3.2: Residue Systems and Euler’s φ-Function
residue of \(a\) modulo \(m\). As a result, we see that any integer is congruent to one of the integers \(0,1,2,...,m-1\) modulo m.
A complete residue system modulo \(m\) is a set of integers such that every integer is congruent modulo \(m\) to exactly one integer of the set.
The easiest complete residue system modulo \(m\) is the set of integers \(0,1,2,...,m-1\). Every integer is congruent to one of these integers modulo \(m\).
The set of integers \(\{0,1,2,3,4\}\) form a complete residue system modulo \(5\). Another complete residue system modulo \(5\) could be \(6,7,8,9,10\).
A reduced residue system modulo \(m\) is a set of integers \(r_i\) such that \((r_i,m)=1\) for all \(i\) and \(r_i \neq r_j (mod \ m)\) if \(i\neq j\).
Notice that, a reduced residue system modulo \(m\) can be obtained by deleting all the elements of the complete residue system set that are not relatively prime to \(m\).
The set of integers \(\{1,5\}\) is a reduced residue system modulo \(6\).
The following lemma will help determine a complete residue system modulo any positive integer \(m\).
LEmma
A set of \(m\) incongruent integers modulo \(m\) forms a complete residue system modulo \(m\).
We will prove this lemma by contradiction. Suppose that the set of \(m\) integers does not form a complete residue system modulo \(m\). Then we can find at least one integer \(a\) that is not congruent to any element in this set. Hence non of the elements of this set is actually congruent to the remainder when \(a\) is divided by \(m\). Thus dividing by \(m\) yields to at most \(m-1\) remainders. Therefore by the pigeonhole principle, at least two integers in the set that have the same remainder modulo \(m\). This is a contradiction since the set of integers is formed of \(m\) integers that are incongruent modulo \(m\).
If \(a_1, a_2,...,a_m\) is a complete residue system modulo \(m\), and if \(k\) is a positive integer with \((k,m)=1\), then \[ka_1+b, ka_2+b,...,ka_m+b\] is another complete residue system modulo \(m\) for any integer \(b\).
Let us prove first that no two elements of the set \(\{ka_1+b, ka_2+b,...,ka_m+b\}\) are congruent modulo \(m\). Suppose there exists \(i\) and \(j\) such that \[ka_i+b\equiv ka_j+b(mod\ m).\] Thus we get that \[ka_i\equiv ka_j(mod \ m).\] Now since \((k,m)=1\), we get \[a_i\equiv a_j(mod\ m)\] But for \(i\neq j\), \(a_i\) is inequivalent to \(a_j\) modulo \(m\). Thus \(i=j\). Now notice that there are \(m\) inequivalent integers modulo m and thus by Lemma 10, the set form a complete residue system modulo \(m\).
Euler’s \(\phi\)-Function
We now present a function that counts the number of positive integers less than a given integer that are relatively prime to that given integer. This function is called Euler \(\phi\)-function. We will discuss the properties of Euler \(\phi\)-function in details in chapter 5. It will be sufficient for our purposes in this chapter to the notation.
The Euler \(\phi\)-function of a positive integer n, denoted by \(\phi(n)\) counts the number of positive integers less than \(n\) that are relatively prime to n.
Since 1 and 3 are the only two integers that are relatively prime to 4 and less than 4, then \(\phi(4)=2\). Also, 1,2,...,6 are the integers that are relatively prime to 7 that are less than 7, thus \(\phi(7)=6\).
Now we can say that the number of elements in a reduced residue system modulo \(n\) is \(\phi(n)\).
If \(a_1,a_2,...,a_{\phi(n)}\) is a reduced residue system modulo \(n\) and \((k,n)=1\), then \(ka_1,ka_2,...,ka_{\phi(n)}\) is a reduced residue system modulo \(n\).
The proof proceeds exactly in the same way as that of Theorem 24.
Exercises
- Give a reduced residue system modulo 12.
- Give a complete residue system modulo 13 consisting only of odd integers.
- Find \(\phi(8)\) and \(\phi(101)\).
Contributors and Attributions
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Dr. Wissam Raji, Ph.D., of the American University in Beirut. His work was selected by the Saylor Foundation’s Open Textbook Challenge for public release under a Creative Commons Attribution ( CC BY ) license.