As we explore equations involving multiplication (e.g., $a⋅x=b$), we discover that not all equations have solutions in $ℤ$, leading us to the set of rational numbers $ℚ$:

Rational numbers are defined as ratios of integers, i.e., $\frac{b}{a}$​ where $b≠0$.

Why is 0⋅x=00⋅x=0 for all xx?
Can you find a consistent explanation that justifies the rule against division by zero?

𝑧=𝑥/𝑦
⇕
𝑥= 𝑦 × 𝑧

𝑥∈ℤ
𝑦∈ℕ⁺
⇒ 𝑧∈ℚ

Dividend ⎣ Divisor
 *        Quotient
  *
   Remainder

Dividend = divisor * quotient + remainder
a = bq + r

0 ≤ |Reminder| < |Divisor|

Proof:
 Let:
  a := Dividend: Integer
  b := Divisor: Integer 
    | b > 0

Consider:
  S: Set of all integers of the form 
     a - b𝑥
     | where 𝑥: Integer
     | a - bx ≥ 0

1: Show S is nonempty
   Find a value of x that satisfies a - bx ≥ 0
      x = |a|
         b ≥ 1
         b|a| ≥ |a|      --  *|a|
         b|a| ≥ -a       --  |a| ≥ -a
         a + b|a| ≥ 0
      Let x = -|a|
         a - bx = a - b(-|a|) = a + b|a| ≥ 0
   ∴ a - bx is in S when x = -|a|
   ∴ S ≠ ∅
   
2: Find q and r | a = bq + r and r ≥ 0     
   By the Well-Ordering Axiom, S contains a smallest element. Call it r
   Since r ∈ S
      r ≥ 0
      r = a - bx  for some x, Say x = q
         ∴ r = a-bq    and r ≥ 0 
           a = bq + r  and r ≥ 0

3: Show that r < b
   Proof by contradiction: r ≥ b
      r - b < r
      a - b(q+1) = r - b < r    -- a-b(q+1)  is an element of S
      a - b(q+1) < r            -- The element cannot be smaller than the smallest element r
   ∴ ¬{r ≥ b }
   ∴ r < b

4: Show that r and q are the only numbers with these properties (unique).
   Suppose q' and r' : a = bq' and 0 ≤ r < b
   Prove that q' = q and r' = r
      Since a = bq + r and a = bq' + r'
         bq + r = bq' + r'
         b (q - q') = r'-r   
         0 ≤ r < b and 0 ≤ r' < b  -- From 3.
         -b < -r ≤ 0     and 0 ≤ r' < b 
         -b < r' - r < b  -- Adding both inequalities
         - b < b(q - q') < b
         -1 < (q - q') < 1     -- :b 
         As q and q' are integers:
           q - q' = 0
              ∴ q = q'
           (r' - r):b = 0
              r' - r = 0
              ∴ r' = r
∎

The set of rational numbers $( \mathbb{Q} )$ is even less likely to be well-ordered than $( \mathbb{Z} )$.

1. No Least Element in Infinitesimal Neighborhoods:

   • Rational numbers are dense on the number line, meaning between any two rational numbers, there exists another rational number.
   • Take any non-empty subset of $( \mathbb{Q} )$ with values above $0$ but arbitrarily close to it, such as $( { x \in \mathbb{Q} \mid 0 < x < 1 } )$.
   • No matter how small a positive rational number we choose, we can always find a smaller positive rational number closer to zero. Thus, this subset does not contain a least element.

2. Implication of Density:

   • The density of $( \mathbb{Q} )$ implies that for any rational number chosen as the "smallest" in a subset, we can always find another rational number between it and zero, showing that $( \mathbb{Q} )$ cannot satisfy the well-ordering property.

The subset $( { x \in \mathbb{Q} \mid 0 \leq x \leq 1 } )$ is also not well-ordered for similar reasons.

1. No Smallest Positive Element:

   • Consider the subset $( (0, 1] \subset \mathbb{Q} )$. This subset does not have a least positive element because we can always find a smaller positive rational number within this interval.
   • For instance, if we take a rational number $( x )$ very close to zero (e.g., $( x = \frac{1}{1000} )$), we can always find another rational number closer to zero, such as $( \frac{1}{1001} )$, meaning there is no well-defined least element in this subset.

2. Contrast with Bounded Intervals in $( \mathbb{N} )$:

   • In $( \mathbb{N} )$, bounded intervals like $( {1, 2, 3, \dots, n})$ have a well-defined smallest element (e.g., $( 1 )$ in this case). However, because rational numbers are dense, this does not apply to bounded subsets in $( \mathbb{Q} )$.

In summary:

• $( \mathbb{Z} )$ and $( \mathbb{Q} )$ lack the well-ordering property because they do not have a smallest element in certain infinite subsets.
• Even subsets of $( \mathbb{Q} )$ (like $( { x \in \mathbb{Q} \mid 0 \leq x \leq 1 } )$) are not well-ordered due to the density of rational numbers, which prevents the existence of a minimum in intervals approaching zero.