section-division-algorithm.ipynb

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Section2.2The Division Algorithm

¶

An application of the Principle of Well-Ordering that we will use often is the division algorithm.

Theorem2.9Division Algorithm

Let $a$ and $b$ be integers, with $b \gt 0\text{.}$ Then there exist unique integers $q$ and $r$ such that

$$\begin{equation*} a = bq + r \end{equation*}$$

where $0 \leq r \lt b\text{.}$

Proof

This is a perfect example of the existence-and-uniqueness type of proof. We must first prove that the numbers $q$ and $r$ actually exist. Then we must show that if $q'$ and $r'$ are two other such numbers, then $q = q'$ and $r = r'\text{.}$

Existence of $q$ and $r\text{.}$ Let

$$\begin{equation*} S = \{ a - bk : k \in {\mathbb Z} \text{ and } a - bk \geq 0 \}. \end{equation*}$$

If $0 \in S\text{,}$ then $b$ divides $a\text{,}$ and we can let $q = a/b$ and $r = 0\text{.}$ If $0 \notin S\text{,}$ we can use the Well-Ordering Principle. We must first show that $S$ is nonempty. If $a \gt 0\text{,}$ then $a - b \cdot 0 \in S\text{.}$ If $a \lt 0\text{,}$ then $a - b(2a) = a(1 - 2b) \in S\text{.}$ In either case $S \neq \emptyset\text{.}$ By the Well-Ordering Principle, $S$ must have a smallest member, say $r = a - bq\text{.}$ Therefore, $a = bq + r\text{,}$ $r \geq 0\text{.}$ We now show that $r \lt b\text{.}$ Suppose that $r \gt b\text{.}$ Then

$$\begin{equation*} a - b(q + 1)= a - bq - b = r - b \gt 0. \end{equation*}$$

In this case we would have $a - b(q + 1)$ in the set $S\text{.}$ But then $a - b(q + 1) \lt a - bq\text{,}$ which would contradict the fact that $r = a - bq$ is the smallest member of $S\text{.}$ So $r \leq b\text{.}$ Since $0 \notin S\text{,}$ $r \neq b$ and so $r \lt b\text{.}$

Uniqueness of $q$ and $r\text{.}$ Suppose there exist integers $r\text{,}$ $r'\text{,}$ $q\text{,}$ and $q'$ such that

$$\begin{equation*} a = bq + r, 0 \leq r \lt b \quad \text{and}\quad a = bq' + r', 0 \leq r' \lt b. \end{equation*}$$

Then $bq + r = bq' + r'\text{.}$ Assume that $r' \geq r\text{.}$ From the last equation we have $b(q - q') = r' - r\text{;}$ therefore, $b$ must divide $r' - r$ and $0 \leq r'- r \leq r' \lt b\text{.}$ This is possible only if $r' - r = 0\text{.}$ Hence, $r = r'$ and $q = q'\text{.}$

Let $a$ and $b$ be integers. If $b = ak$ for some integer $k\text{,}$ we write $a \mid b\text{.}$ An integer $d$ is called a of $a$ and $b$ if $d \mid a$ and $d \mid b\text{.}$ The of integers $a$ and $b$ is a positive integer $d$ such that $d$ is a common divisor of $a$ and $b$ and if $d'$ is any other common divisor of $a$ and $b\text{,}$ then $d' \mid d\text{.}$ We write $d = \gcd(a, b)\text{;}$ for example, $\gcd( 24, 36) = 12$ and $\gcd(120, 102) = 6\text{.}$ We say that two integers $a$ and $b$ are if $\gcd( a, b ) = 1\text{.}$

Theorem2.10

Let $a$ and $b$ be nonzero integers. Then there exist integers $r$ and $s$ such that

$$\begin{equation*} \gcd( a, b) = ar + bs. \end{equation*}$$

Furthermore, the greatest common divisor of $a$ and $b$ is unique.

Proof

Let

$$\begin{equation*} S = \{ am + bn : m, n \in {\mathbb Z} \text{ and } am + bn \gt 0 \}. \end{equation*}$$

Clearly, the set $S$ is nonempty; hence, by the Well-Ordering Principle $S$ must have a smallest member, say $d = ar + bs\text{.}$ We claim that $d = \gcd( a, b)\text{.}$ Write $a = dq + r'$ where $0 \leq r' \lt d\text{.}$ If $r' \gt 0\text{,}$ then

$$\begin{align*} r'& = a - dq\\ & = a - (ar + bs)q\\ & = a - arq - bsq\\ & = a( 1 - rq ) + b( -sq ), \end{align*}$$

which is in $S\text{.}$ But this would contradict the fact that $d$ is the smallest member of $S\text{.}$ Hence, $r' = 0$ and $d$ divides $a\text{.}$ A similar argument shows that $d$ divides $b\text{.}$ Therefore, $d$ is a common divisor of $a$ and $b\text{.}$

Suppose that $d'$ is another common divisor of $a$ and $b\text{,}$ and we want to show that $d' \mid d\text{.}$ If we let $a = d'h$ and $b = d'k\text{,}$ then

$$\begin{equation*} d = ar + bs = d'hr + d'ks = d'(hr + ks). \end{equation*}$$

So $d'$ must divide $d\text{.}$ Hence, $d$ must be the unique greatest common divisor of $a$ and $b\text{.}$

Corollary2.11

Let $a$ and $b$ be two integers that are relatively prime. Then there exist integers $r$ and $s$ such that $ar + bs = 1\text{.}$

SubsectionThe Euclidean Algorithm

¶

Among other things, Theorem 2.10 allows us to compute the greatest common divisor of two integers.

Example2.12

Let us compute the greatest common divisor of $945$ and $2415\text{.}$ First observe that

$$\begin{align*} 2415 & = 945 \cdot 2 + 525\\ 945 & = 525 \cdot 1 + 420\\ 525 & = 420 \cdot 1 + 105\\ 420 & = 105 \cdot 4 + 0. \end{align*}$$

Reversing our steps, 105 divides 420, 105 divides 525, 105 divides 945, and 105 divides 2415. Hence, 105 divides both 945 and 2415. If $d$ were another common divisor of 945 and 2415, then $d$ would also have to divide 105. Therefore, $\gcd( 945, 2415 ) = 105\text{.}$

If we work backward through the above sequence of equations, we can also obtain numbers $r$ and $s$ such that $945 r + 2415 s = 105\text{.}$ Observe that

$$\begin{align*} 105 & = 525 + (-1) \cdot 420\\ & = 525 + (-1) \cdot [945 + (-1) \cdot 525]\\ & = 2 \cdot 525 + (-1) \cdot 945\\ & = 2 \cdot [2415 + (-2) \cdot 945] + (-1) \cdot 945\\ & = 2 \cdot 2415 + (-5) \cdot 945. \end{align*}$$

So $r = -5$ and $s= 2\text{.}$ Notice that $r$ and $s$ are not unique, since $r = 41$ and $s = -16$ would also work.

To compute $\gcd(a,b) = d\text{,}$ we are using repeated divisions to obtain a decreasing sequence of positive integers $r_1 \gt r_2 \gt \cdots \gt r_n = d\text{;}$ that is,

$$\begin{align*} b & = a q_1 + r_1\\ a & = r_1 q_2 + r_2\\ r_1 & = r_2 q_3 + r_3\\ & \vdots \\ r_{n - 2} & = r_{n - 1} q_{n} + r_{n}\\ r_{n - 1} & = r_n q_{n + 1}. \end{align*}$$

To find $r$ and $s$ such that $ar + bs = d\text{,}$ we begin with this last equation and substitute results obtained from the previous equations:

$$\begin{align*} d & = r_n\\ & = r_{n - 2} - r_{n - 1} q_n\\ & = r_{n - 2} - q_n( r_{n - 3} - q_{n - 1} r_{n - 2} )\\ & = -q_n r_{n - 3} + ( 1+ q_n q_{n-1} ) r_{n - 2} \\ & \vdots \\ & = ra + sb. \end{align*}$$

The algorithm that we have just used to find the greatest common divisor $d$ of two integers $a$ and $b$ and to write $d$ as the linear combination of $a$ and $b$ is known as the .

SubsectionPrime Numbers

¶

Let $p$ be an integer such that $p \gt 1\text{.}$ We say that $p$ is a , or simply $p$ is , if the only positive numbers that divide $p$ are 1 and $p$ itself. An integer $n \gt 1$ that is not prime is said to be .

Lemma2.13Euclid

Let $a$ and $b$ be integers and $p$ be a prime number. If $p \mid ab\text{,}$ then either $p \mid a$ or $p \mid b\text{.}$

Proof

Suppose that $p$ does not divide $a\text{.}$ We must show that $p \mid b\text{.}$ Since $\gcd( a, p ) = 1\text{,}$ there exist integers $r$ and $s$ such that $ar + ps = 1\text{.}$ So

$$\begin{equation*} b = b(ar + ps) = (ab)r + p(bs). \end{equation*}$$

Since $p$ divides both $ab$ and itself, $p$ must divide $b = (ab)r + p(bs)\text{.}$

Theorem2.14Euclid

There exist an infinite number of primes.

Proof

We will prove this theorem by contradiction. Suppose that there are only a finite number of primes, say $p_1, p_2, \ldots, p_n\text{.}$ Let $P = p_1 p_2 \cdots p_n + 1\text{.}$ Then $P$ must be divisible by some $p_i$ for $1 \leq i \leq n\text{.}$ In this case, $p_i$ must divide $P - p_1 p_2 \cdots p_n = 1\text{,}$ which is a contradiction. Hence, either $P$ is prime or there exists an additional prime number $p \neq p_i$ that divides $P\text{.}$

Theorem2.15Fundamental Theorem of Arithmetic

Let $n$ be an integer such that $n \gt 1\text{.}$ Then

$$\begin{equation*} n = p_1 p_2 \cdots p_k, \end{equation*}$$

where $p_1, \ldots, p_k$ are primes (not necessarily distinct). Furthermore, this factorization is unique; that is, if

$$\begin{equation*} n = q_1 q_2 \cdots q_l, \end{equation*}$$

then $k = l$ and the $q_i$'s are just the $p_i$'s rearranged.

Proof

Uniqueness. To show uniqueness we will use induction on $n\text{.}$ The theorem is certainly true for $n = 2$ since in this case $n$ is prime. Now assume that the result holds for all integers $m$ such that $1 \leq m \lt n\text{,}$ and

$$\begin{equation*} n = p_1 p_2 \cdots p_k = q_1 q_2 \cdots q_l, \end{equation*}$$

where $p_1 \leq p_2 \leq \cdots \leq p_k$ and $q_1 \leq q_2 \leq \cdots \leq q_l\text{.}$ By Lemma 2.13, $p_1 \mid q_i$ for some $i = 1, \ldots, l$ and $q_1 \mid p_j$ for some $j = 1, \ldots, k\text{.}$ Since all of the $p_i$'s and $q_i$'s are prime, $p_1 = q_i$ and $q_1 = p_j\text{.}$ Hence, $p_1 = q_1$ since $p_1 \leq p_j = q_1 \leq q_i = p_1\text{.}$ By the induction hypothesis,

$$\begin{equation*} n' = p_2 \cdots p_k = q_2 \cdots q_l \end{equation*}$$

has a unique factorization. Hence, $k = l$ and $q_i = p_i$ for $i = 1, \ldots, k\text{.}$

Existence. To show existence, suppose that there is some integer that cannot be written as the product of primes. Let $S$ be the set of all such numbers. By the Principle of Well-Ordering, $S$ has a smallest number, say $a\text{.}$ If the only positive factors of $a$ are $a$ and 1, then $a$ is prime, which is a contradiction. Hence, $a = a_1 a_2$ where $1 \lt a_1 \lt a$ and $1 \lt a_2 \lt a\text{.}$ Neither $a_1\in S$ nor $a_2 \in S\text{,}$ since $a$ is the smallest element in $S\text{.}$ So

$$\begin{align*} a_1 & = p_1 \cdots p_r\\ a_2 & = q_1 \cdots q_s. \end{align*}$$

Therefore,

$$\begin{equation*} a = a_1 a_2 = p_1 \cdots p_r q_1 \cdots q_s. \end{equation*}$$

So $a \notin S\text{,}$ which is a contradiction.

SubsectionHistorical Note

¶

Prime numbers were first studied by the ancient Greeks. Two important results from antiquity are Euclid's proof that an infinite number of primes exist and the Sieve of Eratosthenes, a method of computing all of the prime numbers less than a fixed positive integer $n\text{.}$ One problem in number theory is to find a function $f$ such that $f(n)$ is prime for each integer $n\text{.}$ Pierre Fermat (1601?–1665) conjectured that $2^{2^n} + 1$ was prime for all $n\text{,}$ but later it was shown by Leonhard Euler (1707–1783) that

$$\begin{equation*} 2^{2^5} + 1 = \text{4,294,967,297} \end{equation*}$$

is a composite number. One of the many unproven conjectures about prime numbers is Goldbach's Conjecture. In a letter to Euler in 1742, Christian Goldbach stated the conjecture that every even integer with the exception of 2 seemed to be the sum of two primes: $4 = 2 + 2\text{,}$ $6 = 3 + 3\text{,}$ $8 =3 + 5\text{,}$ $\ldots\text{.}$ Although the conjecture has been verified for the numbers up through $4 \times 10^{18}\text{,}$ it has yet to be proven in general. Since prime numbers play an important role in public key cryptography, there is currently a great deal of interest in determining whether or not a large number is prime.