Ch.2 Solving General Linear Systems

Show that $1+2+3+\cdots+n=\frac{n(n+1)}{2}$

We first show the base step. When $n=1$, $1=\frac{1(2)}{2}=1$, so the statement holds.
Suppose $1+2+\cdots+k=\frac{k(k+1)}{2}$ (inductive hypothesis). Then
$1+2+\cdots+k+k+1=\frac{k(k+1)}{2}+k+1=\frac{k(k+1)+2(k+1)}{2}=\frac{k^2+3k+2}{2}=\frac{(k+1)(k+2)}{2}=\frac{(k+1)(k+1+1)}{2}$Ttherefore it is also true. Thus, $1+2+3+\cdots+n=\frac{n(n+1)}{2}$ for all natural numbers $n$. $_\blacksquare$

Prove there are infinitely many primes

Suppose there are a finitely many primes $p_1,p_2,...,p_k$. Consider the number $p_1p_2...p_k+1$. None of the primes divide this number, as it leaves a remainder of 1, however, every number is the product of primes. That is a contradiction, meaning that the initial proposition that there are finitely many primes must be false. Therefore, there must be infinitely many primes. $_\blacksquare$

Show that $\lnot(P\lor Q)\equiv\lnot P\land \lnot Q$

The left is true if both $P$ and $Q$ are false. The right is true is $P$ is false and $Q$ is false. Clearly, they are equivalent.

Show that $\lnot(P\land Q)=\lnot P \lor \lnot Q$

$\lnot(P\land Q)=\lnot(\lnot(\lnot P\lor\lnot Q))=\lnot\lnot(\lnot P\lor\lnot Q)=\lnot P\lor\lnot Q$

$P\implies Q\equiv\lnot P\lor Q\equiv Q\lor\lnot P\equiv\lnot(\lnot Q)\lor \lnot P\equiv\lnot Q\implies \lnot P$

For the left hand side to be true, $C$ must be true, as well as either $A$ or $B$. Looking at the right side, either $A\land C$ or $B\land C$ must be true. In either case, $C$ must be true. Then, if one of either $A$ or $B$ is true, one of those statements will be satisfied.

We must show $(A\cup B)\cap C\subseteq (A\cap C)\cup(B\cap C)$ and $(A\cap C)\cup(B\cap C)\subseteq (A\cup B)\cap C$
For the first, suppose $x\in(A\cup B)\cap C$. This means $x\in C$, and $x\in A$ or $x\in B$. In other words, $x\in C$ and $x\in A$, or $x\in C$ and $x\in B$, which can be represented as $(A\cap C)\cup(B\cap C)$, so $(A\cup B)\cap C\subseteq (A\cap C)\cup(B\cap C)$.
Next, suppose $x\in(A\cap C)\cup (B\cap C)$. Then $x\in C$ regardless, and $x\in A$ or $x\in B$, so $x\in C\cap(A\cup B)$
Since both sets are subsets of each other,
$(a\cup B)\cap C=(A\cap C)\cup(B\cup C)$

First show $S\subseteq \{\vec{p}+\vec{h}\}$ set.
Take another solution vector $\vec{s}\in S$. Look at $\vec{s}-\vec{p}$ and plug into the $i$-th equation.
$a_{i,1}(s_1-p_1)+\cdots+a_{i,n}(s_n-p_n)=d_i-d_i=0$
Therefore, $\vec{s}-\vec{p}=\vec{h}$, or $\vec{s}=\vec{p}+\vec{h}$, meaning $S\subseteq \{\vec{p}+\vec{h}\}$.

Next show $\vec{p}+\vec{h}\subseteq S$.
Plugging into the $i$-th equation,
$a_{i,1}(p_1+h_1)+\cdots+a_{i,n}(p_n+h_n)=d_i+0=d_i$
Therefore, $\vec{p}+\vec{h}$ solves the system, so $\{\vec{p}+\vec{h}\}\subseteq S$.
Since each are subsets of each other, by the Principle of Extensionality, $S=\{\vec{p}+\vec{h}\}$.

A line in $\mathbb{R}^n$ can be represented by the set
$L=\{\vec{p}+t\vec{v}\space|\space t\in\mathbb{R}\}$
What is the set representation for the line going through $(1,1,1)$ and $(2,3,4)$ in $\mathbb{R}^3$?

The particular solution can be either of the two points. Choose $(1,1,1)$. The vector from $(1,1,1)$ to $(2,3,4)$ shows the direction, i.e. $(1,2,3)$
The line is
$L=\left\{\begin{pmatrix}1\\1\\1\end{pmatrix}+t\begin{pmatrix}1\\2\\3\end{pmatrix}\space\middle|\space t\in\mathbb{R}\right\}$

What is the system for which this is the solution to?

Express this as
$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\1\\1\end{pmatrix}+t\begin{pmatrix}1\\2\\3\end{pmatrix}\rightarrow\begin{pmatrix}x-1\\y-1\\z-1\end{pmatrix}=t\begin{pmatrix}1\\2\\3\end{pmatrix}$
We want to eliminate $t$ to get the system of equations. This can be done by row-reducing
$\left(\begin{array}{c|c}1&x-1\\2&y-1\\3&z-1\end{array}\right)\sim\left(\begin{array}{c|c}1&x-1\\0&y-1-2(x-1)\\0&z-1-3(x-1)\end{array}\right)$
The bottom two equations have no $t$ term, so those are used.
The system is
$\begin{matrix}-2x+y+1=0\\-3x+z+2=0\end{matrix}\space\longrightarrow\space\begin{matrix}2x-y=1\\3x-z=2\end{matrix}$

A plane in $\mathbb{R}^n$ can be represented by the set
$P=\{\vec{p}+t\vec{v}+s\vec{w}\space|\space t,s\in\mathbb{R}\}$
Find the set for the plane going through $(1,1,1)$, $(2,1,2)$ and $(-1,-1,-3)$

The particular solution can be any of these; choose $\vec{p}=(1,1,1)$
The vector from $(1,1,1)$ to $(2,1,2)$ is $(1,0,1)$ and the vector from $(1,1,1)$ to $(-1,-1,-3)$ is $(-2,-2,-4)$, so the set is
$P=\left\{\begin{pmatrix}1\\1\\1\end{pmatrix}+t\begin{pmatrix}1\\0\\1\end{pmatrix}+s\begin{pmatrix}-2\\-2\\-4\end{pmatrix}\space\middle|\space t,s\in\mathbb{R}\right\}$

What is the equation of the plane represented by $P$?

We want to eliminate $t$ and $s$ from the equation.
Since there are 3 total variables and 2 free variables, there will be 1 equation.
Express $P$ as
$\left(\begin{array}{cc|c}1&-2&x-1\\0&-2&y-1\\1&-4&z-1\end{array}\right)$
and row reduce
$\left(\begin{array}{cc|c}1&-2&x-1\\0&-2&y-1\\1&-4&z-1\end{array}\right)\xrightarrow{-\rho_1-\rho_2+\rho_3}\left(\begin{array}{cc|c}1&-2&x-1\\0&-2&y-1\\0&0&z-1-(y-1)-(x-1)\end{array}\right)$
The bottom equation has $0=z-1-(y-1)-(x-1)$, which simplifies to the equation
$x+y-z-1=0$
To check notice that all three of the original vectors satisfy this equation.