Question

1. Let a; b; c; d; n belong to Z with n > 0. Suppose a congruent b (mod n) and c congruent d (mod n). Use the definition

Answer 1

Answer:

Proofs are in the explantion.

Step-by-step explanation:

We are given the following:

1) $a \equi b (mod n) \rightarrow a-b=kn$ for integer $k$.

1) $c \equi d (mod n) \rightarrow c-d=mn$ for integer $m$.

a)

Proof:

We want to show $a+c \equiv b+d (mod n)$.

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 $a+c \equiv b+d (mod n)$.

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, $a+c \equiv b+d (mod n)$.

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b) Proof:

We want to show $ac \equiv bd (mod n)$.

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 $ac \equiv bd (mod n)$.

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, $ac \equiv bd (mod n)$.

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