Modular arithmetic proposition

Question

I am trying to solve this Modular arithmetic problem but I have no idea how to verify if the proposition holds or not.

$\forall a,b \in \mathbb{Z}: (a \equiv b \pmod{5} \Rightarrow 2a \equiv 2b \pmod{5} )$

Answer 1

You can prove this directly.

$$a \equiv b \pmod {5} \iff 5|(b-a) \iff b-a = 5k ~(\text{for some } k \in \Bbb Z)\\ \iff 2b-2a =5(2k) \iff 5|(2b-2a) \iff 2a \equiv 2b \pmod {5}.$$

Answer 2

Try it this way:

$$ a \equiv b \pmod 5 \implies \exists (k_a, k_b, m \in \Bbb N) : \\a = 5k_a+m , \\b = 5k_b+m, \\0 \leq m <5 $$ Okay, now what about comparing $2a$ and $2b$?

One thing to try is to multiply those relations above by 2, getting $$ 2a = 10k_a+2m , \\2b = 10k_b+2m, \\0 \leq 2m <10 $$ If $2m < 5$ it is easy to see that $2a \equiv 2b$ by using multipliers $k'_a = 2k_a$ and $k'_b = 2k_b$. That leaves the headache of the possibility that $2m > 5$.

Can you see how to handle that case and thus complete the proof? (Hint - what happens if you use $k'_a = 2k_a + 1$ ?)