Question

Prove that if a≡b mod n and c≡d mod n, then ac≡bd mod n.

a=b+kn where k is any integer
c=d+ln where l is any integer

(b+kn)(d+ln)
n(b+k)(d+l)=n(b+d)(d+l)????

Answer 1

\[ ac=(b+kn)(d+ln) = n(dk+l(b+kn)) +bd \]\[ \implies ac≡bd \pmod n \]

Answer 2

Yeah dunno what you did there?

Answer 3

dk+l(b+kn) is an integer ... say z
ac = nz + bd

Answer 4

stlil confused.

Answer 5

a = b + nk ... for some integer k, it implies
a≡b mod n

for some integer z
ac = nz + bd
ac≡bd mod n

Answer 6

@experimentX how is dk+l(b+kn) an integer??

Answer 7

because every element is an integer ,,