## Tuesday, September 13, 2016

### Alex's Intro To Analysis Homework Using LaTeX

Song:  White Rabbit

Group:  Jefferson Airplane

This is the second article I have written about Alex using
the computer language to illustrate mathematical symbols called LATEX.

Just last week...
I was watching Alex complete his homework for his UNR class called:
Intro to Analysis [Mathematical].

Alex was typing away using LATEX coding as if he were typing in English.
Alex is very quick using the LATEX programming language.

Alex said that LATEX is a necessary skill to master before a person goes to university...
if they are to take any advanced math classes at all.

Alex also said that anyone going on to university should practice as much as possible while
he or she is still in high school, to allow the student to have as much practice
using it before university, so as to have it be second nature before the skills are needed.

This is a picture of Alex's computer screen as he had used it to complete his homework.

The vertical bar on the left shows all of Alex's previous projects for which he had used LATEX.

The large coded section just to the right of that is the LATEX input symbol area.

As Alex types away using the programming language...
he may then activate the translation process at will...
and it will type up the result on the area just to the right of it.

Alex said that using just 50 - 60 symbols...
he could type up his homework.

In a few days time...
Alex said that a student could easily get used to using Latex...
and with continuing practice...
he or she may type up mathematical papers with speed and ease.

A high school student who types up his homework using LATEX
would be putting himself way ahead of the game...
especially when the transition to university will be complicated enough...
and the load so heavy.

Below is the actual coding as it appeared on his computer screen (screen capture)...
below this is the copied text as the screen capture is too small to show the symbols used...
and below that is the actual result.

This first set is last week's homework.

Then I had put this week's homework below using the same layout.

\documentclass[12pt]{article}
\setlength{\parindent}{0cm}
\usepackage{fancyhdr}
\usepackage{graphicx}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{gensymb}
\usepackage{easytable}
\usepackage{enumitem}
\usepackage{amsthm}
\usepackage{array}

\begin{document}

\raggedleft
Alex  \\~\\
Math 310 \\~\\
September 8, 2016 \\~\\

\centering
Homework \#1 \\~\\

\begin{enumerate}
\item
\begin{enumerate}[label=(\alph*)]
\item $A_1 = (-1,\, 3)$, $A_3 = (-3,\, 5)$; $A_3 \backslash A_1 = (-3,\, -1] \cup [3,\, 5)$
\item $$\bigcap_{n = 1}^{10} A_{\frac{1}{n}} = A_{\frac{1}{10}} = (-\frac{1}{10},\, \frac{21}{10})$$
\item
\begin{proof}
Since $\forall m < n,\, m,\, n > 0,\, A_m \subset A_n$, $$\bigcap_{t > 0} A_t = [\lim_{t\to0} -t,\, \lim_{t\to0} 2 + t] = [0,\, 2]$$ As $\forall t > 0,\, -t \neq 0$, and $\forall t > 0,\, 2 + t \neq 2$, the values 0 and 2 are not included in the interval of intersection, and so it is closed.
\end{proof}
\end{enumerate}
\item
\begin{enumerate}[label=(\alph*)]
\item
\begin{proof}
We use the contrapositive method; specifically, we show that, if $g$ is not onto, then $g \circ f$ is not onto. For $g$ not to be onto, $\exists c \in C,\, b \in B$ such that $g(c) \neq b$. Then $g \circ f$, or $g(f) : (A \rightarrow B) \rightarrow C$, would not be onto, as the codomain remains the same. Thus, if $g \circ f$ is onto, then $g$ is onto as well.
\end{proof}
\item
\begin{proof}
We again use the contrapositive method to show that, if $f$ is not one-to-one (injective), then $g \circ f$ is not injective. In order for $f$ to be injective, $\forall a \in A$, $\exists b \in B$ such that $a$ is mapped to $b$. Should $f$ not be injective, $\exists a \in A$, $\forall b \in B$, $a$ is not mapped to $b$. Consequently, $g(f) : (A \rightarrow B) \rightarrow C$ cannot be injective, since not all elements in the domain are mapped to an element in the codomain. Hence, if $g \circ f$ is injective, then $f$ must be injective.
\end{proof}
\end{enumerate}
\item
\begin{enumerate}[label=(\alph*)]
\item $f^{-1}(X) = (-2,\, 2)$, $f^{-1}(Y) = [-2,\, -1] \cup [1,\, 2]$
\item $f(f^{-1}(X)) = f((-2,\, 2)) = (0,\, 4)$, $f(f^{-1}(Y)) = f([-2,\, -1] \cup [1,\, 2]) = [1,\, 4]$
\end{enumerate}
\item
\begin{proof}
Suppose that $P$ is the statement that $f : D \rightarrow C$ is onto, and $Q$ is the statement $\forall X \subset C,\, f(f^{-1}(X)) = X$. We show that $P \iff Q$ by showing that $P \subset Q$ and $Q \subset P$. First, we prove $P \subset Q$. Assuming $P$ is true, let $c \in f(f^{-1}(X))$. By definition, $\exists d \in f^{-1}(X)$ such that $f(d) = c$. Since $d \subset f^{-1}(X)$ means that $f(d) \subset X$ by definition, $c \subset X$, and so $f(f^{-1}(X)) \subset X$. Furthermore, we prove that $X \subset f(f^{-1}(X))$, and so $f(f^{-1}(X)) = X$. Let $x \in X \subset C$. Since $f$ is surjective, $\exists d \in D$ such that $f(d) = x$. By definition, $d \in f^{-1}(X)$, so $x = f(d) \in f(f^{-1}(X))$. \\~\\
We then assume that $Q$ is true, so that we must prove that the function $f : D \rightarrow C$ is an onto function. Suppose to the contrary that $f : D \rightarrow C$ were not an onto function, which implies that $\exists c \in C$ such that $\forall d \in D$, $f(d) \neq c$. Then, $\exists X \subset C$ such that $f^{-1}(X) = \emptyset$, and $f(f^{-1}(X)) = f(\emptyset) \neq X$, which is a contradiction to the true statement $Q$. Therefore, $f : D \rightarrow C$ must be an onto function, proving that $P$ is true. Conclusively, since $P \subset Q$ and $Q \subset P$, $P \iff Q$.
\end{proof}
\item
\begin{proof}
We prove the given statement by induction. Suppose that $S(n)$ is the statement $$1 + 3 + \ldots + (2n - 1) = n^2$$ Firstly, we show $S(1)$ holds. Since $1 = 1^2$, the base case is true. Next, we show that, if $S(n)$ is true, then $S(n + 1)$ is true for all $n$. We have $$S(n + 1) = 1 + 3 + \ldots + (2n + 1) = (1 + 3 + \ldots + (2n - 1)) + (2n + 1)$$ $$= S(n) + (2n + 1) = n^2 + 2n + 1 = (n + 1)^2$$ Therefore, whenever $S(n)$ holds, $S(n + 1)$ holds as well, and so $S(n)$ holds for all $n \in \mathbb{N}$.
\end{proof}
\item
\begin{proof}
We prove that $x_n < x_{n + 1}$ for all $n \in \mathbb{N}$ using induction. Suppose $S(n)$ is the statement $x_n < x_{n + 1}$; we show that $S(1)$ is true. $x_1 = 0$, $x_2 = \frac{1 + 0}{2} = \frac{1}{2}$; since $0 < \frac{1}{2}$, $S(1)$ holds. To show that $S(n + 1)$ holds whenever $S(n)$ holds, we substitute $n + 1$ for $n$ to obtain $$x_{n + 2} = \frac{1 + x_{n + 1}}{2} = \frac{1 + (\frac{1 + x_n}{2})}{2} = \frac{3 + x_n}{4}$$ Since $x_{n + 1} = \frac{1 + x_n}{2}$ by definition, we have $\frac{1 + x_n}{2} < \frac{3 + x_n}{4}$ for all $n \in \mathbb{N}$. $\frac{3 + x_n}{4} = \frac{1 + x_n}{2} + 1 - x_n$; to show that $S(n + 1)$ is true for all $n \in \mathbb{N}$, we prove that $x_n < 1$ for all $n \in \mathbb{N}$. If we substitute $a = x_n$ and set $x_n = x_{n + 1}$, we obtain $a = \frac{1 + a}{2} \rightarrow a = 1$. Thus, $\lim_{n\to\infty} x_n = 1$, but $x_n \neq 1$, as $x_1 = 0$. Since $\forall n \in \mathbb{N},\, x_n < 1$, $\frac{1 + x_n}{2} < \frac{3 + x_n}{4}$, and so $\forall n x_{n + 1} < x_{n + 2}$. Therefore, $S(n + 1)$ holds true, and $S(n)$ holds for all $n \in \mathbb{N}$.
\end{proof}
A general, non-recursive formula for $x_n$ is $x_n = \frac{2^{n - 1} - 1}{2^{n - 1}}$.
\item
\begin{proof}
We prove this statement by induction upon the terms of $a_n$ for all $n \in \mathbb{N}$. The $a_n$ and $a_{n + 1}$ terms satisfy the equation $a_{n + 1}^2 - 4a_na_{n + 1} + a_n^2 = -2$ for all $n \in \mathbb{N}$. Let $x = a_n$, $y = a_{n + 1}$; we show that the solutions $(x,\, y)$ to the quadratic equation $x^2 - 4xy + y^2 = -2$ are positive integer ordered pairs. If we replace the solution $(x,\, y)$ with the new solution $(y,\, 4y - x)$, we have the ordered pair $(a_{n + 1},\, a_{n + 2})$ where $y^2 - 4y(4y - x) + (4y - x)^2 = -2 \rightarrow y^2 - 16y^2 + 4xy + (16y^2 - 8xy + x^2) = -2 \rightarrow x^2 + y^2 - 4xy = -2$, as desired. In terms of $a_{n + 1}$, $a_{n + 2} = 4a_{n + 1} - a_n$; since $a_1$ is a positive integer, and $a_n$ is an increasing sequence ($\forall n \in \mathbb{N},\, a_n < 4a_{n + 1}$ (as $a_{n + 1} = 4a_n - a_{n - 1}$, and $a_{n - 1} > 0$ for all $n \in \mathbb{N}$)), all of the terms of the sequence $a_n$ are positive integers.
\end{proof}
\end{enumerate}

\end{document}

And...
Alex's second homework using LATEX.

\documentclass[12pt]{article}
\setlength{\parindent}{0cm}
\usepackage{fancyhdr}
\usepackage{graphicx}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{gensymb}
\usepackage{easytable}
\usepackage{enumitem}
\usepackage{amsthm}
\usepackage{array}

\begin{document}

\raggedleft
Alex  \\~\\
Math 310 \\~\\
September 13, 2016 \\~\\

\centering
Homework \#2 \\~\\

\begin{enumerate}
\item
\begin{proof}
We prove the statement that there does not exist a rational solution to the equation $x^2 = 3$ by contradiction. Assume that there exists a rational solution $x = \frac{a}{b}$, so that $x^2 = \frac{a^2}{b^2} = 3$, yielding $a^2 = 3b^2$. For $a^2$ to have 3 as a factor, $a$ must have a factor of 3 as well. We prove this by the contrapositive method; specifically, we show that, whenever $a$ is not a multiple of 3, $a^2$ is not a multiple of 3. Thus, $a \equiv 1 \pmod 2$ or $a \equiv 2 \pmod 2$. In the first case, $a = 3k + 1$ for some $k \in \mathbb{N}$. Then $a^2 = 9k^2 + 6k + 1 = 3(3k^2 + 2k) + 1 \equiv 1 \pmod 3$, so $3 \nmid a^2$. In the second case, $a = 3k + 2$ for some $k \in \mathbb{N}$, so $a^2 = 9k^2 + 12k + 4 = 3(3k^2 + 4k + 1) + 1 \equiv 1 \pmod 3$; again, $3 \nmid a^2$. Hence, $a$ is a multiple of 3, and so $a = 3l$ for some $l \in \mathbb{N}$. In turn, this gives $a^2 = 9l^2 \rightarrow b^2 = 3l^2$, which implies that $3 \mid b^2$, and so $3 \mid b$ by a similar argument as above. However, since $3 \mid a$ and $3 \mid b$, $\gcd(a,\, b) = 3 \neq 1$, which is a contradiction to the original statement, since we are required to have $\gcd(a,\, b) = 1$ by the definition of a rational number $\frac{a}{b}$. Therefore, there does not exist any rational value of $x = \frac{a}{b}$ such that $x^2 = 3$.
\end{proof}
\item
\begin{proof}
An example of irrational numbers $a,\, b$ such that $a + b,\, ab \in \mathbb{Q}$ is $a = \sqrt{2},\, b = -\sqrt{2}$. Here, we have $a + b = 0 \in \mathbb{Q}$, and $ab = -2 \in \mathbb{Q}$.
\end{proof}
\item
\begin{proof}
An example of $a,\, b \notin \mathbb{Q}$ with $a^b \in \mathbb{Q}$ is $a = e,\, b = \ln(2)$, so that $a^b = 2 \in \mathbb{Q}$.
\end{proof}
\item
\begin{enumerate}[label=(\alph*)]
\item The set of upper bounds for the set $\{3,\, 1,\, 0\}$ is $[3,\, \infty)$, since 3 is the least upper bound.
\item The set of upper bounds for $\mathbb{N}$ is $\emptyset$. There does not exist a value $m \in \mathbb{N}$ such that $\forall n \in \mathbb{N},\, m \geq n$, since there is no greatest natural number. \newpage
\item The set of upper bounds for the set $\{e^{-x} \colon x \geq 0\}$ is $[1,\, \infty)$, since the least upper bound is 1, and the function $f \colon \mathbb{R^+} \rightarrow \mathbb{R},\, f(x) = e^{-x}$ is decreasing for $x \geq 0$. (To show that $f(x) = e^{-x} = \frac{1}{e^x}$ is decreasing for non-negative real values of $x$, we show that $\forall x \geq 0,\, f(x + 1) - f(x) < 0$. We have $f(x + 1) - f(x) = \frac{1}{e^{x + 1}} - \frac{1}{e^{x}} = \frac{1 - e}{e^{x + 1}}) < 0$ for $x \geq 0$, since $1 - e < 0$ and $e^{x + 1} > 0$. Thus, $f$ is a decreasing function, and so the least upper bound must be 1.)
\item The set of upper bounds is $(\sqrt{5},\, \infty)$, since $\forall d \in D,\, d < \sqrt{5}$, but $\sqrt{5} \notin D$.
\item We seek the least upper bound of the set $E = \{\frac{2n - 1}{n},\, n \in \mathbb{N}\}$, which yields the set of all upper bounds. We observe that $\lim_{n\to\infty}$ $\frac{2n - 1}{n} = 2$, yielding a set of upper bounds of $[2,\, \infty)$. To show that $\sup E = 2$, we first prove that 2 is indeed an upper bound for the set. Since $\forall n \in \mathbb{N},\, \frac{2n - 1}{n} < 2$, 2 is an upper bound for $E$. Next, we show that 2 is the smallest upper bound. Assume to the contrary that there exists a smaller upper bound, $2 - \epsilon$, where $\epsilon > 0$. This implies that $\forall n \in \mathbb{N},\, 2 - \epsilon \geq \frac{2n - 1}{n}$, or $\epsilon \leq \frac{1}{n} \rightarrow n \leq \frac{1}{\epsilon}$ for all $n \in \mathbb{N}$. However, this is false, as we can set $n = \frac{1}{\lceil \epsilon \rceil - 1}$ - and poses the desired contradiction. Conclusively, 2 is the smallest upper bound of the set, and so the set of upper bounds is $[2,\, \infty)$.
\end{enumerate}
\item
\begin{enumerate}[label=(\alph*)]
\item Set $A$: The set is bounded above at 3, which is the least upper bound.
\item Set $B$: $\mathbb{N}$ is not bounded above, since there is no greatest natural number. Therefore, there is no least upper bound.
\item Set $C$: The set is bounded above at 1, the least upper bound.
\item Set $D$: The set is not bounded above, since $\sqrt{5} \notin D$; hence, the set does not have a least upper bound.
\item Set $E$: The set is bounded above at its supremum/least upper bound of 2.
\end{enumerate} \newpage
\item
\begin{proof}
The set $S$ consists of all values of $x$ such that $x^2 < 2x + 3 \implies x^2 - 2x - 3 < 0 \implies (x - 3)(x + 1) < 0 \implies x \in (-1,\, 3)$. To show that $S = (-1,\, 3)$ is bounded above, we show that $\sup S$ exists, then determine its value (the least upper bound of $S$). In order for $\sup S$ to exist, it must be an upper bound for $S$; i.e. a value $m$ such that, $\forall x \in S,\, m \geq x$. All $m \geq 3$ satisfy this condition, so $\sup S$ necessarily exists; thus, $S$ is bounded from above, with 3 being the least upper bound of $S$.
\end{proof}
\item
\begin{proof}
Suppose that $a,\, b \in \mathbb{R}$, with $a - \frac{1}{n} < b$ for every $n \in \mathbb{N}$. We prove that $a \leq b$ by contradiction. Suppose that $a > b$. Then $0 < a - b < \frac{1}{n} \implies \forall n \in \mathbb{N},\, \frac{1}{a - b} > n$. But for $n = \lfloor \frac{1}{a - b} \rfloor + 1$, we have a contradiction. Therefore, $a \leq b$.
\end{proof}
\item
\begin{proof}
Let $n \in \mathbb{N}$. Then $m - \frac{1}{n} < m = \sup A$, so $m - \frac{1}{n} = \sup A$. Since $\forall a \in A,\, \sup A \geq a$, $\exists a \in A$ such that $m - \frac{1}{n} < a$.
\end{proof}
\end{enumerate}

\end{document}

I can't stress enough how important using LATEX is
while a child is still in high school.

He will then have had plenty of practice using it before getting to university.

Again...
learning and using LATEX is free.

A link:

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