We have the following proposition which we will take as is.

Proposition 1.6.1
  1. Addition on \(\mathbf{Z}\) is commutative and associative.
  2. 0 is an identity element for addition; that is, for all \(a \in \mathbf{Z}\), \(a + 0 = a\).
  3. Every element \(a\) of \(\mathbf{Z}\) has an additive inverse \(-a\), satisfying \(a + (-a) = a\). We write \(a - b\) for \(a + (-b)\).
  4. Multiplication on \(\mathbf{Z}\) is commutative and associative.
  5. 1 is an identity element for multiplication; that is, for all \(a \in \mathbf{Z}\), \(1a = a\).
  6. The distributive law holds; For all \(a, b, c \in \mathbf{Z}\), \(a(b+c) = ab + ac\).
  7. \(\mathbf{N}\) is closed under addition and multiplication. That is, the sum and product of positive integers is positive.
  8. The product of non-zero integers is non-zero. That is \(|ab| \geq \max\{|a|,|b|\} \) for non-zero integers \(a\) and \(b\).


Based on the proposition above, we can now conclude the following

Proposition 1.6.2
  1. If \(uv = 1\), then \(u = v = 1\) or \(u = v = =1\).
  2. If \(a|b\) and \(b|a\), then \(a = \pm b\).
  3. Divisibility is transitive: If \(a|b\) and \(b|c\), then \(a|c\).
  4. If \(a|b\) and \(a|c\), then \(a\) divides all integers that can be expressed in the form \(sb + tc\), where \(s\) and \(t\) are integers.


Proof (a)
We know \(a\) or \(b\) can’t be zero. By Proposition 1.6.1, we know that \(|ab| \geq \max\{|a|,|b|\}\). So

$$ \begin{align*} |uv| &\geq \max\{|u|,|v|\} \\ 1 &\geq \max\{|u|,|v|\} \\ \end{align*} $$

From this we see that \(u = v = 1\) or \(u = v = -1\).

Proof (b)
If \(a|b\), then \(b = ca\) for some integer \(c\). Similarly, if \(b|a\), then \(a = db\) for some integer \(d\). Therefore,

$$ \begin{align*} b &= ca \\ b &= cdb \\ 0 &= (cd - 1)b \end{align*} $$

Since the product of non-zero integers is non-zero, then either \(cd = 1\) or \(b=0\). If \(b = 0\), then \(a\) must be zero. If \(cd = 1\), then by (a), \(c = d = \pm 1\) and so \(c = d\).

Proof (c)
If \(a|b\), then \(b = ua\) for some integer \(u\). Similarly, if \(b|c\), then \(c = vb\) for some integer \(v\). Therefore,

$$ \begin{align*} c &= vb \\ c &= (uv)a \end{align*} $$

From this we see that \(a|c\).

Proof (d)
If \(a|b\), then \(b = ua\) for some integer \(u\). Similarly, if \(a|c\), then \(c = va\) for some integer \(v\). Now let \(s\) and \(t\) be integers. Observe that

$$ \begin{align*} sb + tc &= ua + va \\ &= (u+v)a \end{align*} $$

From this we see that \(a|(sb + tc) \ \blacksquare\).


References