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I have come across a problem in an elementary mathematics book that I am seeking help with.

The problem states :

"Given $m,n \in \mathbb{N}$, with $m > 1$, prove that there exist unique nonnegative integers $k,a_{0},a_{1},...,a_{k}$ s.t. $0 \leq a_{i} < m$ for $0 \leq i \leq k$, $a_{k} \neq 0$ and : \begin{equation} n = a_{k} m^{k} + a_{k-1}m^{k-1} + \dots + a_{2}m^{2} + a_{1}m + a_{0} \end{equation} In this case, we write $n = (a_{k}a_{k-1}...a_{1}a_{0})_{m}$ and say that the right hand side above is the representation of $n$ in base $m$ or the m-adic representation of $n$."

Can anyone help with an explained solution to this ? This is on a chapter on mathematical induction.

scipio
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  • @BillDubuque You're welcome. I've noticed there's quite a few questions that you've closed as being duplicate, in many cases with more than one listed post (such as with this question), so thank you for taking the time & effort do that. – John Omielan Oct 27 '20 at 23:42
  • @John Thanks. If only others helped more in such endeavors. The hope is that doing so will lead to iterative refinements of prior answers (and eventually possibly to all the "proofs from the book"), and that they'd be much easier to locate in searches without being buried like needles in a haystack among poor rushed FGITW answers by users more interested more in rep and badges than in pedagogy. – Bill Dubuque Oct 28 '20 at 00:55
  • @John Also there would be far less bias in dupe selection if more users helped out, e.g. I have the time to quickly find dupes only because I use keywords etc that help me quickly locate them (else the search may be impractical given the weak search tools). But this has the disadvantage that it only locates threads in which I have answered, which, alas, introduces bias. If more users assist in such endeavors then we will have more diverse dupe targets (and probably better targets since the additional cumulative search time will likely locate better targets). – Bill Dubuque Oct 28 '20 at 01:02

1 Answers1

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SKETCH: First prove existence, then worry about uniqueness. Clearly for $n=1$ we may take $k=0$ and $a_0=1$. Now let $n>1$, and suppose as your induction hypothesis that every $\ell\in\Bbb N$ with $\ell<n$ has an $m$-adic representation. There is a largest integer $k$ such that $m^k\le n$.

  • What is an $m$-adic representation of $n$ if $n=m^k$?

If $n>m^k$, there is a largest integer $a$ such that $am^k\le n$.

  • Show that $1\le a<m$.
  • If $n=am^k$, what is an $m$-adic representation of $n$?
  • If $n>am^k$, let $r=n-am^k$, apply the induction hypothesis to $r$, and get an $m$-adic representation of $n$.

For uniqueness, suppose that $n$ is the smallest positive integer with two $m$-adic representation, say

$$n=a_km^k+\ldots+a_1m+a_0=b_\ell m^\ell+\ldots+b_1m+b_0\,.\tag{1}$$

Show that $a_km^k+\ldots+a_1m+a_0<m^{k+1}$, and use this to show that $k=\ell$, so that we can rewrite $(1)$ as

$$n=a_km^k+\ldots+a_1m+a_0=b_km^k+\ldots+b_1m+b_0\,.$$

We may assume that $a_k\ge b_k$. Then

$$(a_k-b_k)m^k+a_{k-1}b^{k-1}+\ldots+a_0=b_{k-1}m^{k-1}+\ldots+b_0\,.$$

  • This is impossible if $a_k-b_k>0$; why?

Thus, $a_k=b_k$, and

$$n-a_km^k=a_{k-1}b^{k-1}+\ldots+a_0=b_{k-1}m^{k-1}+\ldots+b_0\,.$$

Finally, $n-a_km^k<n$, so $n-a_km^k$ has a unique $m$-adic representation, so $a_i=b_i$ for $i=0,\ldots,k-1$, and the two representations of $n$ in $(1)$ are therefore identical, contrary to assumption. This contradiction shows that no such $n$ exists: every positive integer has a unique $m$-adic representation.

Brian M. Scott
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