In this section we present two properties that are equivalent to induction, namely, the well-ordering principle, and strong induction.
Suppose \(S\) is a subset of the natural numbers with the property:
\begin{equation*} (\forall n \in \mathbb{N}) \big( \{ k \in \mathbb{N} \mid k \lt n \} \subseteq S \implies n \in S \big). \end{equation*}Then \(S = \mathbb{N}\text{.}\)
We prove by induction that for every \(n \in \mathbb{N}\text{,}\) \(\{1, \ldots, n\}\)
Notice that by taking \(n=1\text{,}\) we see that \(\{ k \in \mathbb{N} \mid k \lt 1 \} = \emptyset\) which is clearly a subset of \(S\text{.}\) Therefore, by the property of \(S\text{,}\) we find that \(1 \in S\text{,}\) so \(\{1\} \subseteq \mathbb{N}\text{.}\)
Let \(n \in \mathbb{N}\) be arbitrary and suppose that \(\{1,\ldots,n\} \subseteq S\text{.}\) This is the same as the set \(\{ k \in \mathbb{N} \mid k \lt n+1 \}\text{,}\) so by the property of \(S\text{,}\) \(n+1 \in S\text{.}\) Therefore, \(\{1, \ldots, n+1\} \subseteq S\text{.}\)
By induction, for every \(n \in \mathbb{N}\text{,}\) \(\{1,\ldots,n\} \subseteq S\text{.}\) Hence
\begin{equation*} \mathbb{N} = \bigcup_{n \in \mathbb{N}} \{1,\ldots,n\} \subseteq S. \end{equation*}We already knew \(S \subseteq \mathbb{N}\text{,}\) so they must in fact be equal.
Every nonempty subset of the natural numbers has a least element.
Let \(S\) be a subset of natural numbers with no least element. Note that \(1 \notin S\) (i.e., \(1 \in S^c\)) since \(1\) is the smallest natural number. Now let \(n \in \mathbb{N}\) be arbitrary and suppose \(\{1, \ldots, n\} \subseteq S^c\text{.}\) Therefore \(S \subseteq \{n+1, n+2, \ldots \} = \{ k \in \mathbb{N} \mid k \ge n+1 \}\text{.}\) Thus \(n+1 \notin S\) because otherwise it would be a least element. Hence \(\{0, \ldots, n+1\} \subseteq S^c\text{.}\) By Strong Induction, \(S^c = \mathbb{N}\) and hence \(S = \emptyset.\) By contraposition, the desired result follows.
The well-ordering principle implies the principle of mathematical induction.
Suppose \(\mathbb{N}\) has the well-ordering principle. Let \(S \subseteq \mathbb{N}\) be any subset with the property that \(1 \in S\) and for every \(n \in \mathbb{N}\text{,}\) \(n \in S\) implies \(n+1 \in S\text{.}\) We wish to prove that \(S = \mathbb{N}\text{.}\)
Suppose to the contrary that \(S \not= \mathbb{N}\text{.}\) Then \(S^c \not= \emptyset\text{,}\) and so by the well-ordering principle has a least element \(k \in S^c\text{.}\) Since \(1 \in S\text{,}\) \(k \not= 1\text{,}\) so \(k-1 \in \mathbb{N}\text{.}\) Moreover, we must have \(k-1 \in S\) since \(k\) is the minimal element of \(S^c\text{.}\) By the property assumed by \(S\) for the value \(n = k-1\text{,}\) we find \(k \in S\) which is a contradiction.
Therefore, our assumption that \(S \not= \mathbb{N}\) was false, and hence \(S = \mathbb{N}\text{.}\) In other words, \(\mathbb{N}\) has the principle of mathematical induction.
We now recall the division algorithm, but we can provide a proof this time.
For any integers \(a,b\) with \(a \not= 0\text{,}\) there exists unique integers \(q\) and \(r\) for which
\begin{equation*} b = aq + r, \quad 0 \le r \lt \abs{a}. \end{equation*}The intger \(b\) is called the dividend, \(a\) the divisor, \(q\) the quotient, and \(r\) the remainder.
Let \(a,b \in \mathbb{Z}\) with \(a\) nonzero. For simplicity, we will assume that \(a \gt 0\) because the proof when \(a \lt 0\) is similar. Consider the set of integers \(A = \{ b-ak \mid k \in \mathbb{Z}, b-ak \ge 0\text{.}\) This set is clearly nonempty, for if \(b \ge 0\) then \(b-a0 = b \ge 0\) is in \(A\text{,}\) and if \(b \lt 0\) then \(b-ab = b(1-a) \ge 0\) is in \(A\text{.}\)
By the Well-Ordering Principle, \(A\) has a minimum element, which we call \(r\text{,}\) and some integer which we call \(q\) so that \(r = b-aq\text{.}\) Then \(r \ge 0\) and notice that \(r-a = b - aq - a = b-a(q+1)\text{.}\) Since \(a \gt 0\text{,}\) \(r-a \lt r.\) By the minimality of \(r\text{,}\) we know \(r-a \lt 0\) or equivlaently, \(r \lt a\text{.}\)
Now suppose there are some other integers \(q',r'\) with \(0 \le r' \lt a\) and \(b = aq'+r'\text{.}\) Then \(aq+r = aq' + r'\) and hence \(r-r' = aq - aq' = a(q-q')\text{.}\) Now \(-a \lt -r' \le r-r' \le r \lt a\) and hence \(a\abs{q-q'} = \abs{a(q-q')} = \abs{r-r'} \lt a\text{.}\) Dividing through by \(a\text{,}\) we obtain \(\abs{q-q'} \lt 1\) and since \(q-q' \in \mathbb{Z}\) it must be zero. Hence \(q=q'\) and so also \(r = r'\text{.}\)
If \(a,b \in \mathbb{Z}\) are nonzero and relatively prime and if \(a,b\) divide \(c\text{,}\) then \(ab\) divides \(c\text{.}\)
Let \(a,b\) be nonzero relatively prime integers which divide some integer \(c\text{.}\) Since \(a,b\) divides \(c\) there exist integers \(r,s\) for which \(c = ar = bs\text{.}\) Since \(\gcd(a,b) = 1\text{,}\) by Theorem 1.7.5 there exist integers, \(x,y\) for which \(ax+by = 1\text{.}\) Multipliying by \(c\) and using the division properties above we find
\begin{equation*} c = acx + bcy = absx + abry = ab(sx+ry), \end{equation*}so \(ab\) divides \(c\text{.}\)