A description of a successful ABA therapy graduate. Initially diagnosed as Autistic, he is a 2E child (twice exceptional - In the Autistic Spectrum and Profoundly Gifted). A brief description of his path to enrichment and excellence as laid out by the therapists and by, mostly, his Mother - his Chief Instructor.
Both, my mother, and Alex's mother (my wife), are from Japan.
------Alex was born in the year of the Golden Dragon (Nov 2000) ------

Alex had received an email which had been sent out

by a UNR professor of advanced and graduate level mathematics

to all students of higher level mathematics at the university.

He is recruiting advanced mathematicians at the university to attend his

competition math practice to prepare for the Intermountain / Putnam math competitions.

Alex started his first practice session for the Intermountain Math Competition at UNR today.

It is a math competition for university students...

and it is one step below the famed Putnam university math competition.

Alex wishes to compete at the 2016 Intermountain Math Competition.

Addendum: Alex learned that the competition will be held at each university...
and the results will be compiled and sent to each participating university
so placements can be known before the end of the contest.

Alex going in to his practice session at UNR.

Alex is excited about his competing at this competition.

Well...

in a little over one month...

Alex will be competing at UNR at the same time
as the other universities will be taking theirs.

This week's driving lessons included more left turns...
blind curves, pedestrians and cyclists in high numbers and close quarters...
a circumnavigation of the whole city of Reno at night...
and driving Reno with a festival changing the traffic patterns downtown.

We practiced more blind curves.

Once again...
I had him completely stop behind the line...
and then creep out until he could see around the blind curve...
and stop...clear all avenues (left, center, and right)...
and then, and only then...go.

By the way...
this is the kind of lunacy which angers me every time I see it.

Blind curves at intersections are inexcusable because they are preventable
with just common sense at the city planner's office.

There should always be set back from corners to allow drivers to fully see up and down
the cross streets. Not being able to puts pedestrians and other drivers
at a highly preventable and unnecessary risk.

This is just one of the reasons I hate driving in the city.

I get frustrated seeing blind curves...
decreasing radius turns...
poorly placed cross walks...
incredibly stupid light patterns which put pedestrians at risk...
along with drivers...
and so many other things which frustrate me every time I see them...
because I know the reason behind the decisions.

The priority is not one of safety.

It is one of expediency...
temporary cost savings...
and a longer term income increase
because of zoned square footage taxes.

It is one of politicians overruling the engineers.

This is something that should never happen.

Safety can be engineered to a much higher level every where...
you just have to have the last word going to the engineers.

In order to properly clear this blind curve...
the front of the car has to protrude into the cross traffic's number one lane.

As I had Alex drive in the busiest areas of Reno...
I had him constantly scan for pedestrians and cyclists.

We turned on to the main drag between the many casinos.

There were many people crossing where they weren't supposed to...
some were staggering...others were numbskulls.

There were also idiotic cyclists riding in the car lanes...
and an overall area of highly unpredictable behavior from everyone.

I had Alex drive here to practice his peripheral vision scanning.

In the city...
there is too much going on at once to concentrate on any one area.

He has to drive in scan mode...
always on the alert for anomalies of motion
as presented either in speed or direction...
or both.

For a couple of days...
I had Alex combine areas of his driving to include the past two weeks' learning areas.

A downtown festival had changes the traffic patterns...
so I had him make a much larger driving learning area.

On one night...
I had Alex circumnavigate the city of Reno.

There is one expressway which forms a complete circle around Reno.

Alex had a stressful week of city driving...
so, I decided to give him a relaxing night of easy night driving.

--------------

This week's driving lesson will be more city driving.

While Alex is still learning to drive in peripheral scan mode...
I want to reinforce this necessary skill while it is fresh in his mind.

On today's drive...
we drove in the city...
and then to a shopping area.

Within our one hour lesson...
one car had run a red light (next to us)...
two cars in quick succession
had run a yield sign in a roundabout (they should have yielded to Alex)...
and just a few miles from home...
an incredibly stupid person had run a stop sign in front of us
(cross traffic to Alex's right...he would have hit us had he not...
at the last moment, looked to see us AS HE WAS RUNNING THE STOP SIGN...
and then hit his brakes).

This numbskull had made up his mind to run the stop sign from the beginning...
and then, just as an after thought...
decided to look our way.

I had seen he was going too fast and I had told Alex to look out.

Alex had also seen him and had reacted appropriately
by braking and turning left.

As I have been explaining to Alex...
were everyone to follow the traffic rules...
there would be very few accidents.

It is because of the numbskulls that he must constantly guard against
those who's driving defies all logic.

Every light and stop sign is a potential death trap...
and so, he must scan for light or stop sign runners before entering the intersection.

Every yield sign is an opportunity for a numbskull to not yield...
and so, Alex must monitor to ensure the other driver slows and yields...
before entering the merge zone.

Although, directly following each of these incidents...
the movie of my mind involved me playing a drum solo...
with their heads being the drums...
I was glad that these had happened with me being an extra pair of eyes.

This allowed Alex to directly experience the driving habits of a numbskull...
and exactly why he MUST drive defensively.

Simply put...
a safe driver is a great observer...
and a great forward planner.

By keeping a safe following distance...
and by driving as much in the bubble as possible...
he is allowing as much reaction time as possible.

By clearing all blind curves beforehand...
and by using peripheral vision scanning for any speed or directional anomalies
in the city, and at all intersections or merge points...
he can prevent himself from becoming the victim of a numbskull.

--------

This weekend...
Alex will be driving us just beyond Sacramento, California.

We will be attending a symposium.

This will give more mountain and city driving to Alex.

The driving time roundtrip should be about 6 - 7 hours.

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.

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.

(I picked this song twice because I feel so at home in high altitude mountain communities).

As we walked up and down the wooden planked walkways of Virginia City...

we came across this little shop which sold politically incorrect joke items.

Many were very funny...
and yet... so true :)

These were prints of actual wanted posters.

This steakhouse had GREAT steaks.

We had dinner here...and I had a FANTASTIC steak.

This is on the back of the menus.

This steakhouse had a brothel theme.

We had the great fortune to have dinner Al Fresco.

As the altitude kept the temperature mild...

there was also a constant light breeze.

My wife had taken this photo of Alex and I looking out at the surrounding countryside.

Alex had ordered the scallops.

My wife had seared tuna.

I had ordered the Rib Eye steak...cooked to medium well.

I always cross my fingers whenever I order it this way.

Although it is my preferred doneness of steaks...

almost never do the chefs get it right.

On this day...

at this restaurant...

the chef had not only gotten it spot on...

he had made the PERFECT steak.

Never had I had such perfection...
and I fear I never shall again
(unless I go back here).

The meat was not only a true medium well done...

it was the most juicy and well spiced steak I have ever had.

This can only come with the combination of an excellent quality meat...

and excellent preparation from a chef who knows what he is doing...

and cooks from a heart filled with professional pride.

Although I usually eat a steak twice the size of this one...

it was such an exceptional steak...

I cut off a fairly large portion for Alex to eat...

and I gave my wife a bite of my steak to also try.

I especially wanted Alex to know what a truly delicious steak tastes like.

It needed no sauce what so ever.

It was just so tender, juicy, and so bursting with flavor...

it was just so perfect as it was.

After I had stewed in the pleasure of each, almost melt in your mouth, morsel...
everything in the world just felt so right.

I had a big smile on my face...
all the way home.

I was just so satisfied.

I just knew that it was one of those meals which will be remembered for a lifetime.

I simply had never known that it was possible to have a medium well done steak...
and to have it the most juicy and well flavored steak of all I have ever eaten.

I also know that all other restaurants are going to catch hell from me if they don't
get their steaks right...especially if they try to tell me that to have a steak medium well...
always dries it out somewhat.

I now know better :)

And if anyone tries to tell me differently in the future...
I will simply say to them:

'Mustang Ranch Steakhouse in Virginia City, NV.
Eat there...you will know what is possible in a steak.
Emulate them...you will have risen to the few which are World Class'.

As we were leaving the restaurant...
I told the check out waiter that it was simply the most delicious steak I had ever had in my life.

What a deep pleasure it was for me to have eaten there.

After eating...

we walked back to the car...

and Alex drove us home.

We had sat on the upper outside balcony of this yellow building for our dinner.

We are leaving Virginia City here.

Just outside the city...

a herd of feral horses (Mustangs - wild horses)...

blocked the road temporarily.

There were signs depicting Mustangs as crossing the roads wherever they please.

The mountains of Nevada are filled with Mustangs.

Mustangs roam freely...

and belong to no one.

Alex did well with a turnout and with his descent speed control using his mountain gears.