Question

Prove that if a≡b mod n and c≡d mod n then a−c≡b−d mod n??

Answer 1

u start like a=kn + b........

Answer 2

\(a≡b \pmod n \implies a=n.k+b \) where \( k\in \mathbb{N} \)
\(c≡d \pmod n \implies c=n.p+d \) where \( p\in \mathbb{N} \)

\(a-c = nk+b-np-d = n(k-p) +(b-d)\), where \( (k-p)\in \mathbb{N} \). Hence \((a−c)≡(b−d) \pmod n \) QED!

Answer 3

c= pn+d
a-c=n(k-p)+b-d :P......m late

Answer 4

Similarly we can show that \( (a+c)≡(b+d) \pmod n \) and more generally \( (pa+qc)≡(pb+qd) \pmod n \)

Answer 5

so basically we need to show that difference between a and c will be a multiple of n

Answer 6

Yes.