Answer:
Proofs are in the explantion.
Step-by-step explanation:
We are given the following:
1) [tex]a \equi b (mod n) \rightarrow a-b=kn[/tex] for integer [tex]k[/tex].
1) [tex] c \equi d (mod n) \rightarrow c-d=mn[/tex] for integer [tex]m[/tex].
a)
Proof:
We want to show [tex]a+c \equiv b+d (mod n)[/tex].
So we have the two equations:
a-b=kn and c-d=mn and we want to show for some integer r that we have
(a+c)-(b+d)=rn. If we do that we would have shown that [tex]a+c \equiv b+d (mod n)[/tex].
kn+mn = (a-b)+(c-d)
(k+m)n = a-b+ c-d
(k+m)n = (a+c)+(-b-d)
(k+m)n = (a+c)-(b+d)
k+m is is just an integer
So we found integer r such that (a+c)-(b+d)=rn.
Therefore, [tex]a+c \equiv b+d (mod n)[/tex].
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b) Proof:
We want to show [tex]ac \equiv bd (mod n)[/tex].
So we have the two equations:
a-b=kn and c-d=mn and we want to show for some integer r that we have
(ac)-(bd)=tn. If we do that we would have shown that [tex]ac \equiv bd (mod n)[/tex].
If a-b=kn, then a=b+kn.
If c-d=mn, then c=d+mn.
ac-bd = (b+kn)(d+mn)-bd
= bd+bmn+dkn+kmn^2-bd
= bmn+dkn+kmn^2
= n(bm+dk+kmn)
So the integer t such that (ac)-(bd)=tn is bm+dk+kmn.
Therefore, [tex]ac \equiv bd (mod n)[/tex].
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