\begin{document} \begin{titlepage} \newcommand{\HRule}{\rule{\linewidth}{0.5mm}} \center \textsc{\LARGE George Mason University}\\[1.5cm] \textsc{\Large \hmwkClass}\\[0.5cm] \HRule \\[0.4cm] { \huge \bfseries \hmwkTitle}\\[0.4cm] \HRule \\[1.5cm] \begin{minipage}{0.4\textwidth} \begin{flushleft} \large \emph{Author:}\\ \hmwkAuthorName \end{flushleft} \end{minipage} \begin{minipage}{0.4\textwidth} \begin{flushright} \large \emph{Professor:} \\ \hmwkClassInstructor \end{flushright} \end{minipage}\\[4cm] {\large \hmwkDueDate}\\[3cm] \end{titlepage} \topmargin=-0.45in \evensidemargin=0in \oddsidemargin=0in \textwidth=6.5in \textheight=9.0in \headsep=0.25in \linespread{1.1} \pagestyle{fancy} \lhead{\hmwkAuthorName} \chead{\hmwkClass\ (\hmwkClassInstructor)} \rhead{\hmwkTitle} \lfoot{All LaTeX code available upon request.} \cfoot{} \rfoot{Page\ \thepage\ of\ \pageref{LastPage}} \textit{We wish to prove that $\mathbb{R}^n$ satisfies VS 1-8 conditions of a Vector Space [VS 1-8: Pg. 7; Katz]. We will argue each condition separately. W.L.O.G. every vector/set has n elements in it.} \\ \textbf{(VS 1) $\forall$ x, y in V. $x+y=y+x$ (Commutative of Addition).} \begin{proof} Consider x, y $\in$ $\mathbb{R}^n$. We will to prove by method of induction. \\ i. Let $x=y=(1,1,...,1)$ By direct substitution, we have: $$(1,1,...,1)+(1,1,...,1)= (2,2,...,2) = (1,1,...,1) + (1,1,...,1)$$ ii. $x=\left\{x_n \mid r_n+1\right\}$ and $y=\left\{y_n \mid z_n+1\right\}$ By direct substitution, we have: $$(z_1+1,z_2+1,....z_n+1)+(r_1+1,r_2+1...,r_n+1) = (r_1+1,r_2+1...,r_n+1) + (z_1+1,z_2+1,...,z_n+1)$$ $$(z_1+r_1,z_2+r_2,...,z_n+r_n)+(2,2,...,2)=(2,2,...,2)+(r_1+z_1,r_2+z_2,...r_n+z_n)$$ $$=(r_1,r_2...,r_n)+(z_1,z_2,...,z_n)$$ This completes the proof.\\ \end{proof} \textbf{(VS 2) $\forall$ x, y, z $\in$ V, $(x+y)+z=x+(y+z)$ (Associative Property of Addition)} \begin{proof} Consider $x, y, z,$ $\in$ $\mathbb{R}^n$. We will to prove by method of induction.\\ i. Let $x=y=z=(1,1,...,1)$\\ By direct substitution, we have: $$[(1,1....,1)+(1,1....,1)]+(1,1....,1)=(2,2,...,2)+(1,1....,1)$$ Note, by (VS 1): $(2,2,...,2)+(1,1....,1)=(1,1....,1)+(2,2,...,2)$\\ Then, by expansion: $$(1,1....,1)+(2,2,...,2)=(1,1....,1)+[(1,1....,1)+(1,1....,1)] = x+(y+z)$$ ii. $x=\left\{x_n \mid r_n+1\right\}, y=\left\{y_n \mid v_n+1\right\}, z=\left\{z_n \mid t_n+1\right\}$\\ By direct substitution, we have: $$[(r_1+1,r_2+1,...,r_n+1)+(v_1+1,v_2+1,...,v_n+1)]+(t_1+1,t_2+1,...,t_n+1)$$ $$=[(r_1,r_2,...,r_n)+(v_1,v_2,...,v_n)+(2,2,...,2)]+(t_1+1,t_2+1,...,t_n+1)$$ Removing brackets and combining terms yields the equivalent statement: $$=(r_1,r_2,...,r_n)+(v_1,v_2,...,v_n)+(t_1,t_2,...,t_n)+(3,3,...,3)$$ Finally, subtracting $(3,3,...,3)$ and placing brackets gives: $$=(r_1,r_2,...,r_n)+[(v_1,v_2,...,v_n)+(t_1,t_2,...,t_n)]=x+(y+z)$$ This completes the proof.\\ \end{proof} \break \textbf{(VS 3) $\exists$ an element $0$ $\in$ V $\mid$ $x+0=x$ $\forall$ $x$ $\in$ V. (The Empty Set)} \begin{proof} Consider the set of dimension $n$ with all elements being 0. i.e. $\left\{0\right\}=(0,0,...,0)$. This is known as the Empty set.\\ This completes the proof.\\ \end{proof} \textbf{(VS 4) $\forall$ elements $x$ $\in$ V $\exists$ an element $y$ $\in$ V $\mid$ $x+y=0$. (Additive Reciprocal)} \begin{proof} Consider any set $\in$ $\mathbb{R}^n$. We have: $x=(x_1, x_2,...,x_n)$. Consider now the negation of each element in x. i.e. $-x=(-x_1, -x_2,...,-x_n)$. From this, it is clear that $y=-x$ and thus $x+y=0$.\\ This completes the proof.\\ \end{proof} \textbf{(VS 5) $\forall$ $x$ $\in$ V, $1*x=x$. (Multiplicative Identity)} \begin{proof} Consider any set $x$ $\in$ $\mathbb{R}^n$. We have: $$(1*x_1,1*x_2,...,1*x_n)=(x_1,x_2,...,x_n)=x$$. This completes the proof.\\ \end{proof} \textbf{(VS 6) $\forall$ pairs of elements $a,b$ $\in$ F and each element $x$ $\in$ V, $(ab)x=a(bx)$. (Associative Property of Multiplication)} \begin{proof} Consider $x$ $\in$ $\mathbb{R}^n$ and $a,b$ $\in$ F. We have: $$(ab)x=(abx_1,abx_2,...,abx_n)$$ Note that there is a common factor of a in this set. $$=a(bx_1,bx_2,...,bx_n)=a(bx)$$ This completes the proof.\\ \end{proof} \textbf{(VS 7) $\forall$ elements a $\in$ F $\forall$ pairs of elements $\in$ V, $a(x+y)=ax+ab)$. (Distributive Property)} \begin{proof} Consider $x,y$ $\in$ $\mathbb{R}^n$ and $a$ $\in$ F. We have: $$a(x+y)=(ax_1,ax_2,...,ax_n)+(ay_1,ay_2,...ay_n)=ay+ax$$ This completes the proof.\\ \end{proof} \textbf{(VS 8)$\forall$ pairs of elements $a,b$ $\in$ F and $\forall$ x $\in$ V, $(a+b)x=ax+bx$ (Reverse Distributive Property} \begin{proof} Consider $x$ $\in$ $\mathbb{R}^n$ and $a,b$ $\in$ F. We have: $$(a+b)x=((a+b)x_1,(a+b)x_2,...,(a+b)x_n)$$ Separating a and b yields: $$(ax_1,ax_2,...,ax_n)+(bx_1,bx_2,...,bx_n)=ax+bx$$ This completes the proof. \end{proof} This completes the paper.\\ \begin{flushright} X----------------------------------------------------------- \end{flushright} \end{document}