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 of c

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

of congruence to

(a) prove that a + c congruent b + d (mod n).

(b) prove that ac congruent bd (mod n).

Expert Answer

in progress 0
Margaret 2 weeks 2021-09-09T13:09:14+00:00 1 Answer 0

Answers ( )

    0
    2021-09-09T13:10:51+00:00

    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).

    //

    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).

    //

Leave an answer

Browse
Browse

27:3+15-4x7+3-1=? ( )