2.6 More Decimal Expansions
We finish this chapter by using the Remainder Theorem to prove that every positive integer has a decimal expansion.
There is nothing special about the number \(10\) here, we could of course use any base \(b>1\).
Every positive integer \(n\) has a decimal expansion.
Note that the decimal expansion is unique. We leave this as an exercise.
Proof. We will prove this using strong induction on \(n\).
Base cases: If \(n<10\) then \(n\) is its own decimal expansion: \(n=d_0\cdot 10^0\) where \(d_0=n\) and \(0<d_0\leq 9\).
Induction step: We take \(n\geq 10\) and assume inductively that every integer from \(1\) to \(n-1\) has a decimal expansion.
By the Remainder Theorem there exist integers \(q,r\) such that \[n=10q+r\,\text{ and }\,0\leq r<10\] By 2.7 we can say \(r+1\leq 10\) so \(r\leq 9\).
As \(n\geq 10\) we must have \(q>0\) so \(q\geq 1\) by 2.6.
So \(10q=(9+1)q=9q+q\geq 9+q>q\) by (Sc),(Sh) and using \(0<1\).
As \(0\leq r\) we have \(10q\leq n\) and we deduce that \(q<n\).
Hence by the induction hypothesis \(q\) has a decimal expansion: \[q=d_k\cdot 10^k+\dots+d_1\cdot 10^1+d_0\cdot 10^0\] where each digit \(d_i\) is an integer such that \(0\leq d_i\leq 9\) and \(d_k\neq 0\).
By (Dist),(Comm): \[10q=d_k\cdot 10^{k+1}+\dots+d_1\cdot 10^2+d_0\cdot 10^1\] so \(n=d_k\cdot 10^{k+1}+\dots+d_1\cdot 10^2+d_0\cdot 10^1+r\).
Hence we have shown that there exists numbers \(d_0',\dots d_{k+1}'\) defined by \[d'_0=r,\quad d_{i+1}'=d_i\text{ for }i=0,\dots, k\] such that \[n=d_{k+1}'\cdot 10^{k+1}+\dots+d_2'\cdot 10^2+d_1'\cdot 10^1+d_0\cdot 10^0\] where each digit \(d_i'\) is an integer such that \(0\leq d_i'\leq 9\) and \(d_{k+1}'\neq 0\). (For \(i>0\) this holds because we know this for \(d_{i-1}\) while for \(d_0\) we showed that \(0\leq r\leq 9\).)
Hence we have shown that if all positive integers less than \(n\) have decimal expansions then \(n\) also has a decimal expansion.
By induction it follows that all positive integers have decimal expansions.