Lecture 4

Well-Ordering Principle

An element of a subset $s\in S\subseteq \mathbb{N}=({\mathbb{Z}}^{+}\cup \{0\})$ is the first element of $S$ if $s\le x$ for all $x\in S$. For example, $\{11,17,4,12,100\}$ has the first element $4$.

The well-ordering principle is an axiom of mathematics (statement assumed to be true) that all non-empty subsets of non-negative integers contain a first element.

Division and Remainder

Let $n$ and $d$ be integers with $d>0$. Then, there exists integers $q$ and $r$ ($0\le r\le d$) such that $n=qd+r$. For example, if $n=107,d=10$, we would obtain $q=10,r=7$ since $107=10(10)+7$.

To prove this, we define a set $M=\{n-qd:q\in \mathbb{Z}\}$. Clearly, $M\cap \mathbb{N}$ is a non-empty subset of $M$ since we can select a $q$ sufficiently large (positively/negatively) to ensure that there exist positive integers in the set $M$. By the well-ordering principle, there must be a first element which we will define as $r$.

We know $r$ exists and that it belongs to the set $M\cap \mathbb{N}$, so it therefore takes some non-negative integer value. Since we have defined this as the first element, we can also say that $n=qd+r$.

For contradiction, we now assume that $r\ge d$. This then gives that $r-d\ge 0$, and that $r-d=n-(q+1)d$. $(r-d)\in M\cap \mathbb{N}$ contradicts that $r$ is the first element, so we get $r\in \{0,1,2,3,...,d-1\}$.

Next, we prove that the decomposition of $n=qd+r$ is unique (there can only be one $r$ and $q$ for any given $n$ and $d$). For contradiction, assume that this is not the case (they are not unique; $n=q_{1}d+r_1=q_{2}d+r_2$ for $q_1\ne q_2$).

$(r_1-r_2)=d(q_1-q_2)$ so $d$ divides $(r_1-r_2)$. Since both remainders are between $0$ and $d-1$, their difference must also be between $0$ and $d-1$, their difference must also be between $0$ and $d-1$. The only value here divisible by $d$ is $0$, and hence this contradicts the original assumption, so $r_1=r_2$ and $q_1-q_2$.

Mathematical Induction

Used to prove a statement of the form $P(n)$ is true for all natural numbers $n$. It proves by showing two things:

• Establishing $P(n)\to P(n+1)$
• Establishing an initial term $n$, for example $P(0)$ or $P(1)$

Consider the partial sum of the geometric series $S(n)=A+Ar+A{r}^2+A{r}^3+A{r}^4+...+A{r}^n$ for $r\ne 1$. We need to prove that: $S(n)=A[\frac{1-r^{n+1}}{1-r}]$

Trivially, $S(0)$ is true. Assume true for $S(n)$.

We now need to prove that $S(n+1)=S(n)+A{r}^{n+1}$.

\begin{eqnarray} S(n+1)=S(n)+Ar^{n+1} &=& A[\frac{1-r^{n+1}}{1-r}]+Ar^{n+1}\\ &=& A[\frac{1-r^{n+1}}{1-r}+r^{n+1}]\\ &=& A[\frac{1-r^{n+1}}{1-r}+\frac{r^{n+1}-r^{n-2}}{1-r}]\\ &=& A[\frac{1-r^{n+2}}{1-r}] \end{eqnarray}

Another Example

We’re trying to prove that the sum of the the first $n$ positive odd numbers is equal to $n^2$. ${\sum }_{k=1}^n(2k-1)=n^2$ This is trivially true for $n=1$ ($2\times 1-1=1^2$), so we assume true for $n=a$: ${\sum }_{k=1}^a(2k-1)=a^2$ We are then going to prove $n=a+1$: ${\sum }_{k=1}^{a+1}(2k-1)=(a+1{)}^2$ Which goes a little like:

\begin{eqnarray} \sum_{k=1}^{a+1}(2k-1) &=& [\sum_{k=1}^{a}(2k-1)] \ +2(a+1)-1 \\ &=& a^2 +2a +1 \\ &=& (a+1)^2 \\ && \therefore \text{proven true for all }n+1 \end{eqnarray}

Power Set

Correctness

We say a procedure is correct if it:

• Stops after a finite number of steps, and
• Produces an output which is what is claimed

Loop Invariance

A loop invariant is a property of a program loop that is true before and after each iteration of the loop. For example, consider the simple pseudocode:

Input integers m and n
While m + n < 996
	Replace m := m + 5
	Replace n := n - 1
Output m and n

The eventual output for $m=1,n=1$ will be $m=1246,n=-248$. There are several loop invariants, such as $m+n<1000$, and $m>0$. We can prove that $2\mid (m+n)$ is loop invariant. $2$ divides into $(m+n)$, for example $m+n=2k$, and $m+n\ge 996$ (the way that the loop is closed). Either $m$ and $n$ are unchanged (the loop never occurs), or $m+n$ is updated to $(m+5)+(n-1)$…

\begin{eqnarray} (m+5)+(n-1) &=& 2k+4 \\ &=& 2(k+2) \end{eqnarray}

…which is divisible by 2.

Euclidean Algorithm

Keep dividing the remainder of $n$ and $m$ (where $n>m$) until they cleanly divide.

$n$	$m$	$n=q_{i}m+r_i$	$i$
$527$	$341$	$527=1(341)+186$	$1$
$341$	$186$	$341=1(186)+155$	$2$
$186$	$155$	$186=1(155)+31$	$3$
$155$	$31$	$155=5(31)+0$	$4$

”Motivating” (?) Problem (Extended / Reverse Euclidean Algorithm)

Consider the problem: finding a way of expressing an integer as a linear sum of (integer) multiples of other integers. For example, how would we find $a$ and $b$ such that $11a+17b=1$? In this case, $11(-3)+17(2)=1$ however guessing solutions is not trivial, nor always possible. Euclidean’s algorithms calculates the greatest common divisor of any two non-zero integers, however we can reverse the process to find integers $a$ and $b$ such that $am+bn=\text{gcd}(m,n)$ where $m$ and $n$ are two known input integers. If the “target” value $t$ cannot be found such that $am+bn=t$, then no integer solutions can be found.

$n$	$m$	$n=q_{i}m+r_i$	$i$
$17$	$11$	$17=1(11)+6$	$1$
$11$	$6$	$11=1(6)+5$	$2$
$6$	$5$	$6=1(5)+1$	$3$
$5$	$1$	$5=5(1)+0$	$4$
We take the second last row of the table (where the remainder is not $0$). We then represent rearrange so that the subject is the $gcd(11,17)=1$, so that $1=6-1(5)$. Whilst keeping the subject as the gcd, we can substitute values without expanding for the $q_{i}m$ portion once keeping the remainder / $r_i$ as the subject (for example, $11=1(6)+5⟹5=11-1(6)⟹1=6-1(11-1(6))$ which just needs rearranging, and then the same can be done for $6$ until the equation is in the form of $am+bn=gcd(m,n)$).