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If $R$ is a reduced residue system modulo $n$ and $\gcd(a,n)=1$ then $\exists! b\in R$ such that $a\equiv b\pmod n$.

Recall:

$$R=\{a\in C \mid \gcd(a,n)=1\}$$ Where $C$ Is a complete residue system modulo $n$

I didn't quite understand the theorem even when i‘ve flipped the paper so i can see the proof of it, let’s take an example to clarify things.

Example:

Let $C=\{0,1,2,3\}\implies R=\{1,3\}$ but where is $a$ here? If $a=3,n=4$ Then $b$ here is just $3$, and it’s obvious that $3\equiv 3\pmod 4$ , so my question is what is $a$.

PNT
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  • See the edit @WillJagy – PNT Feb 23 '21 at 22:00
  • In the definition, $a$ is any integer coprime with $n$. Although it doesn't change anything, you should use some other name for members of $R$. Like $R={x\in C:(x,n)=1}$. – Shubham Johri Feb 23 '21 at 22:08
  • @ShubhamJohri i’ve Got it, i was supposed that a is in $R$, but actually the theorem didn’t said that, thank you . – PNT Feb 23 '21 at 22:14
  • Sure. If my answer worked for you, consider accepting it by clicking the tick mark button next to it. – Shubham Johri Feb 23 '21 at 22:25
  • $a$ can be anything that is relatively prime to $4$. For example: $a = 5397$. The theorem is simply saying that $5397$ must be congruent to $1$ or $3$ and it can't be congruent to neither or both. And indeed $5397 \equiv 1 \pmod 4$. – fleablood Feb 23 '21 at 23:15
  • Perhaps a less trivial example is $C={4, 9, 27, 50} $ is a complete residue system $\mod 4$. $R = {9, 27}$. So if $\gcd(a, 4) = 1$ the theorem is saying: Either $a \equiv 9\pmod 4$ or $a\equiv 27 \pmod 4$ but not both. And sure if $\gcd(4,a)=1$ then $a$ is odd. So $a=2k+1$ for some $k$. Either $k$ is even or odd. If $k$ is even then $k=2m$ for some $m$ and $a=4k+1\equiv 1 \equiv 9\pmod 4\not \equiv 27\pmod 4$. Or if $k=2m+1$ is odd and $a = 2(2k+1)+1=4k +3\equiv 27\pmod 4\not \equiv 9\pmod 4$. That's what the theorem is saying. – fleablood Feb 23 '21 at 23:22
  • There are various (equivalent) definitions of "reduced residue system". The sought proof depends on which definition you are using. We can't answer (or evaluate answers) until you reveal your definition. Please do so. – Bill Dubuque Feb 25 '21 at 09:31
  • A reduced residue system is a set that contains all elements that are co-prime with $n$ in a complete residue system, or you can define it as $\phi(n)$ @BillDubuque – PNT Feb 25 '21 at 13:19
  • By the dupe $,1 = \gcd(a,n) = \gcd(r,n), $ for $,r = a\bmod n,,$ and $,0\le r < n,$ so $,r \in C\ \ $ – Bill Dubuque Feb 25 '21 at 13:35

2 Answers2

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I believe the proper formulation is the below:

If $R$ is a reduced residue system modulo $n$ and $a\in\Bbb Z$ such that $\gcd(a,n)=1$, then $\exists b\in R$ such that $a\equiv b\mod n$.

Since we have already used up the variable name $a$ in the above statement, we will write $R=\{x\in C:\gcd(x,n)=1\}$.

The proof of this statement is simple: let $a\equiv y\in C$. Then $\exists k\in \Bbb Z|a=kn+y$. Thus $1=\gcd(a,n)=\gcd(kn+y,n)=\gcd(y,n)$ which implies $y\in R$.

Edit: If $\exists!b$ means "one and only one $b$", we should add that if $a\equiv b_1\equiv b_2\in R$, then $b_1=b_2$ as no two distinct elements of $C$ are congruent.

Shubham Johri
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  • Looking again, I now think the notation $\exists!b$ in the question abovemeans uniqueness, there exists one and only one $b$ that works – Will Jagy Feb 23 '21 at 22:46
  • @WillJaggy I have updated my answer but do you have a reference to that notation? – Shubham Johri Feb 23 '21 at 22:51
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    https://math.stackexchange.com/questions/228285/how-can-i-get-the-negation-of-exists-unique-existential-quantification Over the years, I have not seen the notation that often. I had assumed it meant "does not exist;" put together, I would say it is not good notation for day by day mathematics instruction. – Will Jagy Feb 23 '21 at 22:55
  • Hmm, thanks for the information. At first glance I too assumed it meant "does not exist" because of the usage of $!=$ symbol to denote "not equal to" especially in computer science. – Shubham Johri Feb 23 '21 at 22:57
  • https://math.stackexchange.com/questions/73300/negation-of-uniqueness-quantifier is another. I did ask the OP about his text, one would expect – Will Jagy Feb 23 '21 at 22:57
  • I thought that you guys knew about the notation before – PNT Feb 23 '21 at 23:17
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well, as $R \subset C$ and, by definition of complete residue system, then for any $a\in \mathbb Z$ there is a unique $b \in C$ so that $a\equiv b \pmod n$.

So this statement is equivalent to proving, if $a \equiv b \pmod n$ then $\gcd(a,n) = 1 \iff \gcd(b,n) = 1$

And that follows fairly straightforwardly.

$\gcd(a,n) = \gcd(a+kn, n)$ for all integers $k$.

And $a \equiv b \pmod n$ is defined as $n|b-a$ which means that there is an integer $k$ so that $a = b+kn$. So $\gcd(b,n) = \gcd(a + kn, n) =\gcd(a,n)$.

....

Perhaps a stronger and more useful statement would be:

Lemma: If $a\equiv b \pmod n$ then $\gcd(a,n) =\gcd(b,n)$.

fleablood
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