# Lecture 29: Euclidean division

• 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 $$q_1 = q_2$$.