A covering system is a set of congruences $\{a_1 \mod n_1, \dots a_k \mod n_k\}$ covering $\mathbb{Z}$. The Mirsky-Newman theorem says that a covering system cannot cover disjoint sets and have distinct moduli. The proof seems to assume we cover only the natural numbers. Why is this? Is it obvious that a congruence system covers $\mathbb{Z}$ iff it covers $\mathbb{N}$?
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It's not true that a congruence system that covers the naturals necessarily covers the integers. E.g., $0\bmod2,1\bmod4,3\bmod8,7\bmod{16},\dots$ covers the naturals but not $-1$. But a finite congruence system that covers the naturals must cover the integers. Let $m$ be the least common multiple of the moduli, then the whole shebang is periodic mod $m$, so if it covers everything from $1$ to $m$, it covers everything.
Gerry Myerson
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But it doesn't cover $\Bbb N$ since it omits $,15+16\Bbb N.\ \ $ – Bill Dubuque Sep 09 '24 at 23:22
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@Bill, what doesn't cover $\bf N$? or was it just not clear that the three dots meant "and so on, according to the obvious pattern"? – Gerry Myerson Sep 10 '24 at 02:30
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Could you please be more precise. – Bill Dubuque Sep 10 '24 at 02:48
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@Bill, every nonnegative integer satisfies exactly one of $0\bmod2,1\bmod4,3\bmod8,7\bmod{16},15\bmod{32},31\bmod{64},63\bmod{128},127\bmod{256},255\bmod{512},511\bmod{1024},1023\bmod{2048},2047\bmod{4096},4095\bmod{8192},8191\bmod{16384},16383\bmod{32768},32767\bmod{65536},\dotsc$. That is, $x\equiv2^n-1\bmod{2^{n+1}}$ for some nonnegative integer $n$. Precise enough? – Gerry Myerson Sep 10 '24 at 07:11
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Consider the finite set of natural numbers under mod $n$ $$\{0,1,2,\ldots,n-1\}$$
I think it is sufficient if we show that above set is congruent to the set $$\{-0,-1,-2,\ldots,-(n-1)\}$$ in some order.
Take a look at this theorem
Lucenaposition
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AgentS
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