2.3 Order properties of \(\mathbb Z\)
When we say that a number \(n\) is positive we mean that it satisfies the strict inequality \(0<n\); in other words \(0\leq n\) and \(0\neq n\).
The set \(\mathbb N\) of natural numbers is the set of all integers \(n\) such that \(0<n\).
The number \(0\) is in \(\mathbb N\), that is \(0<1\).
We will use proof by contradiction. Recall that we showed that \(\lnot (0<1)\) is \(1\leq 0\).
Proof. Suppose \(1\leq 0\). Then \(1+(-1)\leq 0+(-1)\) by (Sc).
So \(0\leq -1\) by (Neg),(Id).
As \(1\leq 0\) and \(0\leq -1\) we have \[1\cdot(-1)\leq 0\cdot(-1)\] by (Sc). Hence \(-1\leq 0\) by (Id) and Lemma 2.3.
But we also showed \(0\leq -1\) so \(0=-1\) by (ASy).
Hence \(1=0+1=-1+1=0\) by (Id),(Neg).
This contradicts \(0\neq 1\) which is true by (Id).
We deduce that \(1\leq 0\) is false, so \(0< 1\) is true.
We will show now show that \(1\) is the least positive integer.
There must be a least positive integer by the axiom (WO). Recall this axiom says:
The well ordering principle – any non-empty subset of non-negative integers has a unique least element.
\(1\) is the least element of the set \(\mathbb N\) of positive integers.
Proof. Let \(n\) be the least element of \(\mathbb N\), which exists by \((WO)\).
Our strategy is to show that \(n\) cannot be greater than \(1\) or less than \(1\). By (Comp) we know that \(1\leq n\) or \(n\leq 1\). We will show that in each case we deduce that \(n=1\).
Suppose \(1\leq n\). Since \(1\in \mathbb N\) and \(n\) is the least element of \(\mathbb N\) we have \(1=n\) as required.
Suppose \(n\leq 1\). As \(n\in\mathbb N\) we have \(0\leq n\) so by (Sc) \[n\cdot n \leq 1\cdot n = n.\] Now \(0\leq n\cdot n\) by (Sc) since \(0\leq n\).
If \(0=n\cdot n\) then \(n=0\) or \(n=0\) by (ZD) so \(n=0\) (this is the idempotent rule from logic). But \(n\) is in \(\mathbb N\) so \(n\neq 0\). This is a contradiction so we deduce that \(0<n\cdot n\).
Thus \(n\cdot n\in \mathbb N\) and \(n\cdot n\leq n\) so as \(n\) is the least element of \(\mathbb N\) we have \(n\cdot n=n\).
Rearranging this \(n\cdot n=n=1\cdot n\) by (Id) so \[(n-1)\cdot n = n\cdot n+(-1)\cdot n = 1\cdot n+(-1)\cdot n = 0\cdot n\] by (Dist) twice. This is zero by Lemma 2.3 so by (ZD) either \(n-1=0\) or \(n=0\). As before we can rule out the case \(n=0\) so \(n-1=0\), and rearranging \[n=(n-1)+1=0+1=1.\] using Lemma 2.4, (Id).
So we have proved that if \(1\leq n\) then \(n=1\) and if \(n\leq 1\) then \(n=1\). Hence we deduce that the least element \(n\) of \(\mathbb N\) is \(1\).
\[\forall m,n\in \mathbb Z,\; m<n\iff m+1\leq n\]
Proof. Suppose that \(m<n\). Then \(m\leq n\) so \(0=m-m\leq n-m\) by (Neg),(Sh), and if \(n-m=0\) then \(n=(n-m)+m=0+m=m\), by Lemma 2.4, (Id).
So \(n-m\) is strictly positive, therefore by the theorem \(1\leq n-m\). Hence \[1+m\leq (n-m)+m=n\] by (Sh) and Lemma 2.4. Hence \(m+1\leq n\) be (Comm).
Conversely suppose \(m+1\leq n\). Reversing the above argument: \[1=(1+m)-m\leq n-m.\] by Lemma 2.4, (Sh).
As \(0\leq 1\) and \(1\leq n-m\) we have \(0\leq n-m\) by (Tr). Thus \(m=0+m\leq n-m+m=n\) by (Id),(Sh) and our favourite lemma.
If \(m=n\) then \(m+1\leq m\) which implies \(1\leq 0\) by (Sh). This contracts \(0<1\), so \(m\neq n\). Hence the inequality is strict \(m<n\).