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Lecture 29: Euclidean division

• Reading: MCS Chapter 9.1

• We spent most of lecture talking about Turing machines; these notes have been added to the lecture 28 notes

• We proved the Euclidean division algorithm/theorem

• Review exercises
  • state and prove the euclidean division algorithm.
  • "execute" the algorithm contained in the proof for a few steps to see how it works
  • this is a different algorithm than you normally use for division with remainder; try to encode your algorithm for division with remainder as an inductive proof.
  • Prove that if $a|b$ and $a|c$ then for any $s$ and $t$, $a|sb+tc$.

Terminology/Definitions

• $\mathbb{Z}$ is the set of integers $\{\dots, -2, -1, 0, 1, 2, \dots\}$

• If $a,b \in \mathbb{Z}$ then $a$ divides $b$ (written $a | b$) if there exists some $k \in \mathbb{Z}$ such that $b = ak$. In this case, $b$ is a multiple of $a$.

• Unless otherwise specified, for the lectures on number theory, variables are taken from $\mathbb{Z}$. I will tend to use $a$, $b$, $p$, $q$, $r$ to denote integers.

• Don't use fractions for a few weeks: While studying number theory, we will be working exclusively with integers. The division operation doesn't make sense in general for integers; writing fractions in your equations will only lead to confusion.

Euclidean division

Claim (Euclidean division algorithm): For any $a$ and $b > 0$ there exist $q$ and $r$ such that $a = qb+r$ and $0 \leq r \lt b$. Moreover, $q$ and $r$ are unique: if $a = qb + r = q'b + r'$ then $q = q'$ and $r = r'$.

Notation: - $q$ is called the quotient of $a$ by $b$, and is written $quot(a,b)$. Some programming languages write $a/b$ to refer to $q$. Do not do this. $a/b$ should have the property that $b \cdot (a/b) = a$; the quotient does not have this property.

• $r$ is called the remainder of $a$ by $b$, and is written $rem(a,b)$. Some languages write $a \% b$ to refer to $r$, but many languages give the wrong answer when $a$ is negative. Some books write $a~mod~b$, but this notation will lead to confusion when we start talking about modular arithmetic, so it should be avoided.

• Note that the uniqueness criteria in the claim is another way of saying that $quot$ and $rem$ are well-defined functions from $\mathbb{Z} \times \mathbb{Z}$ to $\mathbb{Z}$.

Proof of existence: By induction on $a$. Fix $b$ and let $P(a)$ be the statement "$∃ q, r \in \mathbb{Z}$ such that $a = qb + r$." We must prove $P(0)$ and $P(a+1)$ assuming $P(a)$.

For $P(0)$, let $q = r = 0$. Since $b > 0$, $0 \leq r \lt b$. Moreover, $0 = qb + r$ as required.

To prove $P(a+1)$, assume $P(a)$. Then there exist some $q'$ and $r' \lt b$ with $a = q'b + r'$. Since $r' \lt b$, either $r' = b-1$ or $r' \lt b-1$. We want to show that in either case, there exist $q$ and $r$ with $a + 1 = qb + r$.

In the former case, we have $a = q'b + r' = q'b + b-1$. Therefore $a + 1 = q'b + b = (q'+1)b + 0$. Let $q = q'+1$ and $r = 0$, and we see that $a + 1 = qb + r$ as required.

In the latter case, we can let $q = q'$ and $r = r'+1$. Since $0 \leq r' \lt b-1$, we see that $0 \leq r \lt b$. Moreover, since $a = q'b + r'$, we see that $a + 1 = qb + r$ as required.

Proof of uniqueness: Assume there are two pairs of numbers $q_1, r_1$ and $q_2, r_2$ with $a = q_1b + r_1 = q_2b + r_2$ and with $0 \leq r_i \lt b$. We want to show that $q_1 = q_2$ and $r_1 = r_2$.

Rearranging $q_1b + r_1 = q_2b + r_2$, we see that $(q_1 - q_2)b = r_2 - r_1$. I claim that $-b \lt r_2 - r_1 \lt b$, and the only multiple of $b$ between $-b$ and $b$ is $0$, so $r_2 - r_1 = 0$.

To see this more carefully, we know $0 \leq r_2 \lt b$. The same equation holds for $r_1$; negating it yields $-b \lt -r_1 \leq 0$. Adding these equations together gives $-b \lt r_2 - r_1 \lt b$.

Therefore $r_2 - r_1 = 0$, so $r_2 = r_1$. Since $(q_1 - q_2)b = r_2 - r_1 = 0$ and since $b \neq 0$, we have $q_1 - q_2 = 0$, or $q_1 = q_2$.