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I was going over the proof of the following claim:

Let $b$ be a positive integer greater than one. Then every positive integer $n$ can be written uniquely as

$$n=\sum_{n=0}^k a_{k-n} b^{k-n} $$

where $k $ is a nonnegative integer, $a_j $ is an integer with $0\leq a_j \leq b-1$ for $j=0,1,\dotsc k $, and $a_k\neq 0$.

The proof begins with

$$n=bq_0 + a_0$$

And then successively divides each quotient by $b $, assuming that the quotients do not equal zero, to produce

$$q_0 = bq_1 + a_1$$ $$q_1 = bq_2 + a_2$$ $$q_2 = bq_3 + a_3$$ $$\vdots $$

Then we observe that the sequence of quotients must satisfy

$$n > q_1 > q_2 > \dotsc \geq 0$$

Then this is what the text says: "Because the sequence $q_0$, $q_1$, $q_2$, $\dotsc $ is a decreasing sequence of nonnegative integers that continues as long as it's terms are positive, there are at most $q_0$ terms in this sequence, and the last term equals $0$."

I have some questions regarding this quote:

  1. How do we know that there are at most $q_0$ terms?

  2. Why can't there be more?

  3. What contributes to there being less than $q_0$ terms?

  4. Given that there are at most $q_0$ terms in the decreasing sequence of nonnegative integers $q_0$, $q_1$, $q_2$, $\dotsc $, how are we then able to conclude that 0 is the last quotient term in the sequence? I grant that the penultimate quotient will be positive, but based on the nature of the sequence, how can we conclude that the next, the final, quotient is 0?

I've been thinking about answers to these questions all day, but I only have probable answers and not answers deduced from the facts. Please help. Thanks.

J. Dunivin
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2 Answers2

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Hint: (for your items 1, 2 and 3, which are different ways of expressing the same question): a decreasing sequence of integers decreases by at least $1$ on each step, so if $q_0 > q_1 > \ldots > q_n$, $q_n$ is at most what you get by decreasing $q_0$ by $n$, i.e., $q_n \le q_0 - n$. For item 4, you will have to think about what could prevent you constructing $q_{i+1}$ given $q_i$.

Rob Arthan
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Note that the $q$ sequence is strictly decreasing because $1<b$.
Starting from a finite amount, it must end at $0$, because if it was more you could always divide by $b$. And you stop when the remaining is less than $b$.
Being strictly decreasing, each term must be, at least, $1$ lower than the preceding term, so you can establish a bijection $0 \to q_0, 1 \to q_1, \cdots $ and count.

G Cab
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