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


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

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

One thing to try is to multiply those relations above by 2, getting
$$
2a = 10k_a+2m , \\2b = 10k_b+2m,  \\0 \leq 2m &lt;10 
$$
If $2m &lt; 5$ 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 $2m &gt; 5$.

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

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