Multivariable Functions
Functions that feature multiple independent variables.
Graphing
- Always one additional dimension for graphing
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\chapter{Real number}\section{Ordered Fields}The property of ordered fields\subsection{Theorem}
\begin{enumerate}\item A1. $\forall x,y\in\mathbb{R}$ and if $x=w$ and $y=z$, then $x+y=w+z$.%-------------------\item A2. $\forall x,y\in\mathbb{R}$, $x+y=y+x$.%-------------------\item A3. $\forall x,y,z\in\mathbb{R}$, $x+(y+z)=(x+y)+z$.%-------------------\item A4. $\exists$ unique real number 0 $\ni x+0=x$ for all $x\in\mathbb{R}$.%-------------------
\subsection{Exercise}Q3. Let $x,y,z\in\mathbb{R}$. Prove the following.\\Q3(a) $-(-x)=x$:\\\by M1:\\Let $x=-1$, $y=-x$, then\\\begin{align}x\cdot y&=(-1)(-x)\nonumber\\&=(-1\cdot -1)x\leftarrow\text(from,M3)\nonumber\\&=x\end{align}
(b) $(-x)\cdot y=-(xy)$ and $(-x)\cdot(-y)=xy$:\\\by M3:\\\Let $x=-1$, $y=x$,$z=y$ then\\\begin{align}(-x).y&=(-1.x).y\nonumber\\&=-(x,y)\leftarrow \text{from\, M3}\end{align}
To prove second partby M3:\\Let $x=-1$, $y=-x$,$z=-y$ then\\\begin{align}(-x)\cdot (-y)&=(-1\cdot x)(-1\cdot y)\\&=(-1).x.(-1)y\nonumber\\&=xy\leftarrow \text{from\, M3}\end{align}
(e) if $x\neq 0$, then $x^2>0$\\consider $x>0$:\\Let $x=2$:\\\begin{align}x^2&=(2)^2\nonumber\\&=4\end{align}consider $x<0$:\\Let $x=-2$:\\\begin{align}x^2&=(-2)^2\nonumber\\&=4\end{align}
\end{enumerate}
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