a,b,nZ;\forall a,b,n\in\Z;
ab(modn):    kZ;ab=kna\equiv b\pmod{n}:\iff \exists k\in\Bbb{Z};a-b=kn
aabbnn
便
a,b,na,b,n

abbc    ac(modn)a\equiv b\land b\equiv c\implies a\equiv c\pmod{n}
a,b,c,nZ\forall a,b,c,n\in\Z
abc(modn)a\equiv b\equiv c\pmod{n}
    k,lZ;ab=knbc=ln\iff\exists k,l\in\Z;a-b=kn\land b-c=ln
    k,lZ;ac=(k+l)n\implies \exists k,l\in\Z;a-c=(k+l)n
    kZ;ac=kn\implies \exists k\in\Z;a-c=kn
k+lk+lmmmmkk
    ac(modn)\implies a\equiv c\pmod{n}

#2022-02-24 18:04:35
#2021-12-14 20:20:59
#2021-06-14 16:21:52