Problem in proof concerning division algorithm.

Question

This is taken from the book titled: Numbers, Sequences and Series, by Keith Hirst.
I am unable to understand the logic given in Prop. 3 in sec. 2.4 as shown below (with $t$ replaced by $q$)

Let $T$ denote the set of numbers of the form $a - bq$, where $q\in \mathbb{I}$. If $a\ge 0$, then $q = 0$ gives $a - bq = a\ge 0$. If $a \lt 0$ then $q = a$ gives $a - bq = a(1 - b) =0$.

I am unable to understand the following, given $a$ is dividend, $b$ is a positive integer serving as divisor, $q(t)$ is quotient:

(i) Why $q=0$ for the $a\ge 0$ case, which means $b\gt a$.
(ii) Why taken the value $q=a$ for the $a\lt 0$ case.
What is the significance here of taking $b=1$? I mean if $a=-2, b=1$, then only $q= a$.

Answer 1

• Why do you replace $q$ by $t$? I am going to assume you have a good reason and follow your convention.

• No one says anything about $q$ is the quotient at the first part. The goal is to be able to pick a $q$ to illustrate that $T=\{a-bq: q \in \mathbb{Z}\}$ contains a nonnegative element. $q$ is any integer that we pick to illustrate that.

• We want to show that $T$ contains a nonnegative element in order to use well -ordering principle.

• $q$ is chosen to be $0$ if $a \ge 0$ because that choice allows us to show that $a \in T$ and $a \ge 0$.

• Let $a=5$, $b=1$, why would picking $q=0$ lead you conclude that $b > a$?

• If $a<0$, picking $q=a$ would allow us to show that $a-bq=a(1-b) \ge 0$ since $a<0$ and $1-b \le 0$.

• You are given $a$ and $b$. You do not get to choose $b$.

• If $a=-2, b=1$. then $-2-(1)(-2)=0 \in T$

Answer 2

The aim is to show that $T$ contains non-negative numbers of the form $a-bt$ where $a,b$ are given with $b>0$ and $t$ varies of the integers.

The author considers two cases:

1. Let $a\geq 0$. Then take $t=0$ and so $a-bt = a\geq 0$.

2. Let $a<0$. Then take $t=a$ and so $a-tb = a(1-b)$. But $b>0$ is an integer and so $1-b\leq 0$. This gives $a(1-b)\geq 0$.