# Proof

• We have to show that these operations for the congruence classes modulo a positive integer $m$ are well-defined, i.e. do not depend on the special choice of the representatives $a$ and $b$.
• Assume, $a\neq a'$ and $b\neq b'$ are all integers with equal congruence classes $a\equiv a'(m)$ and $b\equiv b'(m).$
• By definition $m\mid (a-a')$ and $m\mid (b-b').$
• Using the addition and/or subtraction "$\pm$" of integers:
• I.e. $m\mid ((a-a')\pm (b-b')).$
• I.e. $m\mid ((a\pm b)-(a'\pm b').$
• I.e. $m\mid ((a\pm b)$ and $m\mid (a'\pm b').$
• I.e. $(a\pm b)(m)=(a'\pm b')(m).$
• Using the multiplication of integers:
• I.e. $a=a'+km$ and $b=b'+lm$ for some integers $k,l.$
• I.e. $ab=a'b'+(a'l+kb'+klm)m$
• I.e. $ab(m)\equiv a'b'(m).$
• Therefore, the set $(\mathbb Z_m,\cdot,+)$ is a commutative unit ring, (a detailed proof is left for the reader as an exercise) since:
• $(\mathbb Z,+)$ is a commutative group with the neutral element $0(m),$ i.e. $a(m)+0(m)\equiv 0(m)+a(m)\equiv a(m)$ and the inverse elements $(-a)(m)+a(m)=0(m)$ for all $a(m)\in\mathbb Z_m.$
• $(\mathbb Z,\cdot)$ is a commutative semigroup with the neutral element $1(m),$ i.e. $a(m)\cdot 1(m)\equiv 1(m)\cdot a(m)\equiv a(m)$ for all $a(m)\in\mathbb Z_m.$

Github: ### References

#### Bibliography

1. Jones G., Jones M.: "Elementary Number Theory (Undergraduate Series)", Springer, 1998
2. Kraetzel, E.: "Studienbücherei Zahlentheorie", VEB Deutscher Verlag der Wissenschaften, 1981