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A set S of integers $\{s_1, s_2, . . . , s_k\}$ such that 0 ∈ S is called a set of digits modulo m if and only if any integer b can be written as $$b = f_nm^n + f_{n−1}m^{n−1} + · · · + f_1m + f_0$$ for some $n \geq 0$ and some $f_i$ ∈ S, moreover such expression is unique assuming that $f_n \neq 0$. Show that:

a) $k = m$ and the set of $s_i$ is a complete set of representatives modulo m

b) S = {0, 1, −1} is a set of digits modulo 3

c) S = {0, 2, −2} is NOT a set of digits modulo 3

I can't really seem to get anywhere with this. I've tried representing b in a few ways using the division algorithm and then trying to convert to the expression above but this doesn't really seem to get me anywhere. I'd appreciate any help, thanks!

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    This is very similar (but not exactly) like expressing any integer in base m notation. The only difference is that you may replace the digit (m-1) with the "digit" -1 (or other things). Notice any $n \sum s_i 3^i; s_i \in \{0,2,-2\}$ will have $n$ being even. That should give you a hint with c).2017-02-11

2 Answers 2

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1) $S$ must contain a complete representative of modulo $m$ and $k \ge n$.

Suppose not: Let $a\in \mathbb Z$ be such that $a \not \equiv s_i \mod m$ for any $s_i$ in $S$. Then $a = \sum_{i=0}^n s_i m^i; s_i \in S$ is impossible as $\sum_{i=0}^n s_i m^i \equiv s_0 \mod m$ but $a \not \equiv s_0 \mod m$. So $a$ can not be represented in terms of $S$.

2) $S$ may not contain any two integers, $s \ne t$ but $s \equiv t \mod m$.

Suppose $S$ could: If $s, t \in S$ and $s \ne t$ and $s \equiv t \mod m$. So $s = t + km$ for some k. Let $k = \sum k_i m^i;k_i \in S$ be the unique representation of $k$ using terms of $S$. Then $s = t + \sum k_i m^{i+1}$ is one representation of $s$ using terms of $S$. And $s = s$ is another. That contradicts that all representations are unique.

So 1) and 2) mean $S$ must be a completer modulo $m$ representation.

b) Will show by induction.

We want to find $n = \sum s_i 3^i; s_i \in \{0,1,-1\}$.

So we will need $n \equiv \sum s_i 3^i \mod 3 \equiv s_0 \mod 3$.

Now $n \equiv 0, -1, 1$ so we can and we must choose $s_0 = 0, -1, 1$ so that $n \equiv s_0 \mod 3$.

Now $n - s_0 \equiv 0 \mod 3$. Let $n_1 = n/3$. We can, and must, repeat these steps inductively. To get $n = \sum_{i = 0}^N s_i 3^i$ being a unique representation.

c) Note if $n = \sum_{i=0}^m s_i 3^i; s_i = 0, -2,2$ then $n$ is even. No odd number can be thusly represented. So this fails to be a set of digits modulo $3$.

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I feel we are asking the exact same problem and the rest of it goes this way

d) S = {0,1,2} is a set of digits modulo -3

e) S = {0,1,-1,2,-2} is a set of digits modulo 5

f) S = {0,1,-1,8,-8} is NOT a set of digits modulo 5

g) S={0,1,-1,17,-17} is a set of digits modulo 5

h) S={0,1,-2,(-2)^2, (-2)^3, (-2)^4, (-2)^5} is a set of digits modulo 7

I am still kind of confused about how to proceed for the rest of the question. Can someone give a hint? Thank you so much!

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    This is not even a very good question. It certainly shouldn't be posted as an answer.2017-02-12
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    I don't have enough reputation to add a comment yet2017-02-12
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    I believe this does not prevent you from asking your own question... with your own context (efforts to understand the question).2017-02-12