I usually pay very little attention to my age and sometimes have trouble exactly telling it off the top of my head. Recently, however, I had a series of conversations that burned into my mind the number 38. Nevertheless, the immediate reaction was “so what?” I did not think of what to do with it or how to interpret it.

Although I still hesitate to make any general statement of getting older, I can at least compare my own thirty eight and twenties. The biggest difference is that I am better at identifying and working for things important for me. In my twenties, I really could not tell whether I wanted to cherish someone or be cherished by everyone, or whether I wanted to recognize something great or be recognized as great. Even after bumping into what mattered to me, I did not know how to take risks and dedicate myself to them.

Now, I am at thirty eight and lucky enough to have found a few things I know important and love to serve.

## Aug 10, 2017

## Aug 4, 2017

### Unqualified to be helpful

In the early afternoon, one of the colleagues in the department asked me for some help on machine learning. I gave some advice and promised to get back to him with more information. Since I had some clues, I was confident of further help and naturally felt good.

A few hours later, a woman who I recently had got acquainted with called me and said she had to cancel the lunch next day. There was a quiver in her voice. While she tried to tell me the reason, I did not quite get it but learned only the fact that something terrible had happened to her. She hung up and I was left texting “Please let me know anything I can help you with. Please.” I knew that I was not close enough to be asked for any help in her emergency. The text message looked pointless and made me feel useless.

I must be prepared, think, and take actions in order to be helpful for those who I try to help.

## May 26, 2017

### A conference poster of BTABM

Here is a conference poster of the ABM project that I have been working on with Paul Mitchell.

Bacillus thuringiensis (Bt) corn is a transgenic seed variety, used for controlling pest damage, whose adoption involves many stakeholders. We think that it is appropriate to model the situation as a complex system. Like many others, I like Python and so use Mesa as a platform. We try to get it done during the summer!

(linked to the original file)

Bacillus thuringiensis (Bt) corn is a transgenic seed variety, used for controlling pest damage, whose adoption involves many stakeholders. We think that it is appropriate to model the situation as a complex system. Like many others, I like Python and so use Mesa as a platform. We try to get it done during the summer!

## May 15, 2017

### The irrationality of √2

When we start to learn analysis or in particular real numbers, a proof for the irrationality of $\sqrt{2}$ typically goes like this.

Assume, for contradiction, there exists $p, q \in \mathbb{Z}$ satisfying $$ \left( \frac{p}{q} \right)^2 = 2 $$ where $p$ and $q$ are relatively prime and $q \neq 0$. Then, rearranging, we have $p^2 = 2q^2$ and so $p$ is even or $p = 2r$ where $r\in \mathbb{Z}$. But, substituting it above, we also have $q^2 = 2r^2$ and $q$ is even as well. This is a contradiction to being relatively prime.

In the lecture on Field automorphism (at 6:35), Professor Macauley gives another proof based on the following proposition. If $\phi$ is an automorphism of $\mathbb{Q}(\sqrt{2})$ ($\mathbb{Q}$ adjoining $\sqrt{2}$), then, $$ \phi(q) = q, \quad \forall q\in \mathbb{Q} .$$ One of such automorphisms is $\phi(a+b\sqrt{2}) = a-b\sqrt{2}$. But, now, $\phi(\sqrt{2}) = -\sqrt{2} \neq \sqrt{2}$. Thus, $\sqrt{2}$ is not rational. Constructing the extension field $\mathbb{Q}(\sqrt{2})$, it is tautological but nonetheless cute.

Assume, for contradiction, there exists $p, q \in \mathbb{Z}$ satisfying $$ \left( \frac{p}{q} \right)^2 = 2 $$ where $p$ and $q$ are relatively prime and $q \neq 0$. Then, rearranging, we have $p^2 = 2q^2$ and so $p$ is even or $p = 2r$ where $r\in \mathbb{Z}$. But, substituting it above, we also have $q^2 = 2r^2$ and $q$ is even as well. This is a contradiction to being relatively prime.

In the lecture on Field automorphism (at 6:35), Professor Macauley gives another proof based on the following proposition. If $\phi$ is an automorphism of $\mathbb{Q}(\sqrt{2})$ ($\mathbb{Q}$ adjoining $\sqrt{2}$), then, $$ \phi(q) = q, \quad \forall q\in \mathbb{Q} .$$ One of such automorphisms is $\phi(a+b\sqrt{2}) = a-b\sqrt{2}$. But, now, $\phi(\sqrt{2}) = -\sqrt{2} \neq \sqrt{2}$. Thus, $\sqrt{2}$ is not rational. Constructing the extension field $\mathbb{Q}(\sqrt{2})$, it is tautological but nonetheless cute.

## May 4, 2017

### RR[x]/(x²+1) isomorphic to CC

It is a very basic problem in ring theory, but I keep forgetting the key ideas involved. The trick is to identify the following evaluation homomorphism:
$$ \varphi_i: \mathbb{R}[x] \to \mathbb{C},\quad \varphi_i(p(x)) = p(i) .$$
To me, this tells a lot. First, it maps all the real coefficients of $\mathbb{R}[x]$ into $\mathbb{C}$, reducing $\mathbb{R}^n$ to $\mathbb{R}^2$. Second, it maps $x$ into $i$ by essentially evaluating each polynomial $p(x)\in\mathbb{R}[x]$ at $i\in\mathbb{C}$. Although $i \notin \mathbb{R}$, it seems okay because the codomain $\mathbb{C}$ is an extension field of $\mathbb{R}$. Finally, it is surjective because for any $a+bi \in \mathbb{C}$ there exists $a+bx \in \mathbb{R}[x]$ such that $\varphi(a+bx) = a+bi$. The kernel captures the very motive for going to this trouble, i.e. no roots of $x^2+1$ in $\mathbb{R}$.
$$ \ker(\varphi) = \{p(x)\in\mathbb{R}[x] \mid p(i)=0\} = \{(x^2+1)q(x) \mid q(x)\in\mathbb{R}[x]\} = \langle x^2+1 \rangle .$$
Thus, by the first isomorphism theorem, $\mathbb{R}[x]/\langle x^2+1 \rangle \cong \mathbb{C}$.

## Apr 27, 2017

### Being a unit in ZZn

$$ \text{$\overline{a}$ is invertible in $\mathbb{Z}_n$ iff $a$ and $n$ are relatively prime.}$$
The following proof from Pinter (p.228) is nice. Use a notation: $\overline{x} = x \pmod n$. Then,
\begin{align*}
& \text{$a$ and $n$ are relatively prime} \\
\Leftrightarrow\quad& \text{gcd}(a,n) = 1 \\
\Leftrightarrow\quad& sa + tn = 1 \quad (\exists s,t \in \mathbb{Z})\\
\Leftrightarrow\quad& 1 - sa = tn \\
\Leftrightarrow\quad& \overline{sa} = \overline{1} \\
\Leftrightarrow\quad& \overline{s}\:\overline{a} = \overline{1} \\
\end{align*}

## Mar 9, 2017

### Normality of the Commutator Subgroup

The problem is:

Let $g \in G$ and $n,n' \in N$. Then, $$ gng^{-1} = gng^{-1}(n^{-1}n) = (gng^{-1}n^{-1})n = n'n \in N .$$

Given a group $G$, show that a subgroup $N = \{xyx^{-1}y^{-1} \mid x,y \in G \}$ is normal.

Since it was difficult for me, I searched online and found a proof on Stack Exchange. As many others, I think it is slick.Let $g \in G$ and $n,n' \in N$. Then, $$ gng^{-1} = gng^{-1}(n^{-1}n) = (gng^{-1}n^{-1})n = n'n \in N .$$

## Feb 24, 2017

### Fiber and Preimage

In the algebra class, I encountered a concept called fiber, which sounded similar to preimage, and I didn't get the difference. According to Wikipedia,
given $f: X \to Y$, the preimage of a set $B \subseteq Y$ under $f$ is a subset of $X$ defined by $f^{-1}(B)=\{\,x\in X\,|\,f(x)\in B\}$. If $B$ is a singleton set, i.e. $B = \{y\}$, the preimage $f^{−1}(y)$ is also called the fiber of $y$.