161 lines
5.1 KiB
Markdown
161 lines
5.1 KiB
Markdown
# Lecture 1
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## Toy example (RSA encryption)
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$Enc$
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1. Choose a letter, count to number of letters in the alphabet.
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2. Calculate $s^7$, then divide by 33, take the remainder.
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3. Send the remainder.
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$Dec: s = r^3 \mod 33$
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To build up such system.
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Step 1: Understanding the arithmetic of remainders.
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## Part 1: Divisibility and prime numbers
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### Divisibility and division algorithm
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Let $a, b\in \mathbb{Z}$, with $b>0$. There are unique integers, $q$ (quotient) and $r$ (remainder), such that $a = bq + r$ and $0 \leq r < b$.
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Example: $31=7\times 4 + 3$
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Proof:
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The quotient and remainder are unique.
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(1) Existence:
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Let $S = \{a - bk \mid k \in \mathbb{Z}, a - bk \geq 0\}$. Choose $r$ to be the smallest non-negative element of $S$. This means $r = a - bq$ for some $q \in \mathbb{Z}$. i.e. $a=bq+r$.
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> New notion: $\triangle FSOC$ means For the sake of contradiction.
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Notice that $r \geq 0$, by contradiction, suppose $r \geq b$, then $r-b \geq 0$ and $r-b \in S$, but $r-b < r$, which contradicts the minimality of $r$.
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Therefore, $r \geq 0$ and $r < b$.
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Example: $a=31, b=7$, $S = \{31-7k \mid k \in \mathbb{Z}, 31-7k \geq 0\}=\{\cdots, -32, -25, -18, -11, -4, 3, 10, 17, 24, 31, \cdots\}$, $r=3$.
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So We choose $q=4$ and $r=3$.
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(2) Uniqueness:
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Suppose we have two pairs $(q, r)$ and $(q', r')$ such that $a = bq + r = bq' + r'$ Suppose $q \neq q'$, without loss of generality, suppose $q > q'$, $q-q' \geq 1$. Then $b(q-q') = r'-r$.
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Since $r'=b(q-q')+r \geq b(q-q') \geq b$, which contradicts that $r' < b$.
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Therefore, $q=q'$ and $r=r'$.
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QED
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#### Definition: Divisibility
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Let $a, b \in \mathbb{Z}$, we say $b$ divides $a$ and write $b \mid a$ if there exists $k\in \mathbb{Z}$ such that $a = bk$.
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Example: $3 \mid 12$ because $12 = 3 \times 4$.
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#### Properties of divisibility
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Let $a, b, c \in \mathbb{Z}$.
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(1) $b \mid a \iff r=0$ in the division algorithm.
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(2) If $a \mid b$ and $b \mid c$, then $a \mid c$.
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(3) If $a \mid b$ and $b \mid a$, then $a = \pm b$.
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(4) If $a \mid b$ and $a \mid c$, then $a \mid bx + cy$ for all $x, y \in \mathbb{Z}$. (We call such $bx+cy$ a linear combination of $b$ and $c$.)
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(5) If $c\neq 0$ and $a \mid b \iff ac \mid bc$.
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Some proof examples:
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(2) Since $a \mid b$ and $b \mid c$, there exist $k, l \in \mathbb{Z}$ such that $b = ak$ and $c = bl$. Then $c = bl = (ak)l = a(kl)$, so $a \mid c$.
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QED
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(3) If $a \mid b$ and $b \mid a$, then there exist $k, l \in \mathbb{Z}$ such that $b = ak$ and $a = bl$. Then $a = bl = (ak)l = a(kl)$, so $a(1-kl) = 0$.
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Case 1: $a=0$, then $b=0$, so $a=b$.
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Case 2: $a \neq 0$, then $1-kl=0$, so $kl=1$. Since $k, l \in \mathbb{Z}$, $k=l=\pm 1$, so $a = \pm b$.
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QED
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#### Definition: Divisor
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Let $a\in \mathbb{Z}$, we define $D(a) = \{d\in \mathbb{Z} \mid d \mid a\}$.
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**Note that $D(0) = \mathbb{Z}$.**
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Example: $D(12) = \{\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12\}$.
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#### Definition: Greatest common divisor
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Let $a, b \in \mathbb{Z}$, where $a,b$ not both zero, we define the greatest common divisor of $a$ and $b$ to be the largest element in $D(a) \cap D(b)$. It is denoted by $(a,b)$.
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> Terrible, I really hate this notation. But professor said it's unlikely to be confused with the interval $(a,b)$ since they don't show up in the same context usually.
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Example:
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$(12, 18) = 6$.
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**Note that $(0,0)$ is not defined. (there is no largest element in $D(0) \cap D(0)$.)**
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but it is okay that one of $a, b$ is zero. For example, $(0, 18) = 18$.
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$(n,n) = |n|$ for all $n \in \mathbb{Z}$.
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In general, if $(a,b)=0$ we say $a$ and $b$ are relatively prime, or coprime.
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$\forall a, b \in \mathbb{Z}$, $(a,b) \geq 1$.
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#### Theorem for calculating gcd
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Let $a, b \in \mathbb{Z}$, with $b\neq 0$, then for any $k\in \mathbb{Z}$, $(a,b) = (b,a-bk)$.
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Example: $(12, 18) = (18, 12-18) = (18, -6) = 6$.
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$(938,210)=(210,938-210\times 4)=(210,938-840)=(210,98)$.
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Proof:
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We will prove that $D(a) \cap D(b) = D(b) \cap D(a-bk)$.
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(1) $D(a) \cap D(b) \subseteq D(b) \cap D(a-bk)$:
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Let $d \in D(a) \cap D(b)$, then $d \mid a$ and $d \mid b$.
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By property of divisibility (4), If $a\mid b$ and $b\mid c$, then for all $x,y\in \mathbb{Z}$, $a\mid bx+cy$.
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So $d\mid a+b\cdot (-k) = a-bk$.
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Therefore, $d \in D(b) \cap D(a-bk)$.
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(2) $D(b) \cap D(a-bk) \subseteq D(a) \cap D(b)$:
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Let $d \in D(b) \cap D(a-bk)$, then $d \mid b$ and $d \mid a-bk$.
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By property of divisibility (4), $d \mid bk + (a-bk) = a$.
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Therefore, $d \in D(a) \cap D(b)$.
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QED
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This theorem gives rise to the Euclidean algorithm which is a efficient way to compute the greatest common divisor of two integers. $2\Theta(\log n)+1=O(\log n)$ ([Proof in CSE 442T Lecture 7](https://notenextra.trance-0.com/CSE442T/CSE442T_L7#euclidean-algorithm)).
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### Euclidean algorithm
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We will skip this part, it's already the third time we see this algorithm in wustl.
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#### Theorem: Euclidean algorithm returns correct gcd
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Let $a>b>0$, be integers. Using the Euclidean algorithm, we can find $b>r_0>r_1>r_2>\cdots>r_n$ such that $a=bq_0+r_0, b=r_0q_1+r_1, \cdots, r_{n-1}=r_nq_{n+1}+r_{n+1}, r_n=0$. Then $(a,b)=r_n$.
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Proof:
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(a) This process terminates. $b>r_0>r_1>r_2>\cdots>r_n$ is a strictly decreasing sequence of positive integers, so it must terminate.
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