- $a = q_1d + r_1$
- $b = q_2d + r_2$
- $r_1 = a - q_1d = a - q_1(as + bt) = a(1 - q_1s) + b(-q_1t)$ which must equal $0$ since $r_1 < d$ but $d$ is the least element in $S$. The same argument applies to $r_2$ so that $r_2 = 0$
Theorem 2: Euclid's Division Lemma
Given two integers $a$ and $b$ with $b \ne 0$, there exist unique integers $q$ and $r$ such that:
Note: Where $|b|$ is the absolute value of $b$.
Proof:
(1) Case 1: $a \ge 0$
(2) Let $S$ be the set of all integers $a - bk$ where $k$ is an integer and $a - bk \ge 0$
(3) $S$ is non-empty since either $a = a - b\times0 \ge 0$ which is in $S$
(4) By the Well-Ordering Principle, there exists an integer $q$ such that $a - bq$ is the least element in $S$
(5) Let $r = a - bq$. $0 \le r$ by definition of $S$. Further $r < b$ since if $r \ge b$, then it is not the least element since $r - b = a - bq - b = a - b(q+1) \ge 0$ would be less than $r$ and an element in $S$.
(6) $r,q$ must be unique. Assume that there exists $r', q'$ such that $a = qb + r = q'b + r'$ and $|r - r'| > 0$ and $|q - q'| > 0$. Then, $b > |r - r'| = b|q - q'| > b$ which is impossible.
(7) Case 2: $a < 0$ In this case, let $a' = -a$. It now follows that $a' \ge 0$ and there exists unique $r,q$. It now follows that $a = -(bq +r) = b(-q) -r$ If $r > 0$, then $a = b(-q+1) + (b-r)$ with $0 < b-r < b$
Theorem 1: Well-Ordering Princple
Every non-empty set of non-negative integers contains a least element
Proof:
(1) Assume that there exists a non-empty set $S$ of positive integers that does not contain a least element.
(2) Let $P(n)$ be true if $n$ is an element of $S$
(3) $P(0)$ is false. If true, then $0$ would be the least element for $S$
(4) Assume that for all $i \le n, P(i)$ is false.
(5) It follows that $P(n+1)$ is false. If not, then $n+1$ would be the least element for $S$.
(6) But, then by the Principle of Mathematical Induction, there is no positive integer in $S$
(7) This contradicts our assumption in (1) that $S$ is non-empty. Therefore, we can reject our assumption in step (1).
Note: This is an example of a proof by contradiction.
Axiom 1: Principle of Mathematical Induction
A proof by induction consists of 2 steps:
(1) Base Case: Proves that the statement is true for one case independent of all other cases.
(2) Inductive Case: Proves that if the statement holds for any case $n=k$, then it must also hold for $n=k+1$
The principle can be understood with the analogy of a ladder. if we can climb on the bottom rung of a ladder and for each rung, we can climb to the next, then it follows that we can climb as high as we like up to the top of the ladder.
This principle is very important in the establishment of advanced mathematical arguments.