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 $b>1$.

Theorem 2.15

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.