Problem A finite commutative ring $R$ without zero divisors is a field. Here we don’t assume that $R$ has an identity. Proof $R$ has an identity For each $a \neq 0$ in $R$, consider the map $$\phi_a: x \mapsto ax$$ Since $a$ is not a zero divisor, $\phi_a$ is injective. Because $R$ is finite, $\phi_a$ is also surjective. Hence, there is $e \in R$ such that $$a = \phi_a(e) = ea$$ ...
Probability of collapsing $a |0 \rangle + b |1 \rangle$ 變成 0 的機率是 $|a|^2$ 變成 $1$ 的機率是 $|b|^2$ Unitary operator 保長保角 $M_{\text{NOT}} = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}$ $W_2 = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \ \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \end{pmatrix}$ fair coin toss Tensor product Entangled state v.s. decomposable state $\langle a \otimes b, c \otimes d \rangle = \langle a, c \rangle \langle b, d \rangle$ $$\begin{align*}M_{\text{CNOT}}: &|00\rangle \mapsto |00\rangle, |01\rangle \mapsto |01\rangle, \&|10\rangle \mapsto |11\rangle, |11\rangle \mapsto |10\rangle \end{align*}$$ EPR Pair $$M_{\text{CNOT}}\left((\frac{1}{\sqrt{2}} | 0 \rangle +\frac{1} {\sqrt{2}} |1\rangle) \otimes |0\rangle\right) = \frac{1}{\sqrt{2}} |00 \rangle +\frac{1} {\sqrt{2}} |11\rangle$$ ...
Theorem Suppose $K$ is a compact set of $\mathbb{R}^n$, and $\{V_\alpha\}$ is an open cover of $K$. Then there exists functions $\psi_1, …, \psi_s \in C(\mathbb{R}^n)$, such that $0 \leq \psi_i \leq 1$; each $\psi_i$ has its support in some $V_\alpha$; (subordinate) $\psi_1 + … + \psi_s = 1$ on $K$. (partition of unity) Proof Associate with each $x \in K$ an index $\alpha(x)$ such that $x \in V_{\alpha(x)}$. Then there are open balls $B(x)$ and $W(x)$ centered at $x$ with ...
Property The continued fraction representation of a rational is unique. Proof Let $r \in \mathbb{Q}$ and $[p_0, p_1, …, p_n], [q_0, q_1, …, q_m]$ be continued fraction representations of $r$. Now, process induction on $\min(n, m)$. The case $\min(n, m) = 0$ is trivial. By definition, we know that $p_0 = q_0 = \lfloor r \rfloor$. Then, $$(r - \lfloor r \rfloor)^{-1} = [p_1, …, p_n] = [q_1, …, q_m]$$ ...
Theorem Give 2 sequences $\{a_n\}$ and $\{b_n\}$, put $$ A_n = \sum_{k=0}^n a_n$$ if $n \geq 0$; put $A_{-1} = 0$. Then, if $0 \leq p \leq q$, we have $$\sum_{n=p}^q a_n b_n = \sum_{n=p}^{q-1} A_n(b_n - b_{n+1}) + A_qb_q - A_{p-1}b_p$$ Proof $$ \sum_{n=p}^q a_n b_n = \sum_{n=p}^q (A_n - A_{n-1}) b_n = \sum_{n=p}^q A_nb_n - \sum_{n=p-1}^{q-1}A_nb_{n+1}$$ Simplify the last expression. Note 每次我要用這東西的時候都會忘記，就這樣記下來。 ...
Problem The Quaternion group $G = Q_8$ is not isomorphic to any subgroup of $S_7$. Proof Let $A$ be a set with 7 elements. Any homomorphism $\alpha: G \to S_7$ can be viewed as a group action $G$ on $A$. For each $a \in A$, consider the size of its orbit $Ga$ and stabilizer $G_a$. Orbit-stabilizer theorem gives the following equality: $$ |Ga| = \frac{|G|}{|G_a|} $$ Since $Ga \subseteq A$, we have $|G_a| \leq 7$. Then ...
Theorem Every vector space $V$ has a basis. Proof The case $V$ is zero space is trivial, so we just consider the case $V$ is not zero. We will use “Zorn’s lemma” in the following proof. Let $P$ be the collection of all linearly independent subset of $V$. Note that $P$ is partially ordered with respect to “inclusion”. Also note that $P$ is not empty, since it must contain $\{v\}$ for some $v \in V$. ...
Problem If $S$ is an uncountable collection of positive real numbers, then the summation of $S$ diverges. Proof Suppose otherwise, the summation converges. For $n = 1, 2, …$, denote $A_n = \{x \in S \mid x \geq \frac{1}{n}\}$. Notice that $A_n$ is a subset of $S$, so its summation must converge. Since each element of $A_n$ has a lower bound $1/n > 0$, the cardinality of $A_n$ must be finite. (Otherwise $\sum A_n$ diverges.) ...