## anonymous 4 years ago Prove that if a≡b mod n and c≡d mod n then a−c≡b−d mod n??

1. anonymous

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

2. anonymous

$$a≡b \pmod n \implies a=n.k+b$$ 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!

3. anonymous

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

4. anonymous

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

5. anonymous

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

6. anonymous

Yes.