txt

@@ -0,0 +1,41 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -8 \\
+ -\frac{44}{7} \\
+ -\frac{38}{7} \\
+ -\frac{50}{7} \\
+ -\frac{17}{7} \\
+ \frac{38}{7} \\
+ -7 \\
+ -\frac{8}{7} \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, \left(-8,-\frac{44}{7},-\frac{38}{7},-\frac{50}{7},-\frac{17}{7},\frac{38}{7},-7,-\frac{8}{7}\right)\, : \\
+ \left\| \, \left(-8,-\frac{44}{7},-\frac{38}{7},-\frac{50}{7},-\frac{17}{7},\frac{38}{7},-7,-\frac{8}{7}\right)\, \right\|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{8-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{8-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2+v_8^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, \left(-8,-\frac{44}{7},-\frac{38}{7},-\frac{50}{7},-\frac{17}{7},\frac{38}{7},-7,-\frac{8}{7}\right)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \left\| \, \left(-8,-\frac{44}{7},-\frac{38}{7},-\frac{50}{7},-\frac{17}{7},\frac{38}{7},-7,-\frac{8}{7}\right)\, \right\| =\sqrt{(-8)^2+\left(\frac{-44}{7}\right)^2+\left(\frac{-38}{7}\right)^2+\left(\frac{-50}{7}\right)^2+\left(\frac{-17}{7}\right)^2+\left(\frac{38}{7}\right)^2+(-7)^2+\left(\frac{-8}{7}\right)^2}=\frac{\sqrt{13214}}{7} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/1741.txt

@@ -0,0 +1,40 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -\frac{83}{16} \\
+ -\frac{25}{8} \\
+ \frac{59}{8} \\
+ -\frac{23}{8} \\
+ -\frac{43}{8} \\
+ -\frac{55}{16} \\
+ \frac{13}{16} \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, \left(-\frac{83}{16},-\frac{25}{8},\frac{59}{8},-\frac{23}{8},-\frac{43}{8},-\frac{55}{16},\frac{13}{16}\right)\, : \\
+ \left\| \, \left(-\frac{83}{16},-\frac{25}{8},\frac{59}{8},-\frac{23}{8},-\frac{43}{8},-\frac{55}{16},\frac{13}{16}\right)\, \right\|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{7-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{7-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, \left(-\frac{83}{16},-\frac{25}{8},\frac{59}{8},-\frac{23}{8},-\frac{43}{8},-\frac{55}{16},\frac{13}{16}\right)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \left\| \, \left(-\frac{83}{16},-\frac{25}{8},\frac{59}{8},-\frac{23}{8},-\frac{43}{8},-\frac{55}{16},\frac{13}{16}\right)\, \right\| =\sqrt{\left(\frac{-83}{16}\right)^2+\left(\frac{-25}{8}\right)^2+\left(\frac{59}{8}\right)^2+\left(\frac{-23}{8}\right)^2+\left(\frac{-43}{8}\right)^2+\left(\frac{-55}{16}\right)^2+\left(\frac{13}{16}\right)^2}=\frac{\sqrt{36019}}{16} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/1791.txt

@@ -0,0 +1,41 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ 7 \\
+ -\frac{121}{16} \\
+ -\frac{91}{16} \\
+ \frac{51}{8} \\
+ 3 \\
+ -\frac{41}{8} \\
+ -\frac{131}{16} \\
+ -\frac{15}{2} \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, \left(7,-\frac{121}{16},-\frac{91}{16},\frac{51}{8},3,-\frac{41}{8},-\frac{131}{16},-\frac{15}{2}\right)\, : \\
+ \left\| \, \left(7,-\frac{121}{16},-\frac{91}{16},\frac{51}{8},3,-\frac{41}{8},-\frac{131}{16},-\frac{15}{2}\right)\, \right\|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{8-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{8-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2+v_8^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, \left(7,-\frac{121}{16},-\frac{91}{16},\frac{51}{8},3,-\frac{41}{8},-\frac{131}{16},-\frac{15}{2}\right)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \left\| \, \left(7,-\frac{121}{16},-\frac{91}{16},\frac{51}{8},3,-\frac{41}{8},-\frac{131}{16},-\frac{15}{2}\right)\, \right\| =\sqrt{7^2+\left(\frac{-121}{16}\right)^2+\left(\frac{-91}{16}\right)^2+\left(\frac{51}{8}\right)^2+3^2+\left(\frac{-41}{8}\right)^2+\left(\frac{-131}{16}\right)^2+\left(\frac{-15}{2}\right)^2}=\frac{\sqrt{86459}}{16} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/18.txt

@@ -0,0 +1,41 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ 1 \\
+ 9 \\
+ -4 \\
+ 1 \\
+ 7 \\
+ -4 \\
+ -2 \\
+ 5 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, (1,9,-4,1,7,-4,-2,5)\, : \\
+ \| \, (1,9,-4,1,7,-4,-2,5)\, \|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{8-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{8-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2+v_8^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, (1,9,-4,1,7,-4,-2,5)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \| \, (1,9,-4,1,7,-4,-2,5)\, \| =\sqrt{1^2+9^2+(-4)^2+1^2+7^2+(-4)^2+(-2)^2+5^2}=\sqrt{193} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/1944.txt

@@ -0,0 +1,40 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -1 \\
+ 5 \\
+ -7 \\
+ 4 \\
+ 5 \\
+ -6 \\
+ -2 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, (-1,5,-7,4,5,-6,-2)\, : \\
+ \| \, (-1,5,-7,4,5,-6,-2)\, \|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{7-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{7-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, (-1,5,-7,4,5,-6,-2)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \| \, (-1,5,-7,4,5,-6,-2)\, \| =\sqrt{(-1)^2+5^2+(-7)^2+4^2+5^2+(-6)^2+(-2)^2}=2 \sqrt{39} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/1949.txt

@@ -0,0 +1,41 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -\frac{7}{4} \\
+ \frac{23}{4} \\
+ \frac{13}{4} \\
+ \frac{35}{4} \\
+ 3 \\
+ \frac{25}{4} \\
+ 5 \\
+ \frac{37}{4} \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, \left(-\frac{7}{4},\frac{23}{4},\frac{13}{4},\frac{35}{4},3,\frac{25}{4},5,\frac{37}{4}\right)\, : \\
+ \left\| \, \left(-\frac{7}{4},\frac{23}{4},\frac{13}{4},\frac{35}{4},3,\frac{25}{4},5,\frac{37}{4}\right)\, \right\|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{8-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{8-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2+v_8^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, \left(-\frac{7}{4},\frac{23}{4},\frac{13}{4},\frac{35}{4},3,\frac{25}{4},5,\frac{37}{4}\right)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \left\| \, \left(-\frac{7}{4},\frac{23}{4},\frac{13}{4},\frac{35}{4},3,\frac{25}{4},5,\frac{37}{4}\right)\, \right\| =\sqrt{\left(\frac{-7}{4}\right)^2+\left(\frac{23}{4}\right)^2+\left(\frac{13}{4}\right)^2+\left(\frac{35}{4}\right)^2+3^2+\left(\frac{25}{4}\right)^2+5^2+\left(\frac{37}{4}\right)^2}=\frac{\sqrt{\frac{2255}{2}}}{2} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/1959.txt

@@ -0,0 +1,40 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -5 \\
+ -\frac{3}{2} \\
+ -6 \\
+ -\frac{3}{2} \\
+ -4 \\
+ 4 \\
+ \frac{13}{2} \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, \left(-5,-\frac{3}{2},-6,-\frac{3}{2},-4,4,\frac{13}{2}\right)\, : \\
+ \left\| \, \left(-5,-\frac{3}{2},-6,-\frac{3}{2},-4,4,\frac{13}{2}\right)\, \right\|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{7-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{7-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, \left(-5,-\frac{3}{2},-6,-\frac{3}{2},-4,4,\frac{13}{2}\right)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \left\| \, \left(-5,-\frac{3}{2},-6,-\frac{3}{2},-4,4,\frac{13}{2}\right)\, \right\| =\sqrt{(-5)^2+\left(\frac{-3}{2}\right)^2+(-6)^2+\left(\frac{-3}{2}\right)^2+(-4)^2+4^2+\left(\frac{13}{2}\right)^2}=\frac{\sqrt{559}}{2} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/202.txt

@@ -0,0 +1,40 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -\frac{96}{25} \\
+ \frac{531}{100} \\
+ \frac{19}{5} \\
+ -\frac{437}{50} \\
+ -\frac{59}{100} \\
+ \frac{141}{50} \\
+ -\frac{473}{50} \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, \left(-\frac{96}{25},\frac{531}{100},\frac{19}{5},-\frac{437}{50},-\frac{59}{100},\frac{141}{50},-\frac{473}{50}\right)\, : \\
+ \left\| \, \left(-\frac{96}{25},\frac{531}{100},\frac{19}{5},-\frac{437}{50},-\frac{59}{100},\frac{141}{50},-\frac{473}{50}\right)\, \right\|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{7-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{7-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, \left(-\frac{96}{25},\frac{531}{100},\frac{19}{5},-\frac{437}{50},-\frac{59}{100},\frac{141}{50},-\frac{473}{50}\right)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \left\| \, \left(-\frac{96}{25},\frac{531}{100},\frac{19}{5},-\frac{437}{50},-\frac{59}{100},\frac{141}{50},-\frac{473}{50}\right)\, \right\| =\sqrt{\left(\frac{-96}{25}\right)^2+\left(\frac{531}{100}\right)^2+\left(\frac{19}{5}\right)^2+\left(\frac{-437}{50}\right)^2+\left(\frac{-59}{100}\right)^2+\left(\frac{141}{50}\right)^2+\left(\frac{-473}{50}\right)^2}=\frac{\sqrt{\frac{1157807}{2}}}{50} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/2094.txt

@@ -0,0 +1,41 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -8 \\
+ 8 \\
+ 0 \\
+ 4 \\
+ 8 \\
+ 2 \\
+ -1 \\
+ -6 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, (-8,8,0,4,8,2,-1,-6)\, : \\
+ \| \, (-8,8,0,4,8,2,-1,-6)\, \|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{8-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{8-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2+v_8^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, (-8,8,0,4,8,2,-1,-6)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \| \, (-8,8,0,4,8,2,-1,-6)\, \| =\sqrt{(-8)^2+8^2+0^2+4^2+8^2+2^2+(-1)^2+(-6)^2}=\sqrt{249} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/212.txt

@@ -0,0 +1,41 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -\frac{643}{100} \\
+ \frac{99}{25} \\
+ -\frac{273}{100} \\
+ -\frac{989}{100} \\
+ \frac{127}{25} \\
+ -\frac{68}{25} \\
+ -\frac{561}{100} \\
+ \frac{229}{25} \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, \left(-\frac{643}{100},\frac{99}{25},-\frac{273}{100},-\frac{989}{100},\frac{127}{25},-\frac{68}{25},-\frac{561}{100},\frac{229}{25}\right)\, : \\
+ \left\| \, \left(-\frac{643}{100},\frac{99}{25},-\frac{273}{100},-\frac{989}{100},\frac{127}{25},-\frac{68}{25},-\frac{561}{100},\frac{229}{25}\right)\, \right\|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{8-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{8-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2+v_8^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, \left(-\frac{643}{100},\frac{99}{25},-\frac{273}{100},-\frac{989}{100},\frac{127}{25},-\frac{68}{25},-\frac{561}{100},\frac{229}{25}\right)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \left\| \, \left(-\frac{643}{100},\frac{99}{25},-\frac{273}{100},-\frac{989}{100},\frac{127}{25},-\frac{68}{25},-\frac{561}{100},\frac{229}{25}\right)\, \right\| =\sqrt{\left(\frac{-643}{100}\right)^2+\left(\frac{99}{25}\right)^2+\left(\frac{-273}{100}\right)^2+\left(\frac{-989}{100}\right)^2+\left(\frac{127}{25}\right)^2+\left(\frac{-68}{25}\right)^2+\left(\frac{-561}{100}\right)^2+\left(\frac{229}{25}\right)^2}=\frac{\sqrt{\frac{155437}{5}}}{10} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/2198.txt

@@ -0,0 +1,40 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -4 \\
+ 8 \\
+ 2 \\
+ -5 \\
+ -6 \\
+ 1 \\
+ -4 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, (-4,8,2,-5,-6,1,-4)\, : \\
+ \| \, (-4,8,2,-5,-6,1,-4)\, \|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{7-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{7-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, (-4,8,2,-5,-6,1,-4)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \| \, (-4,8,2,-5,-6,1,-4)\, \| =\sqrt{(-4)^2+8^2+2^2+(-5)^2+(-6)^2+1^2+(-4)^2}=9 \sqrt{2} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/2236.txt

@@ -0,0 +1,40 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ \frac{8}{3} \\
+ -\frac{20}{3} \\
+ \frac{41}{6} \\
+ \frac{3}{2} \\
+ -\frac{1}{3} \\
+ -\frac{26}{3} \\
+ -\frac{13}{3} \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, \left(\frac{8}{3},-\frac{20}{3},\frac{41}{6},\frac{3}{2},-\frac{1}{3},-\frac{26}{3},-\frac{13}{3}\right)\, : \\
+ \left\| \, \left(\frac{8}{3},-\frac{20}{3},\frac{41}{6},\frac{3}{2},-\frac{1}{3},-\frac{26}{3},-\frac{13}{3}\right)\, \right\|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{7-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{7-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, \left(\frac{8}{3},-\frac{20}{3},\frac{41}{6},\frac{3}{2},-\frac{1}{3},-\frac{26}{3},-\frac{13}{3}\right)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \left\| \, \left(\frac{8}{3},-\frac{20}{3},\frac{41}{6},\frac{3}{2},-\frac{1}{3},-\frac{26}{3},-\frac{13}{3}\right)\, \right\| =\sqrt{\left(\frac{8}{3}\right)^2+\left(\frac{-20}{3}\right)^2+\left(\frac{41}{6}\right)^2+\left(\frac{3}{2}\right)^2+\left(\frac{-1}{3}\right)^2+\left(\frac{-26}{3}\right)^2+\left(\frac{-13}{3}\right)^2}=\sqrt{\frac{389}{2}} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/2246.txt

@@ -0,0 +1,41 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ 10 \\
+ 2 \\
+ 9 \\
+ -8 \\
+ -2 \\
+ -6 \\
+ 0 \\
+ -5 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, (10,2,9,-8,-2,-6,0,-5)\, : \\
+ \| \, (10,2,9,-8,-2,-6,0,-5)\, \|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{8-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{8-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2+v_8^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, (10,2,9,-8,-2,-6,0,-5)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \| \, (10,2,9,-8,-2,-6,0,-5)\, \| =\sqrt{10^2+2^2+9^2+(-8)^2+(-2)^2+(-6)^2+0^2+(-5)^2}=\sqrt{314} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/2336.txt

@@ -0,0 +1,41 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ 2 \\
+ -1 \\
+ 5 \\
+ -9 \\
+ \frac{15}{2} \\
+ \frac{15}{2} \\
+ \frac{17}{2} \\
+ -7 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, \left(2,-1,5,-9,\frac{15}{2},\frac{15}{2},\frac{17}{2},-7\right)\, : \\
+ \left\| \, \left(2,-1,5,-9,\frac{15}{2},\frac{15}{2},\frac{17}{2},-7\right)\, \right\|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{8-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{8-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2+v_8^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, \left(2,-1,5,-9,\frac{15}{2},\frac{15}{2},\frac{17}{2},-7\right)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \left\| \, \left(2,-1,5,-9,\frac{15}{2},\frac{15}{2},\frac{17}{2},-7\right)\, \right\| =\sqrt{2^2+(-1)^2+5^2+(-9)^2+\left(\frac{15}{2}\right)^2+\left(\frac{15}{2}\right)^2+\left(\frac{17}{2}\right)^2+(-7)^2}=\frac{\sqrt{1379}}{2} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/2414.txt

@@ -0,0 +1,40 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ 5 \\
+ 10 \\
+ 10 \\
+ 6 \\
+ 1 \\
+ 5 \\
+ -8 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, (5,10,10,6,1,5,-8)\, : \\
+ \| \, (5,10,10,6,1,5,-8)\, \|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{7-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{7-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, (5,10,10,6,1,5,-8)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \| \, (5,10,10,6,1,5,-8)\, \| =\sqrt{5^2+10^2+10^2+6^2+1^2+5^2+(-8)^2}=3 \sqrt{39} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/2430.txt

@@ -0,0 +1,40 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -8 \\
+ -7 \\
+ 4 \\
+ -5 \\
+ -3 \\
+ -2 \\
+ 2 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, (-8,-7,4,-5,-3,-2,2)\, : \\
+ \| \, (-8,-7,4,-5,-3,-2,2)\, \|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{7-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{7-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, (-8,-7,4,-5,-3,-2,2)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \| \, (-8,-7,4,-5,-3,-2,2)\, \| =\sqrt{(-8)^2+(-7)^2+4^2+(-5)^2+(-3)^2+(-2)^2+2^2}=3 \sqrt{19} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/2436.txt

@@ -0,0 +1,40 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ \frac{19}{2} \\
+ \frac{39}{4} \\
+ 1 \\
+ 6 \\
+ -5 \\
+ -6 \\
+ \frac{11}{2} \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, \left(\frac{19}{2},\frac{39}{4},1,6,-5,-6,\frac{11}{2}\right)\, : \\
+ \left\| \, \left(\frac{19}{2},\frac{39}{4},1,6,-5,-6,\frac{11}{2}\right)\, \right\|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{7-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{7-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, \left(\frac{19}{2},\frac{39}{4},1,6,-5,-6,\frac{11}{2}\right)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \left\| \, \left(\frac{19}{2},\frac{39}{4},1,6,-5,-6,\frac{11}{2}\right)\, \right\| =\sqrt{\left(\frac{19}{2}\right)^2+\left(\frac{39}{4}\right)^2+1^2+6^2+(-5)^2+(-6)^2+\left(\frac{11}{2}\right)^2}=\frac{\sqrt{5017}}{4} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/2723.txt

@@ -0,0 +1,41 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -3 \\
+ -9 \\
+ 1 \\
+ 2 \\
+ -3 \\
+ 9 \\
+ 3 \\
+ -6 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, (-3,-9,1,2,-3,9,3,-6)\, : \\
+ \| \, (-3,-9,1,2,-3,9,3,-6)\, \|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{8-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{8-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2+v_8^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, (-3,-9,1,2,-3,9,3,-6)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \| \, (-3,-9,1,2,-3,9,3,-6)\, \| =\sqrt{(-3)^2+(-9)^2+1^2+2^2+(-3)^2+9^2+3^2+(-6)^2}=\sqrt{230} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/2997.txt

@@ -0,0 +1,41 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ \frac{17}{2} \\
+ -\frac{11}{2} \\
+ \frac{3}{2} \\
+ -3 \\
+ -5 \\
+ -10 \\
+ 6 \\
+ -\frac{13}{2} \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, \left(\frac{17}{2},-\frac{11}{2},\frac{3}{2},-3,-5,-10,6,-\frac{13}{2}\right)\, : \\
+ \left\| \, \left(\frac{17}{2},-\frac{11}{2},\frac{3}{2},-3,-5,-10,6,-\frac{13}{2}\right)\, \right\|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{8-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{8-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2+v_8^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, \left(\frac{17}{2},-\frac{11}{2},\frac{3}{2},-3,-5,-10,6,-\frac{13}{2}\right)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \left\| \, \left(\frac{17}{2},-\frac{11}{2},\frac{3}{2},-3,-5,-10,6,-\frac{13}{2}\right)\, \right\| =\sqrt{\left(\frac{17}{2}\right)^2+\left(\frac{-11}{2}\right)^2+\left(\frac{3}{2}\right)^2+(-3)^2+(-5)^2+(-10)^2+6^2+\left(\frac{-13}{2}\right)^2}=\sqrt{317} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/3074.txt

@@ -0,0 +1,41 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -9 \\
+ -3 \\
+ 7 \\
+ 6 \\
+ -2 \\
+ -2 \\
+ 2 \\
+ -8 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, (-9,-3,7,6,-2,-2,2,-8)\, : \\
+ \| \, (-9,-3,7,6,-2,-2,2,-8)\, \|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{8-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{8-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2+v_8^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, (-9,-3,7,6,-2,-2,2,-8)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \| \, (-9,-3,7,6,-2,-2,2,-8)\, \| =\sqrt{(-9)^2+(-3)^2+7^2+6^2+(-2)^2+(-2)^2+2^2+(-8)^2}=\sqrt{251} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/3083.txt

@@ -0,0 +1,41 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -\frac{101}{16} \\
+ -1 \\
+ -\frac{1}{16} \\
+ -\frac{87}{16} \\
+ -\frac{21}{8} \\
+ \frac{147}{16} \\
+ \frac{23}{8} \\
+ -1 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, \left(-\frac{101}{16},-1,-\frac{1}{16},-\frac{87}{16},-\frac{21}{8},\frac{147}{16},\frac{23}{8},-1\right)\, : \\
+ \left\| \, \left(-\frac{101}{16},-1,-\frac{1}{16},-\frac{87}{16},-\frac{21}{8},\frac{147}{16},\frac{23}{8},-1\right)\, \right\|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{8-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{8-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2+v_8^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, \left(-\frac{101}{16},-1,-\frac{1}{16},-\frac{87}{16},-\frac{21}{8},\frac{147}{16},\frac{23}{8},-1\right)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \left\| \, \left(-\frac{101}{16},-1,-\frac{1}{16},-\frac{87}{16},-\frac{21}{8},\frac{147}{16},\frac{23}{8},-1\right)\, \right\| =\sqrt{\left(\frac{-101}{16}\right)^2+(-1)^2+\left(\frac{-1}{16}\right)^2+\left(\frac{-87}{16}\right)^2+\left(\frac{-21}{8}\right)^2+\left(\frac{147}{16}\right)^2+\left(\frac{23}{8}\right)^2+(-1)^2}=\frac{\sqrt{10943}}{8} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/3091.txt

@@ -0,0 +1,41 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -7 \\
+ -7 \\
+ 10 \\
+ 6 \\
+ -5 \\
+ -3 \\
+ 9 \\
+ 5 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, (-7,-7,10,6,-5,-3,9,5)\, : \\
+ \| \, (-7,-7,10,6,-5,-3,9,5)\, \|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{8-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{8-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2+v_8^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, (-7,-7,10,6,-5,-3,9,5)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \| \, (-7,-7,10,6,-5,-3,9,5)\, \| =\sqrt{(-7)^2+(-7)^2+10^2+6^2+(-5)^2+(-3)^2+9^2+5^2}=\sqrt{374} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/3290.txt

@@ -0,0 +1,41 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ \frac{2}{5} \\
+ -6 \\
+ \frac{19}{5} \\
+ -\frac{21}{5} \\
+ \frac{17}{5} \\
+ \frac{4}{5} \\
+ -\frac{22}{5} \\
+ -\frac{26}{5} \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, \left(\frac{2}{5},-6,\frac{19}{5},-\frac{21}{5},\frac{17}{5},\frac{4}{5},-\frac{22}{5},-\frac{26}{5}\right)\, : \\
+ \left\| \, \left(\frac{2}{5},-6,\frac{19}{5},-\frac{21}{5},\frac{17}{5},\frac{4}{5},-\frac{22}{5},-\frac{26}{5}\right)\, \right\|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{8-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{8-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2+v_8^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, \left(\frac{2}{5},-6,\frac{19}{5},-\frac{21}{5},\frac{17}{5},\frac{4}{5},-\frac{22}{5},-\frac{26}{5}\right)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \left\| \, \left(\frac{2}{5},-6,\frac{19}{5},-\frac{21}{5},\frac{17}{5},\frac{4}{5},-\frac{22}{5},-\frac{26}{5}\right)\, \right\| =\sqrt{\left(\frac{2}{5}\right)^2+(-6)^2+\left(\frac{19}{5}\right)^2+\left(\frac{-21}{5}\right)^2+\left(\frac{17}{5}\right)^2+\left(\frac{4}{5}\right)^2+\left(\frac{-22}{5}\right)^2+\left(\frac{-26}{5}\right)^2}=\frac{\sqrt{3171}}{5} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/3341.txt

@@ -0,0 +1,41 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -7 \\
+ -7 \\
+ 8 \\
+ 9 \\
+ -8 \\
+ 5 \\
+ 10 \\
+ -1 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, (-7,-7,8,9,-8,5,10,-1)\, : \\
+ \| \, (-7,-7,8,9,-8,5,10,-1)\, \|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{8-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{8-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2+v_8^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, (-7,-7,8,9,-8,5,10,-1)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \| \, (-7,-7,8,9,-8,5,10,-1)\, \| =\sqrt{(-7)^2+(-7)^2+8^2+9^2+(-8)^2+5^2+10^2+(-1)^2}=\sqrt{433} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/3409.txt

@@ -0,0 +1,41 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -\frac{45}{8} \\
+ \frac{35}{16} \\
+ \frac{95}{16} \\
+ \frac{117}{16} \\
+ \frac{73}{8} \\
+ \frac{43}{16} \\
+ \frac{159}{16} \\
+ -\frac{37}{16} \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, \left(-\frac{45}{8},\frac{35}{16},\frac{95}{16},\frac{117}{16},\frac{73}{8},\frac{43}{16},\frac{159}{16},-\frac{37}{16}\right)\, : \\
+ \left\| \, \left(-\frac{45}{8},\frac{35}{16},\frac{95}{16},\frac{117}{16},\frac{73}{8},\frac{43}{16},\frac{159}{16},-\frac{37}{16}\right)\, \right\|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{8-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{8-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2+v_8^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, \left(-\frac{45}{8},\frac{35}{16},\frac{95}{16},\frac{117}{16},\frac{73}{8},\frac{43}{16},\frac{159}{16},-\frac{37}{16}\right)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \left\| \, \left(-\frac{45}{8},\frac{35}{16},\frac{95}{16},\frac{117}{16},\frac{73}{8},\frac{43}{16},\frac{159}{16},-\frac{37}{16}\right)\, \right\| =\sqrt{\left(\frac{-45}{8}\right)^2+\left(\frac{35}{16}\right)^2+\left(\frac{95}{16}\right)^2+\left(\frac{117}{16}\right)^2+\left(\frac{73}{8}\right)^2+\left(\frac{43}{16}\right)^2+\left(\frac{159}{16}\right)^2+\left(\frac{-37}{16}\right)^2}=\frac{\sqrt{\frac{40927}{2}}}{8} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/341.txt

@@ -0,0 +1,40 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -4 \\
+ 4 \\
+ 7 \\
+ -5 \\
+ -7 \\
+ -10 \\
+ -8 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, (-4,4,7,-5,-7,-10,-8)\, : \\
+ \| \, (-4,4,7,-5,-7,-10,-8)\, \|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{7-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{7-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, (-4,4,7,-5,-7,-10,-8)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \| \, (-4,4,7,-5,-7,-10,-8)\, \| =\sqrt{(-4)^2+4^2+7^2+(-5)^2+(-7)^2+(-10)^2+(-8)^2}=\sqrt{319} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/344.txt

@@ -0,0 +1,40 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -5 \\
+ -8 \\
+ 4 \\
+ -8 \\
+ 1 \\
+ -2 \\
+ 9 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, (-5,-8,4,-8,1,-2,9)\, : \\
+ \| \, (-5,-8,4,-8,1,-2,9)\, \|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{7-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{7-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, (-5,-8,4,-8,1,-2,9)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \| \, (-5,-8,4,-8,1,-2,9)\, \| =\sqrt{(-5)^2+(-8)^2+4^2+(-8)^2+1^2+(-2)^2+9^2}=\sqrt{255} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/345.txt

@@ -0,0 +1,40 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ 4 \\
+ 9 \\
+ -2 \\
+ 6 \\
+ 7 \\
+ 1 \\
+ -2 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, (4,9,-2,6,7,1,-2)\, : \\
+ \| \, (4,9,-2,6,7,1,-2)\, \|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{7-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{7-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, (4,9,-2,6,7,1,-2)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \| \, (4,9,-2,6,7,1,-2)\, \| =\sqrt{4^2+9^2+(-2)^2+6^2+7^2+1^2+(-2)^2}=\sqrt{191} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/3502.txt

@@ -0,0 +1,41 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ \frac{13}{2} \\
+ \frac{5}{2} \\
+ \frac{13}{2} \\
+ -2 \\
+ 10 \\
+ \frac{3}{2} \\
+ -\frac{9}{2} \\
+ \frac{11}{2} \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, \left(\frac{13}{2},\frac{5}{2},\frac{13}{2},-2,10,\frac{3}{2},-\frac{9}{2},\frac{11}{2}\right)\, : \\
+ \left\| \, \left(\frac{13}{2},\frac{5}{2},\frac{13}{2},-2,10,\frac{3}{2},-\frac{9}{2},\frac{11}{2}\right)\, \right\|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{8-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{8-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2+v_8^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, \left(\frac{13}{2},\frac{5}{2},\frac{13}{2},-2,10,\frac{3}{2},-\frac{9}{2},\frac{11}{2}\right)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \left\| \, \left(\frac{13}{2},\frac{5}{2},\frac{13}{2},-2,10,\frac{3}{2},-\frac{9}{2},\frac{11}{2}\right)\, \right\| =\sqrt{\left(\frac{13}{2}\right)^2+\left(\frac{5}{2}\right)^2+\left(\frac{13}{2}\right)^2+(-2)^2+10^2+\left(\frac{3}{2}\right)^2+\left(\frac{-9}{2}\right)^2+\left(\frac{11}{2}\right)^2}=3 \sqrt{\frac{55}{2}} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/3664.txt

@@ -0,0 +1,40 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ \frac{15}{2} \\
+ \frac{17}{4} \\
+ \frac{33}{4} \\
+ -6 \\
+ -4 \\
+ -7 \\
+ \frac{17}{4} \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, \left(\frac{15}{2},\frac{17}{4},\frac{33}{4},-6,-4,-7,\frac{17}{4}\right)\, : \\
+ \left\| \, \left(\frac{15}{2},\frac{17}{4},\frac{33}{4},-6,-4,-7,\frac{17}{4}\right)\, \right\|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{7-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{7-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, \left(\frac{15}{2},\frac{17}{4},\frac{33}{4},-6,-4,-7,\frac{17}{4}\right)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \left\| \, \left(\frac{15}{2},\frac{17}{4},\frac{33}{4},-6,-4,-7,\frac{17}{4}\right)\, \right\| =\sqrt{\left(\frac{15}{2}\right)^2+\left(\frac{17}{4}\right)^2+\left(\frac{33}{4}\right)^2+(-6)^2+(-4)^2+(-7)^2+\left(\frac{17}{4}\right)^2}=\frac{\sqrt{4183}}{4} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/3707.txt

@@ -0,0 +1,41 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -\frac{31}{16} \\
+ -\frac{67}{16} \\
+ -\frac{21}{4} \\
+ -\frac{95}{16} \\
+ -\frac{17}{4} \\
+ -6 \\
+ \frac{27}{16} \\
+ \frac{93}{16} \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, \left(-\frac{31}{16},-\frac{67}{16},-\frac{21}{4},-\frac{95}{16},-\frac{17}{4},-6,\frac{27}{16},\frac{93}{16}\right)\, : \\
+ \left\| \, \left(-\frac{31}{16},-\frac{67}{16},-\frac{21}{4},-\frac{95}{16},-\frac{17}{4},-6,\frac{27}{16},\frac{93}{16}\right)\, \right\|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{8-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{8-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2+v_8^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, \left(-\frac{31}{16},-\frac{67}{16},-\frac{21}{4},-\frac{95}{16},-\frac{17}{4},-6,\frac{27}{16},\frac{93}{16}\right)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \left\| \, \left(-\frac{31}{16},-\frac{67}{16},-\frac{21}{4},-\frac{95}{16},-\frac{17}{4},-6,\frac{27}{16},\frac{93}{16}\right)\, \right\| =\sqrt{\left(\frac{-31}{16}\right)^2+\left(\frac{-67}{16}\right)^2+\left(\frac{-21}{4}\right)^2+\left(\frac{-95}{16}\right)^2+\left(\frac{-17}{4}\right)^2+(-6)^2+\left(\frac{27}{16}\right)^2+\left(\frac{93}{16}\right)^2}=\frac{\sqrt{44749}}{16} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/3838.txt

@@ -0,0 +1,40 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -5 \\
+ 3 \\
+ -\frac{13}{3} \\
+ -\frac{2}{3} \\
+ -\frac{20}{3} \\
+ \frac{29}{3} \\
+ \frac{4}{3} \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, \left(-5,3,-\frac{13}{3},-\frac{2}{3},-\frac{20}{3},\frac{29}{3},\frac{4}{3}\right)\, : \\
+ \left\| \, \left(-5,3,-\frac{13}{3},-\frac{2}{3},-\frac{20}{3},\frac{29}{3},\frac{4}{3}\right)\, \right\|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{7-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{7-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, \left(-5,3,-\frac{13}{3},-\frac{2}{3},-\frac{20}{3},\frac{29}{3},\frac{4}{3}\right)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \left\| \, \left(-5,3,-\frac{13}{3},-\frac{2}{3},-\frac{20}{3},\frac{29}{3},\frac{4}{3}\right)\, \right\| =\sqrt{(-5)^2+3^2+\left(\frac{-13}{3}\right)^2+\left(\frac{-2}{3}\right)^2+\left(\frac{-20}{3}\right)^2+\left(\frac{29}{3}\right)^2+\left(\frac{4}{3}\right)^2}=\frac{2 \sqrt{434}}{3} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/3880.txt

@@ -0,0 +1,41 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -3 \\
+ -9 \\
+ 1 \\
+ 7 \\
+ 1 \\
+ 2 \\
+ -2 \\
+ 0 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, (-3,-9,1,7,1,2,-2,0)\, : \\
+ \| \, (-3,-9,1,7,1,2,-2,0)\, \|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{8-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{8-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2+v_8^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, (-3,-9,1,7,1,2,-2,0)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \| \, (-3,-9,1,7,1,2,-2,0)\, \| =\sqrt{(-3)^2+(-9)^2+1^2+7^2+1^2+2^2+(-2)^2+0^2}=\sqrt{149} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/3887.txt

@@ -0,0 +1,41 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ \frac{63}{8} \\
+ \frac{59}{8} \\
+ \frac{13}{8} \\
+ -\frac{11}{8} \\
+ -\frac{5}{2} \\
+ 7 \\
+ \frac{21}{8} \\
+ -\frac{35}{4} \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, \left(\frac{63}{8},\frac{59}{8},\frac{13}{8},-\frac{11}{8},-\frac{5}{2},7,\frac{21}{8},-\frac{35}{4}\right)\, : \\
+ \left\| \, \left(\frac{63}{8},\frac{59}{8},\frac{13}{8},-\frac{11}{8},-\frac{5}{2},7,\frac{21}{8},-\frac{35}{4}\right)\, \right\|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{8-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{8-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2+v_8^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, \left(\frac{63}{8},\frac{59}{8},\frac{13}{8},-\frac{11}{8},-\frac{5}{2},7,\frac{21}{8},-\frac{35}{4}\right)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \left\| \, \left(\frac{63}{8},\frac{59}{8},\frac{13}{8},-\frac{11}{8},-\frac{5}{2},7,\frac{21}{8},-\frac{35}{4}\right)\, \right\| =\sqrt{\left(\frac{63}{8}\right)^2+\left(\frac{59}{8}\right)^2+\left(\frac{13}{8}\right)^2+\left(\frac{-11}{8}\right)^2+\left(\frac{-5}{2}\right)^2+7^2+\left(\frac{21}{8}\right)^2+\left(\frac{-35}{4}\right)^2}=\frac{\sqrt{16617}}{8} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/4008.txt

@@ -0,0 +1,40 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -2 \\
+ 0 \\
+ -8 \\
+ 0 \\
+ 6 \\
+ 0 \\
+ 8 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, (-2,0,-8,0,6,0,8)\, : \\
+ \| \, (-2,0,-8,0,6,0,8)\, \|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{7-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{7-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, (-2,0,-8,0,6,0,8)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \| \, (-2,0,-8,0,6,0,8)\, \| =\sqrt{(-2)^2+0^2+(-8)^2+0^2+6^2+0^2+8^2}=2 \sqrt{42} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/4075.txt

@@ -0,0 +1,40 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ \frac{11}{10} \\
+ -\frac{7}{2} \\
+ \frac{14}{5} \\
+ \frac{61}{10} \\
+ \frac{27}{5} \\
+ -\frac{19}{2} \\
+ \frac{67}{10} \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, \left(\frac{11}{10},-\frac{7}{2},\frac{14}{5},\frac{61}{10},\frac{27}{5},-\frac{19}{2},\frac{67}{10}\right)\, : \\
+ \left\| \, \left(\frac{11}{10},-\frac{7}{2},\frac{14}{5},\frac{61}{10},\frac{27}{5},-\frac{19}{2},\frac{67}{10}\right)\, \right\|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{7-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{7-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, \left(\frac{11}{10},-\frac{7}{2},\frac{14}{5},\frac{61}{10},\frac{27}{5},-\frac{19}{2},\frac{67}{10}\right)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \left\| \, \left(\frac{11}{10},-\frac{7}{2},\frac{14}{5},\frac{61}{10},\frac{27}{5},-\frac{19}{2},\frac{67}{10}\right)\, \right\| =\sqrt{\left(\frac{11}{10}\right)^2+\left(\frac{-7}{2}\right)^2+\left(\frac{14}{5}\right)^2+\left(\frac{61}{10}\right)^2+\left(\frac{27}{5}\right)^2+\left(\frac{-19}{2}\right)^2+\left(\frac{67}{10}\right)^2}=\frac{\sqrt{22281}}{10} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/414.txt

@@ -0,0 +1,40 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ 6 \\
+ 2 \\
+ -1 \\
+ -2 \\
+ -5 \\
+ -4 \\
+ 6 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, (6,2,-1,-2,-5,-4,6)\, : \\
+ \| \, (6,2,-1,-2,-5,-4,6)\, \|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{7-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{7-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, (6,2,-1,-2,-5,-4,6)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \| \, (6,2,-1,-2,-5,-4,6)\, \| =\sqrt{6^2+2^2+(-1)^2+(-2)^2+(-5)^2+(-4)^2+6^2}=\sqrt{122} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/lp_norm_w_steps/4237.txt

@@ -0,0 +1,41 @@
+Problem:
+Find the $\ell_2$ norm of the following vector:
+$\left(
+\begin{array}{c}
+ -\frac{14}{3} \\
+ -\frac{16}{3} \\
+ \frac{20}{3} \\
+ \frac{11}{3} \\
+ -\frac{50}{9} \\
+ -\frac{71}{9} \\
+ -\frac{16}{3} \\
+ 3 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the norm of the vector }\, \left(-\frac{14}{3},-\frac{16}{3},\frac{20}{3},\frac{11}{3},-\frac{50}{9},-\frac{71}{9},-\frac{16}{3},3\right)\, : \\
+ \left\| \, \left(-\frac{14}{3},-\frac{16}{3},\frac{20}{3},\frac{11}{3},-\frac{50}{9},-\frac{71}{9},-\frac{16}{3},3\right)\, \right\|  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{formula }\text{for }\text{the }\text{8-dimensional }\text{Euclidean }\text{norm }\text{comes }\text{from }\text{the }\text{8-dimensional }\text{Pythagorean }\text{theorem}: \\
+ \left\| \, \left(v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8\right)\, \right\| =\sqrt{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2+v_6^2+v_7^2+v_8^2} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Substitute }\, \left(-\frac{14}{3},-\frac{16}{3},\frac{20}{3},\frac{11}{3},-\frac{50}{9},-\frac{71}{9},-\frac{16}{3},3\right)\,  \text{into }\text{the }\text{formula}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \left\| \, \left(-\frac{14}{3},-\frac{16}{3},\frac{20}{3},\frac{11}{3},-\frac{50}{9},-\frac{71}{9},-\frac{16}{3},3\right)\, \right\| =\sqrt{\left(\frac{-14}{3}\right)^2+\left(\frac{-16}{3}\right)^2+\left(\frac{20}{3}\right)^2+\left(\frac{11}{3}\right)^2+\left(\frac{-50}{9}\right)^2+\left(\frac{-71}{9}\right)^2+\left(\frac{-16}{3}\right)^2+3^2}=\frac{\sqrt{19331}}{9} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}