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)????

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)????

anonymous · Answer

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

anonymous · Answer

Yeah dunno what you did there?

experimentx · Answer

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

anonymous · Answer

stlil confused.

experimentx · Answer

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

anonymous · Answer

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

experimentx · Answer

because every element is an integer ,,